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Check if a large number is divisible by 8 or not | C ++ program to find if a number is divisible by 8 or not ; Function to find that number divisible by 8 or not ; Empty string ; If there is single digit ; If there is double digit ; If number formed by last three digits is divisible by 8. ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool check ( string str ) { int n = str . length ( ) ; if ( n == 0 ) return false ; if ( n == 1 ) return ( ( str [ 0 ] - '0' ) % 8 == 0 ) ; if ( n == 2 ) return ( ( ( str [ n - 2 ] - '0' ) * 10 + ( str [ n - 1 ] - '0' ) ) % 8 == 0 ) ; int last = str [ n - 1 ] - '0' ; int second_last = str [ n - 2 ] - '0' ; int third_last = str [ n - 3 ] - '0' ; return ( ( third_last * 100 + second_last * 10 + last ) % 8 == 0 ) ; } int main ( ) { string str = "76952" ; check ( str ) ? cout << " Yes " : cout << " No ▁ " ; return 0 ; } |
Prime points ( Points that split a number into two primes ) | C ++ program to print all prime points ; Function to count number of digits ; Function to check whether a number is prime or not . Returns 0 if prime else - 1 ; Corner cases ; This is checked so that we can skip middle five numbers in below loop ; Function to print prime points ; counting digits ; As single and double digit numbers do not have left and right number pairs ; Finding all left and right pairs . Printing the prime points accordingly . Discarding first and last index point ; Calculating left number ; Calculating right number ; Prime point condition ; No prime point found ; Driver Program | #include <bits/stdc++.h> NEW_LINE using namespace std ; int countDigits ( int n ) { int count = 0 ; while ( n > 0 ) { count ++ ; n = n / 10 ; } return count ; } int checkPrime ( int n ) { if ( n <= 1 ) return -1 ; if ( n <= 3 ) return 0 ; if ( n % 2 == 0 n % 3 == 0 ) return -1 ; for ( int i = 5 ; i * i <= n ; i = i + 6 ) if ( n % i == 0 || n % ( i + 2 ) == 0 ) return -1 ; return 0 ; } void printPrimePoints ( int n ) { int count = countDigits ( n ) ; if ( count == 1 count == 2 ) { cout << " - 1" ; return ; } bool found = false ; for ( int i = 1 ; i < ( count - 1 ) ; i ++ ) { int left = n / ( ( int ) pow ( 10 , count - i ) ) ; int right = n % ( ( int ) pow ( 10 , count - i - 1 ) ) ; if ( checkPrime ( left ) == 0 && checkPrime ( right ) == 0 ) { cout << i << " ▁ " ; found = true ; } } if ( found == false ) cout << " - 1" ; } int main ( ) { int n = 2317 ; printPrimePoints ( n ) ; return 0 ; } |
Find ways an Integer can be expressed as sum of n | C ++ program to count number of ways any given integer x can be expressed as n - th power of unique natural numbers . ; Function to calculate and return the power of any given number ; Function to check power representations recursively ; Initialize number of ways to express x as n - th powers of different natural numbers ; Calling power of ' i ' raised to ' n ' ; Recursively check all greater values of i ; If sum of powers is equal to x then increase the value of result . ; Return the final result ; Driver Code . | #include <bits/stdc++.h> NEW_LINE using namespace std ; int power ( int num , unsigned int n ) { if ( n == 0 ) return 1 ; else if ( n % 2 == 0 ) return power ( num , n / 2 ) * power ( num , n / 2 ) ; else return num * power ( num , n / 2 ) * power ( num , n / 2 ) ; } int checkRecursive ( int x , int n , int curr_num = 1 , int curr_sum = 0 ) { int results = 0 ; int p = power ( curr_num , n ) ; while ( p + curr_sum < x ) { results += checkRecursive ( x , n , curr_num + 1 , p + curr_sum ) ; curr_num ++ ; p = power ( curr_num , n ) ; } if ( p + curr_sum == x ) results ++ ; return results ; } int main ( ) { int x = 10 , n = 2 ; cout << checkRecursive ( x , n ) ; return 0 ; } |
Number Theory | Generators of finite cyclic group under addition | A simple C ++ program to find all generators ; Function to return gcd of a and b ; Print generators of n ; 1 is always a generator ; A number x is generator of GCD is 1 ; Driver program to test above function | #include <bits/stdc++.h> NEW_LINE using namespace std ; int gcd ( int a , int b ) { if ( a == 0 ) return b ; return gcd ( b % a , a ) ; } int printGenerators ( unsigned int n ) { cout << "1 ▁ " ; for ( int i = 2 ; i < n ; i ++ ) if ( gcd ( i , n ) == 1 ) cout << i << " ▁ " ; } int main ( ) { int n = 10 ; printGenerators ( n ) ; return 0 ; } |
Check if a large number is divisible by 3 or not | C ++ program to find if a number is divisible by 3 or not ; Function to find that number divisible by 3 or not ; Compute sum of digits ; Check if sum of digits is divisible by 3. ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int check ( string str ) { int n = str . length ( ) ; int digitSum = 0 ; for ( int i = 0 ; i < n ; i ++ ) digitSum += ( str [ i ] - '0' ) ; return ( digitSum % 3 == 0 ) ; } int main ( ) { string str = "1332" ; check ( str ) ? cout << " Yes " : cout << " No ▁ " ; return 0 ; } |
Count all perfect divisors of a number | Below is C ++ code to count total perfect divisors ; Pre - compute counts of all perfect divisors of all numbers upto MAX . ; Iterate through all the multiples of i * i ; Increment all such multiples by 1 ; Returns count of perfect divisors of n . ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; #define MAX 100001 NEW_LINE int perfectDiv [ MAX ] ; void precomputeCounts ( ) { for ( int i = 1 ; i * i < MAX ; ++ i ) { for ( int j = i * i ; j < MAX ; j += i * i ) ++ perfectDiv [ j ] ; } } int countPerfectDivisors ( int n ) { return perfectDiv [ n ] ; } int main ( ) { precomputeCounts ( ) ; int n = 16 ; cout << " Total ▁ perfect ▁ divisors ▁ of ▁ " << n << " ▁ = ▁ " << countPerfectDivisors ( n ) << " STRNEWLINE " ; n = 12 ; cout << " Total ▁ perfect ▁ divisors ▁ of ▁ " << n << " ▁ = ▁ " << countPerfectDivisors ( n ) ; return 0 ; } |
Prime Factorization using Sieve O ( log n ) for multiple queries | C ++ program to find prime factorization of a number n in O ( Log n ) time with precomputation allowed . ; stores smallest prime factor for every number ; Calculating SPF ( Smallest Prime Factor ) for every number till MAXN . Time Complexity : O ( nloglogn ) ; marking smallest prime factor for every number to be itself . ; separately marking spf for every even number as 2 ; checking if i is prime ; marking SPF for all numbers divisible by i ; marking spf [ j ] if it is not previously marked ; A O ( log n ) function returning primefactorization by dividing by smallest prime factor at every step ; driver program for above function ; precalculating Smallest Prime Factor ; calling getFactorization function | #include " bits / stdc + + . h " NEW_LINE using namespace std ; #define MAXN 100001 NEW_LINE int spf [ MAXN ] ; void sieve ( ) { spf [ 1 ] = 1 ; for ( int i = 2 ; i < MAXN ; i ++ ) spf [ i ] = i ; for ( int i = 4 ; i < MAXN ; i += 2 ) spf [ i ] = 2 ; for ( int i = 3 ; i * i < MAXN ; i ++ ) { if ( spf [ i ] == i ) { for ( int j = i * i ; j < MAXN ; j += i ) if ( spf [ j ] == j ) spf [ j ] = i ; } } } vector < int > getFactorization ( int x ) { vector < int > ret ; while ( x != 1 ) { ret . push_back ( spf [ x ] ) ; x = x / spf [ x ] ; } return ret ; } int main ( int argc , char const * argv [ ] ) { sieve ( ) ; int x = 12246 ; cout << " prime ▁ factorization ▁ for ▁ " << x << " ▁ : ▁ " ; vector < int > p = getFactorization ( x ) ; for ( int i = 0 ; i < p . size ( ) ; i ++ ) cout << p [ i ] << " ▁ " ; cout << endl ; return 0 ; } |
Nearest element with at | C ++ program to print nearest element with at least one common prime factor . ; Pre - computation of smallest prime divisor of all numbers ; Prime number will have same divisor ; Function to calculate all divisors of input array ; Pre - compute all the divisors of array element by using prime factors ; Traverse all elements , ; For every divisor of current element , find closest element . ; Visit divisor if not visited ; Fetch the index of visited divisor ; Update the divisor index to current index ; Set the minimum distance ; Set the min distance of current index ' i ' to nearest one ; Add 1 as indexing starts from 0 ; Set the min distance of found index ' ind ' ; Add 1 as indexing starts from 0 ; Driver code ; Simple sieve to find smallest prime divisor of number from 2 to MAX ; function to calculate nearest distance of every array elements ; Print the nearest distance having GDC > 1 | #include <bits/stdc++.h> NEW_LINE using namespace std ; const int MAX = 100001 ; const int INF = INT_MAX ; int primedivisor [ MAX ] , dist [ MAX ] , pos [ MAX ] , divInd [ MAX ] ; vector < int > divisors [ MAX ] ; void sieveOfEratosthenes ( ) { for ( int i = 2 ; i * i < MAX ; ++ i ) { if ( ! primedivisor [ i ] ) for ( int j = i * i ; j < MAX ; j += i ) primedivisor [ j ] = i ; } for ( int i = 1 ; i < MAX ; ++ i ) if ( ! primedivisor [ i ] ) primedivisor [ i ] = i ; } void findDivisors ( int arr [ ] , int n ) { for ( int i = 0 ; i < MAX ; ++ i ) pos [ i ] = divInd [ i ] = -1 , dist [ i ] = INF ; for ( int i = 0 ; i < n ; ++ i ) { int num = arr [ i ] ; while ( num > 1 ) { int div = primedivisor [ num ] ; divisors [ i ] . push_back ( div ) ; while ( num % div == 0 ) num /= div ; } } } void nearestGCD ( int arr [ ] , int n ) { findDivisors ( arr , n ) ; for ( int i = 0 ; i < n ; ++ i ) { for ( auto & div : divisors [ i ] ) { if ( divInd [ div ] == -1 ) divInd [ div ] = i ; else { int ind = divInd [ div ] ; divInd [ div ] = i ; if ( dist [ i ] > abs ( ind - i ) ) { dist [ i ] = abs ( ind - i ) ; pos [ i ] = ind + 1 ; } if ( dist [ ind ] > abs ( ind - i ) ) { dist [ ind ] = abs ( ind - i ) ; pos [ ind ] = i + 1 ; } } } } } int main ( ) { sieveOfEratosthenes ( ) ; int arr [ ] = { 2 , 9 , 4 , 3 , 13 } ; int n = sizeof ( arr ) / sizeof ( arr [ 0 ] ) ; nearestGCD ( arr , n ) ; for ( int i = 0 ; i < n ; ++ i ) cout << pos [ i ] << " ▁ " ; return 0 ; } |
Largest subsequence having GCD greater than 1 | Simple C ++ program to find length of the largest subsequence with GCD greater than 1. ; Returns length of the largest subsequence with GCD more than 1. ; Finding the Maximum value in arr [ ] ; Iterate from 2 to maximum possible divisor of all give values ; If we found divisor , increment count ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int largestGCDSubsequence ( int arr [ ] , int n ) { int ans = 0 ; int maxele = * max_element ( arr , arr + n ) ; for ( int i = 2 ; i <= maxele ; ++ i ) { int count = 0 ; for ( int j = 0 ; j < n ; ++ j ) { if ( arr [ j ] % i == 0 ) ++ count ; } ans = max ( ans , count ) ; } return ans ; } int main ( ) { int arr [ ] = { 3 , 6 , 2 , 5 , 4 } ; int size = sizeof ( arr ) / sizeof ( arr [ 0 ] ) ; cout << largestGCDSubsequence ( arr , size ) ; return 0 ; } |
Minimum step to reach one | C ++ program to get minimum step to reach 1 under given constraints ; structure represent one node in queue ; method returns minimum step to reach one ; set is used to visit numbers so that they won 't be pushed in queue again ; loop until we reach to 1 ; if current data value is 1 , return its steps from N ; check curr - 1 , only if it not visited yet ; loop from 2 to sqrt ( value ) for its divisors ; check divisor , only if it is not visited yet if i is divisor of val , then val / i will be its bigger divisor ; Driver code to test above methods | #include <bits/stdc++.h> NEW_LINE using namespace std ; struct data { int val ; int steps ; data ( int val , int steps ) : val ( val ) , steps ( steps ) { } } ; int minStepToReachOne ( int N ) { queue < data > q ; q . push ( data ( N , 0 ) ) ; set < int > st ; while ( ! q . empty ( ) ) { data t = q . front ( ) ; q . pop ( ) ; if ( t . val == 1 ) return t . steps ; if ( st . find ( t . val - 1 ) == st . end ( ) ) { q . push ( data ( t . val - 1 , t . steps + 1 ) ) ; st . insert ( t . val - 1 ) ; } for ( int i = 2 ; i * i <= t . val ; i ++ ) { if ( t . val % i == 0 && st . find ( t . val / i ) == st . end ( ) ) { q . push ( data ( t . val / i , t . steps + 1 ) ) ; st . insert ( t . val / i ) ; } } } } int main ( ) { int N = 17 ; cout << minStepToReachOne ( N ) << endl ; } |
Queries on the sum of prime factor counts in a range | C ++ program to find sum prime factors in given range . ; using sieve method to evaluating the prime factor of numbers ; if i is prime ; setting number of prime factor of a prime number . ; Returns sum of counts of prime factors in range from l to r . This function mainly uses count [ ] which is filled by Sieve ( ) ; finding the sum of number of prime factor of numbers in a range . ; Driven Program | #include <bits/stdc++.h> NEW_LINE #define MAX 1000006 NEW_LINE using namespace std ; void sieve ( int count [ ] ) { for ( int i = 2 ; i * i <= MAX ; i ++ ) { if ( count [ i ] == 0 ) { for ( int j = 2 * i ; j < MAX ; j += i ) count [ j ] ++ ; count [ i ] = 1 ; } } } int query ( int count [ ] , int l , int r ) { int sum = 0 ; for ( int i = l ; i <= r ; i ++ ) sum += count [ i ] ; return sum ; } int main ( ) { int count [ MAX ] ; memset ( count , 0 , sizeof count ) ; sieve ( count ) ; cout << query ( count , 6 , 10 ) << endl << query ( count , 1 , 5 ) ; return 0 ; } |
Generation of n numbers with given set of factors | C ++ program to generate n numbers with given factors ; Generate n numbers with factors in factor [ ] ; array of k to store next multiples of given factors ; Prints n numbers int output = 0 ; Next number to print as output ; Find the next smallest multiple ; Printing minimum in each iteration print the value if output is not equal to current value ( to avoid the duplicates ) ; incrementing the current value by the respective factor ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; void generateNumbers ( int factor [ ] , int n , int k ) { int next [ k ] = { 0 } ; for ( int i = 0 ; i < n ; ) { int toincrement = 0 ; for ( int j = 0 ; j < k ; j ++ ) if ( next [ j ] < next [ toincrement ] ) toincrement = j ; if ( output != next [ toincrement ] ) { output = next [ toincrement ] ; printf ( " % d ▁ " , next [ toincrement ] ) ; i ++ ; } next [ toincrement ] += factor [ toincrement ] ; } } int main ( ) { int factor [ ] = { 3 , 5 , 7 } ; int n = 10 ; int k = sizeof ( factor ) / sizeof ( factor [ 0 ] ) ; generateNumbers ( factor , n , k ) ; return 0 ; } |
Reversible numbers | C ++ program to check if a given number is reversible or not ; Function to check reversible number ; Calculate reverse of n ; Calculate sum of number and its reverse ; Check for reverse number reach digit must be odd ; Driver function | #include <iostream> NEW_LINE #include <cmath> NEW_LINE using namespace std ; void checkReversible ( int n ) { int rev = 0 , rem ; int flag = n ; while ( flag ) { rem = flag % 10 ; rev *= 10 ; rev += rem ; flag /= 10 ; } int sum = rev + n ; while ( sum && ( rem % 2 != 0 ) ) { rem = sum % 10 ; sum /= 10 ; } if ( sum == 0 ) cout << " Reversible ▁ Number " ; else cout << " Non - Reversible ▁ Number " ; } int main ( ) { int n = 36 ; checkReversible ( n ) ; return 0 ; } |
Reversible numbers | C ++ program to find the count of reversible numbers upto a given number n ; Function to calculate the count of reversible number ; Calculate counts of reversible number of 1 to n - digits ; All four possible cases and their formula ; for i of form 2 k ; for i of form 4 k + 3 ; for i of form 4 k + 1 no solution ; Driver function ; count up - to 9 digit numbers ( 1 billion ) | #include <iostream> NEW_LINE #include <cmath> NEW_LINE using namespace std ; void countReversible ( int n ) { int count = 0 ; for ( int i = 1 ; i <= n ; i ++ ) { switch ( i % 4 ) { case 0 : case 2 : count += 20 * pow ( 30 , ( i / 2 - 1 ) ) ; break ; case 3 : count += 100 * pow ( 500 , i / 4 ) ; break ; case 1 : break ; } } cout << count ; } int main ( ) { int n = 9 ; countReversible ( n ) ; return 0 ; } |
Multiplicative order | C ++ program to implement multiplicative order ; function for GCD ; Function return smallest + ve integer that holds condition A ^ k ( mod N ) = 1 ; result store power of A that rised to the power N - 1 ; modular arithmetic ; return smallest + ve integer ; increment power ; driver program to test above function | #include <bits/stdc++.h> NEW_LINE using namespace std ; int GCD ( int a , int b ) { if ( b == 0 ) return a ; return GCD ( b , a % b ) ; } int multiplicativeOrder ( int A , int N ) { if ( GCD ( A , N ) != 1 ) return -1 ; unsigned int result = 1 ; int K = 1 ; while ( K < N ) { result = ( result * A ) % N ; if ( result == 1 ) return K ; K ++ ; } return -1 ; } int main ( ) { int A = 4 , N = 7 ; cout << multiplicativeOrder ( A , N ) ; return 0 ; } |
Sum of product of x and y such that floor ( n / x ) = y | C ++ program to find sum of product of x and y such that n / x = y ( Integer Division ) ; Return the sum of natural number in a range . ; n * ( n + 1 ) / 2. ; Return the sum of product x * y . ; Iterating i from 1 to sqrt ( n ) ; Finding the upper limit . ; Finding the lower limit . ; Driven Program | #include <bits/stdc++.h> NEW_LINE using namespace std ; int sumOfRange ( int a , int b ) { int i = ( a * ( a + 1 ) ) >> 1 ; int j = ( b * ( b + 1 ) ) >> 1 ; return ( i - j ) ; } int sumofproduct ( int n ) { int sum = 0 ; int root = sqrt ( n ) ; for ( int i = 1 ; i <= root ; i ++ ) { int up = n / i ; int low = max ( n / ( i + 1 ) , root ) ; sum += ( i * sumOfRange ( up , low ) ) ; sum += ( i * ( n / i ) ) ; } return sum ; } int main ( ) { int n = 10 ; cout << sumofproduct ( n ) << endl ; return 0 ; } |
Primitive root of a prime number n modulo n | C ++ program to find primitive root of a given number n ; Returns true if n is prime ; Corner cases ; This is checked so that we can skip middle five numbers in below loop ; Iterative Function to calculate ( x ^ n ) % p in O ( logy ) ; x = x % p ; Update x if it is more than or equal to p ; If y is odd , multiply x with result ; y must be even now y = y >> 1 ; y = y / 2 ; Utility function to store prime factors of a number ; Print the number of 2 s that divide n ; n must be odd at this point . So we can skip one element ( Note i = i + 2 ) ; While i divides n , print i and divide n ; This condition is to handle the case when n is a prime number greater than 2 ; Function to find smallest primitive root of n ; Check if n is prime or not ; Find value of Euler Totient function of n Since n is a prime number , the value of Euler Totient function is n - 1 as there are n - 1 relatively prime numbers . ; Find prime factors of phi and store in a set ; Check for every number from 2 to phi ; Iterate through all prime factors of phi . and check if we found a power with value 1 ; Check if r ^ ( ( phi ) / primefactors ) mod n is 1 or not ; If there was no power with value 1. ; If no primitive root found ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool isPrime ( int n ) { if ( n <= 1 ) return false ; if ( n <= 3 ) return true ; if ( n % 2 == 0 n % 3 == 0 ) return false ; for ( int i = 5 ; i * i <= n ; i = i + 6 ) if ( n % i == 0 || n % ( i + 2 ) == 0 ) return false ; return true ; } int power ( int x , unsigned int y , int p ) { while ( y > 0 ) { if ( y & 1 ) res = ( res * x ) % p ; x = ( x * x ) % p ; } return res ; } void findPrimefactors ( unordered_set < int > & s , int n ) { while ( n % 2 == 0 ) { s . insert ( 2 ) ; n = n / 2 ; } for ( int i = 3 ; i <= sqrt ( n ) ; i = i + 2 ) { while ( n % i == 0 ) { s . insert ( i ) ; n = n / i ; } } if ( n > 2 ) s . insert ( n ) ; } int findPrimitive ( int n ) { unordered_set < int > s ; if ( isPrime ( n ) == false ) return -1 ; int phi = n - 1 ; findPrimefactors ( s , phi ) ; for ( int r = 2 ; r <= phi ; r ++ ) { bool flag = false ; for ( auto it = s . begin ( ) ; it != s . end ( ) ; it ++ ) { if ( power ( r , phi / ( * it ) , n ) == 1 ) { flag = true ; break ; } } if ( flag == false ) return r ; } return -1 ; } int main ( ) { int n = 761 ; cout << " ▁ Smallest ▁ primitive ▁ root ▁ of ▁ " << n << " ▁ is ▁ " << findPrimitive ( n ) ; return 0 ; } |
Minimum number of power terms with sum equal to n | C ++ program to calculate minimum number of powers of x to make sum equal to n . ; Return minimum power terms of x required ; if x is 1 , return n since any power of 1 is 1 only . ; Consider n = a * x + b where a = n / x and b = n % x . ; Update count of powers for 1 's added ; Repeat the process for reduced n ; Driven Program | #include <bits/stdc++.h> NEW_LINE using namespace std ; int minPower ( int n , int x ) { if ( x == 1 ) return n ; int ans = 0 ; while ( n > 0 ) { ans += ( n % x ) ; n /= x ; } return ans ; } int main ( ) { int n = 5 , x = 3 ; cout << minPower ( n , x ) << endl ; return 0 ; } |
Sum of Perrin Numbers | C ++ program to calculate sum of Perrin Numbers ; function for sum of first n Perrin number . ; if ( n == 0 ) n = 0 ; if ( n == 1 ) n = 1 ; if ( n == 2 ) n = 2 ; calculate k = 5 sum of three previous step . ; Sum remaining numbers ; int d = a + b ; calculate next term ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int calSum ( int n ) { int a = 3 , b = 0 , c = 2 ; return 3 ; return 3 ; return 5 ; int sum = 5 ; while ( n > 2 ) { sum += d ; a = b ; b = c ; c = d ; n -- ; } return sum ; } int main ( ) { int n = 9 ; cout << calSum ( n ) ; return 0 ; } |
Print the kth common factor of two numbers | C ++ program to find kth common factor of two numbers ; Returns k 'th common factor of x and y. ; Find smaller of two numbers ; Count common factors until we either reach small or count becomes k . ; If we reached small ; Driver code | #include <iostream> NEW_LINE using namespace std ; int findKCF ( int x , int y , int k ) { int small = min ( x , y ) ; int count = 1 ; for ( int i = 2 ; i <= small ; i ++ ) { if ( x % i == 0 && y % i == 0 ) count ++ ; if ( count == k ) return i ; } return -1 ; } int main ( ) { int x = 4 , y = 24 , k = 3 ; cout << findKHCF ( x , y , k ) ; return 0 ; } |
Find minimum number to be divided to make a number a perfect square | C ++ program to find minimum number which divide n to make it a perfect square . ; Return the minimum number to be divided to make n a perfect square . ; Since 2 is only even prime , compute its power separately . ; If count is odd , it must be removed by dividing n by prime number . ; If count is odd , it must be removed by dividing n by prime number . ; Driven Program | #include <bits/stdc++.h> NEW_LINE using namespace std ; int findMinNumber ( int n ) { int count = 0 , ans = 1 ; while ( n % 2 == 0 ) { count ++ ; n /= 2 ; } if ( count % 2 ) ans *= 2 ; for ( int i = 3 ; i <= sqrt ( n ) ; i += 2 ) { count = 0 ; while ( n % i == 0 ) { count ++ ; n /= i ; } if ( count % 2 ) ans *= i ; } if ( n > 2 ) ans *= n ; return ans ; } int main ( ) { int n = 72 ; cout << findMinNumber ( n ) << endl ; return 0 ; } |
Program to implement Collatz Conjecture | C ++ program to implement Collatz Conjecture ; Function to find if n reaches to 1 or not . ; If there is a cycle formed , we can 't r reach 1. ; If n is odd then pass n = 3 n + 1 else n = n / 2 ; Wrapper over isToOneRec ( ) ; To store numbers visited using recursive calls . ; Drivers code | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool isToOneRec ( int n , unordered_set < int > & s ) { if ( n == 1 ) return true ; if ( s . find ( n ) != s . end ( ) ) return false ; return ( n % 2 ) ? isToOneRec ( 3 * n + 1 , s ) : isToOneRec ( n / 2 , s ) ; } bool isToOne ( int n ) { unordered_set < int > s ; return isToOneRec ( n , s ) ; } int main ( ) { int n = 5 ; isToOne ( n ) ? cout << " Yes " : cout << " No " ; return 0 ; } |
Program to implement Collatz Conjecture | C ++ program to implement Collatz Conjecture ; Function to find if n reaches to 1 or not . ; Return true if n is positive ; Drivers code | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool isToOne ( int n ) { return ( n > 0 ) ; } int main ( ) { int n = 5 ; isToOne ( n ) ? cout << " Yes " : cout << " No " ; return 0 ; } |
GCD of two numbers formed by n repeating x and y times | C ++ program to print Greatest Common Divisor of number formed by N repeating x times and y times . ; Return the Greatest common Divisor of two numbers . ; Prints Greatest Common Divisor of number formed by n repeating x times and y times . ; Finding GCD of x and y . ; Print n , g times . ; Driven Program | #include <bits/stdc++.h> NEW_LINE using namespace std ; int gcd ( int a , int b ) { if ( a == 0 ) return b ; return gcd ( b % a , a ) ; } void findgcd ( int n , int x , int y ) { int g = gcd ( x , y ) ; for ( int i = 0 ; i < g ; i ++ ) cout << n ; } int main ( ) { int n = 123 , x = 5 , y = 2 ; findgcd ( n , x , y ) ; return 0 ; } |
Count natural numbers whose factorials are divisible by x but not y | C ++ program to count natural numbers whose factorials are divisible by x but not y . ; GCD function to compute the greatest divisor among a and b ; Returns first number whose factorial is divisible by x . ; int i = 1 ; Result ; Remove common factors ; We found first i . ; Count of natural numbers whose factorials are divisible by x but not y . ; Return difference between first natural number whose factorial is divisible by y and first natural number whose factorial is divisible by x . ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int gcd ( int a , int b ) { if ( ( a % b ) == 0 ) return b ; return gcd ( b , a % b ) ; } int firstFactorialDivisibleNumber ( int x ) { int new_x = x ; for ( i = 1 ; i < x ; i ++ ) { new_x /= gcd ( i , new_x ) ; if ( new_x == 1 ) break ; } return i ; } int countFactorialXNotY ( int x , int y ) { return ( firstFactorialDivisibleNumber ( y ) - firstFactorialDivisibleNumber ( x ) ) ; } int main ( void ) { int x = 15 , y = 25 ; cout << countFactorialXNotY ( x , y ) ; return 0 ; } |
Find the first natural number whose factorial is divisible by x | A simple C ++ program to find first natural number whose factorial divides x . ; Returns first number whose factorial divides x . ; int i = 1 ; Result ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int firstFactorialDivisibleNumber ( int x ) { int fact = 1 ; for ( i = 1 ; i < x ; i ++ ) { fact = fact * i ; if ( fact % x == 0 ) break ; } return i ; } int main ( void ) { int x = 16 ; cout << firstFactorialDivisibleNumber ( x ) ; return 0 ; } |
Find two prime numbers with given sum | C ++ program to find a prime number pair whose sum is equal to given number C ++ program to print super primes less than or equal to n . ; Generate all prime numbers less than n . ; Initialize all entries of boolean array as true . A value in isPrime [ i ] will finally be false if i is Not a prime , else true bool isPrime [ n + 1 ] ; ; If isPrime [ p ] is not changed , then it is a prime ; Update all multiples of p ; Prints a prime pair with given sum ; Generating primes using Sieve ; Traversing all numbers to find first pair ; Driven program | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool SieveOfEratosthenes ( int n , bool isPrime [ ] ) { isPrime [ 0 ] = isPrime [ 1 ] = false ; for ( int i = 2 ; i <= n ; i ++ ) isPrime [ i ] = true ; for ( int p = 2 ; p * p <= n ; p ++ ) { if ( isPrime [ p ] == true ) { for ( int i = p * p ; i <= n ; i += p ) isPrime [ i ] = false ; } } } void findPrimePair ( int n ) { bool isPrime [ n + 1 ] ; SieveOfEratosthenes ( n , isPrime ) ; for ( int i = 0 ; i < n ; i ++ ) { if ( isPrime [ i ] && isPrime [ n - i ] ) { cout << i << " ▁ " << ( n - i ) ; return ; } } } int main ( ) { int n = 74 ; findPrimePair ( n ) ; return 0 ; } |
Count numbers with same first and last digits | C ++ program to get count of numbers with same start and end digit in an interval ; Utility method to get first digit of x ; method to return count of numbers with same starting and ending digit from 1 upto x ; get ten - spans from 1 to x ; add 9 to consider all 1 digit numbers ; Find first and last digits ; If last digit is greater than first digit then decrease count by 1 ; Method to return count of numbers with same starting and ending digit between start and end ; Driver code to test above methods | #include <bits/stdc++.h> NEW_LINE using namespace std ; int getFirstDigit ( int x ) { while ( x >= 10 ) x /= 10 ; return x ; } int getCountWithSameStartAndEndFrom1 ( int x ) { if ( x < 10 ) return x ; int tens = x / 10 ; int res = tens + 9 ; int firstDigit = getFirstDigit ( x ) ; int lastDigit = x % 10 ; if ( lastDigit < firstDigit ) res -- ; return res ; } int getCountWithSameStartAndEnd ( int start , int end ) { return getCountWithSameStartAndEndFrom1 ( end ) - getCountWithSameStartAndEndFrom1 ( start - 1 ) ; } int main ( ) { int start = 5 , end = 40 ; cout << getCountWithSameStartAndEnd ( start , end ) ; return 0 ; } |
Right | Program to check whether a given number is right - truncatable prime or not . ; Generate all prime numbers less than n . ; Initialize all entries of boolean array as true . A value in isPrime [ i ] will finally be false if i is Not a prime , else true bool isPrime [ n + 1 ] ; ; If isPrime [ p ] is not changed , then it is a prime ; Update all multiples of p ; Returns true if n is right - truncatable , else false ; Generating primes using Sieve ; Checking whether the number remains prime when the last ( " right " ) digit is successively removed ; Driver program | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool sieveOfEratosthenes ( int n , bool isPrime [ ] ) { isPrime [ 0 ] = isPrime [ 1 ] = false ; for ( int i = 2 ; i <= n ; i ++ ) isPrime [ i ] = true ; for ( int p = 2 ; p * p <= n ; p ++ ) { if ( isPrime [ p ] == true ) { for ( int i = p * 2 ; i <= n ; i += p ) isPrime [ i ] = false ; } } } bool rightTruPrime ( int n ) { bool isPrime [ n + 1 ] ; sieveOfEratosthenes ( n , isPrime ) ; while ( n ) { if ( isPrime [ n ] ) n = n / 10 ; else return false ; } return true ; } int main ( ) { int n = 59399 ; if ( rightTruPrime ( n ) ) cout << " Yes " << endl ; else cout << " No " << endl ; return 0 ; } |
Mersenne Prime | Program to generate mersenne primes ; Generate all prime numbers less than n . ; Initialize all entries of boolean array as true . A value in prime [ i ] will finally be false if i is Not a prime , else true bool prime [ n + 1 ] ; ; If prime [ p ] is not changed , then it is a prime ; Update all multiples of p ; Function to generate mersenne primes less than or equal to n ; Create a boolean array " prime [ 0 . . n ] " ; Generating primes using Sieve ; Generate all numbers of the form 2 ^ k - 1 and smaller than or equal to n . ; Checking whether number is prime and is one less then the power of 2 ; Driven program | #include <bits/stdc++.h> NEW_LINE using namespace std ; void SieveOfEratosthenes ( int n , bool prime [ ] ) { for ( int i = 0 ; i <= n ; i ++ ) prime [ i ] = true ; for ( int p = 2 ; p * p <= n ; p ++ ) { if ( prime [ p ] == true ) { for ( int i = p * 2 ; i <= n ; i += p ) prime [ i ] = false ; } } } void mersennePrimes ( int n ) { bool prime [ n + 1 ] ; SieveOfEratosthenes ( n , prime ) ; for ( int k = 2 ; ( ( 1 << k ) - 1 ) <= n ; k ++ ) { long long num = ( 1 << k ) - 1 ; if ( prime [ num ] ) cout << num << " ▁ " ; } } int main ( ) { int n = 31 ; cout << " Mersenne ▁ prime ▁ numbers ▁ smaller ▁ " << " than ▁ or ▁ equal ▁ to ▁ " << n << endl ; mersennePrimes ( n ) ; return 0 ; } |
Find sum of modulo K of first N natural number | C ++ program to find sum of modulo K of first N natural numbers . ; Return sum of modulo K of first N natural numbers . ; Iterate from 1 to N && evaluating and adding i % K . ; Driver Program | #include <bits/stdc++.h> NEW_LINE using namespace std ; int findSum ( int N , int K ) { int ans = 0 ; for ( int i = 1 ; i <= N ; i ++ ) ans += ( i % K ) ; return ans ; } int main ( ) { int N = 10 , K = 2 ; cout << findSum ( N , K ) << endl ; return 0 ; } |
Find sum of modulo K of first N natural number | C ++ program to find sum of modulo K of first N natural numbers . ; Return sum of modulo K of first N natural numbers . ; Counting the number of times 1 , 2 , . . , K - 1 , 0 sequence occurs . ; Finding the number of elements left which are incomplete of sequence Leads to Case 1 type . ; adding multiplication of number of times 1 , 2 , . . , K - 1 , 0 sequence occurs and sum of first k natural number and sequence from case 1. ; Driver program | #include <bits/stdc++.h> NEW_LINE using namespace std ; int findSum ( int N , int K ) { int ans = 0 ; int y = N / K ; int x = N % K ; ans = ( K * ( K - 1 ) / 2 ) * y + ( x * ( x + 1 ) ) / 2 ; return ans ; } int main ( ) { int N = 10 , K = 2 ; cout << findSum ( N , K ) << endl ; return 0 ; } |
Smallest number to multiply to convert floating point to natural | C ++ program to find the smallest number to multiply to convert a floating point number into natural number . ; Finding GCD of two number ; Returns smallest integer k such that k * str becomes natural . str is an input floating point number ; Find size of string representing a floating point number . ; Below is used to find denominator in fraction form . ; Used to find value of count_after_dot ; To find numerator in fraction form of given number . For example , for 30.25 , numerator would be 3025. ; If there was no dot , then number is already a natural . ; Find denominator in fraction form . For example , for 30.25 , denominator is 100 ; Result is denominator divided by GCD - of - numerator - and - denominator . For example , for 30.25 , result is 100 / GCD ( 3025 , 100 ) = 100 / 25 = 4 ; Driven Program | #include <bits/stdc++.h> NEW_LINE using namespace std ; int gcd ( int a , int b ) { if ( b == 0 ) return a ; return gcd ( b , a % b ) ; } int findnum ( string & str ) { int n = str . length ( ) ; int count_after_dot = 0 ; bool dot_seen = false ; int num = 0 ; for ( int i = 0 ; i < n ; i ++ ) { if ( str [ i ] != ' . ' ) { num = num * 10 + ( str [ i ] - '0' ) ; if ( dot_seen == true ) count_after_dot ++ ; } else dot_seen = true ; } if ( dot_seen == false ) return 1 ; int dem = ( int ) pow ( 10 , count_after_dot ) ; return ( dem / gcd ( num , dem ) ) ; } int main ( ) { string str = "5.125" ; cout << findnum ( str ) << endl ; return 0 ; } |
Find the maximum number of handshakes | C ++ program to find maximum number of handshakes . ; Calculating the maximum number of handshake using derived formula . ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int maxHandshake ( int n ) { return ( n * ( n - 1 ) ) / 2 ; } int main ( ) { int n = 10 ; cout << maxHandshake ( n ) << endl ; return 0 ; } |
Count digits in given number N which divide N | C ++ program to find number of digits in N that divide N . ; Utility function to check divisibility by digit ; ( N [ i ] - '0' ) gives the digit value and form the number ; Function to count digits which appears in N and divide N divide [ 10 ] -- > array which tells that particular digit divides N or not count [ 10 ] -- > counts frequency of digits which divide N ; We initialize all digits of N as not divisible by N . ; start checking divisibility of N by digits 2 to 9 ; if digit divides N then mark it as true ; Now traverse the number string to find and increment result whenever a digit divides N . ; Driver program to run the case | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool divisible ( string N , int digit ) { int ans = 0 ; for ( int i = 0 ; i < N . length ( ) ; i ++ ) { ans = ( ans * 10 + ( N [ i ] - '0' ) ) ; ans %= digit ; } return ( ans == 0 ) ; } int allDigits ( string N ) { bool divide [ 10 ] = { false } ; for ( int digit = 2 ; digit <= 9 ; digit ++ ) { if ( divisible ( N , digit ) ) divide [ digit ] = true ; } int result = 0 ; for ( int i = 0 ; i < N . length ( ) ; i ++ ) { if ( divide [ N [ i ] - '0' ] == true ) result ++ ; } return result ; } int main ( ) { string N = "122324" ; cout << allDigits ( N ) ; return 0 ; } |
Aliquot Sequence | C ++ implementation of Optimized approach to generate Aliquot Sequence ; Function to calculate sum of all proper divisors ; Note that this loop runs till square root of n ; If divisors are equal , take only one of them ; else Otherwise take both ; calculate sum of all proper divisors only ; Function to print Aliquot Sequence for an input n . ; Print the first term ; Calculate next term from previous term ; Print next term ; Driver program to test above function | #include <bits/stdc++.h> NEW_LINE using namespace std ; int getSum ( int n ) { for ( int i = 1 ; i <= sqrt ( n ) ; i ++ ) { if ( n % i == 0 ) { if ( n / i == i ) sum = sum + i ; { sum = sum + i ; sum = sum + ( n / i ) ; } } } return sum - n ; } void printAliquot ( int n ) { printf ( " % d ▁ " , n ) ; unordered_set < int > s ; s . insert ( n ) ; int next = 0 ; while ( n > 0 ) { n = getSum ( n ) ; if ( s . find ( n ) != s . end ( ) ) { cout << " Repeats with " break ; } cout << n << " ▁ " ; s . insert ( n ) ; } } int main ( ) { printAliquot ( 12 ) ; return 0 ; } |
Find the prime numbers which can written as sum of most consecutive primes | C ++ program to find Longest Sum of consecutive primes ; utility function for sieve of sundaram ; In general Sieve of Sundaram , produces primes smaller than ( 2 * x + 2 ) for a number given number x . Since we want primes smaller than MAX , we reduce MAX to half This array is used to separate numbers of the form i + j + 2 ij from others where 1 <= i <= j ; Main logic of Sundaram . Mark all numbers which do not generate prime number by doing 2 * i + 1 ; Since 2 is a prime number ; Print other primes . Remaining primes are of the form 2 * i + 1 such that marked [ i ] is false . ; function find the prime number which can be written as the sum of the most consecutive primes ; To store maximum length of consecutive primes that can sum to a limit ; The prime number ( or result ) that can be represented as sum of maximum number of primes . ; Consider all lengths of consecutive primes below limit . ; if we cross the limit , then break the loop ; sum_prime [ i ] - sum_prime [ j ] is prime number or not ; Check if sum of current length of consecutives is prime or not . ; update the length and prime number ; Returns the prime number that can written as sum of longest chain of consecutive primes . ; Store prime number in vector ; Calculate sum of prime numbers and store them in sum_prime array . sum_prime [ i ] stores sum of prime numbers from primes [ 0 ] to primes [ i - 1 ] ; Process all queries one by one ; Driver program | #include <bits/stdc++.h> NEW_LINE using namespace std ; const int MAX = 10000 ; void sieveSundaram ( vector < int > & primes ) { bool marked [ MAX / 2 + 1 ] = { 0 } ; for ( int i = 1 ; i <= ( sqrt ( MAX ) - 1 ) / 2 ; i ++ ) for ( int j = ( i * ( i + 1 ) ) << 1 ; j <= MAX / 2 ; j = j + 2 * i + 1 ) marked [ j ] = true ; primes . push_back ( 2 ) ; for ( int i = 1 ; i <= MAX / 2 ; i ++ ) if ( marked [ i ] == false ) primes . push_back ( 2 * i + 1 ) ; } int LSCPUtil ( int limit , vector < int > & prime , long long int sum_prime [ ] ) { int max_length = -1 ; int prime_number = -1 ; for ( int i = 0 ; prime [ i ] <= limit ; i ++ ) { for ( int j = 0 ; j < i ; j ++ ) { if ( sum_prime [ i ] - sum_prime [ j ] > limit ) break ; long long int consSum = sum_prime [ i ] - sum_prime [ j ] ; if ( binary_search ( prime . begin ( ) , prime . end ( ) , consSum ) ) { if ( max_length < i - j + 1 ) { max_length = i - j + 1 ; prime_number = consSum ; } } } } return prime_number ; } void LSCP ( int arr [ ] , int n ) { vector < int > primes ; sieveSundaram ( primes ) ; long long int sum_prime [ primes . size ( ) + 1 ] ; sum_prime [ 0 ] = 0 ; for ( int i = 1 ; i <= primes . size ( ) ; i ++ ) sum_prime [ i ] = primes [ i - 1 ] + sum_prime [ i - 1 ] ; for ( int i = 0 ; i < n ; i ++ ) cout << LSCPUtil ( arr [ i ] , primes , sum_prime ) << " ▁ " ; } int main ( ) { int arr [ ] = { 10 , 30 , 40 , 50 , 1000 } ; int n = sizeof ( arr ) / sizeof ( arr [ 0 ] ) ; LSCP ( arr , n ) ; return 0 ; } |
Count numbers which can be constructed using two numbers | C ++ program to count all numbers that can be formed using two number numbers x an y ; Returns count of numbers from 1 to n that can be formed using x and y . ; Create an auxiliary array and initialize it as false . An entry arr [ i ] = true is going to mean that i can be formed using x and y ; x and y can be formed using x and y . ; Initialize result ; Traverse all numbers and increment result if a number can be formed using x and y . ; If i can be formed using x and y ; Then i + x and i + y can also be formed using x and y . ; Increment result ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int countNums ( int n , int x , int y ) { vector < bool > arr ( n + 1 , false ) ; if ( x <= n ) arr [ x ] = true ; if ( y <= n ) arr [ y ] = true ; int result = 0 ; for ( int i = min ( x , y ) ; i <= n ; i ++ ) { if ( arr [ i ] ) { if ( i + x <= n ) arr [ i + x ] = true ; if ( i + y <= n ) arr [ i + y ] = true ; result ++ ; } } return result ; } int main ( ) { int n = 15 , x = 5 , y = 7 ; cout << countNums ( n , x , y ) ; return 0 ; } |
Emirp numbers | Program to print Emirp numbers less than n ; Function to find reverse of any number ; Sieve method used for generating emirp number ( use of sieve of Eratosthenes ) ; Create a boolean array " prime [ 0 . . n ] " and initialize all entries it as true . A value in prime [ i ] will finally be false if i is Not a prime , else true . ; If prime [ p ] is not changed , then it is a prime ; Update all multiples of p ; Traverse all prime numbers ; Find reverse a number ; A number is emrip if it is not a palindrome number and its reverse is also prime . ; Mark reverse prime as false so that it 's not printed again ; Driver program | #include <bits/stdc++.h> NEW_LINE using namespace std ; int reverse ( int x ) { int rev = 0 ; while ( x > 0 ) { rev = ( rev * 10 ) + x % 10 ; x = x / 10 ; } return rev ; } void printEmirp ( int n ) { bool prime [ n + 1 ] ; memset ( prime , true , sizeof ( prime ) ) ; for ( int p = 2 ; p * p <= n ; p ++ ) { if ( prime [ p ] == true ) { for ( int i = p * 2 ; i <= n ; i += p ) prime [ i ] = false ; } } for ( int p = 2 ; p <= n ; p ++ ) { if ( prime [ p ] ) { int rev = reverse ( p ) ; if ( p != rev && rev <= n && prime [ rev ] ) { cout << p << " ▁ " << rev << " ▁ " ; prime [ rev ] = false ; } } } } int main ( ) { int n = 40 ; printEmirp ( n ) ; return 0 ; } |
Abundant Number | An Optimized Solution to check Abundant Number ; Function to calculate sum of divisors ; Note that this loop runs till square root of n ; If divisors are equal , take only one of them ; else Otherwise take both ; calculate sum of all proper divisors only ; Function to check Abundant Number ; Return true if sum of divisors is greater than n . ; Driver program to test above function | #include <bits/stdc++.h> NEW_LINE using namespace std ; int getSum ( int n ) { int sum = 0 ; for ( int i = 1 ; i <= sqrt ( n ) ; i ++ ) { if ( n % i == 0 ) { if ( n / i == i ) sum = sum + i ; { sum = sum + i ; sum = sum + ( n / i ) ; } } } sum = sum - n ; return sum ; } bool checkAbundant ( int n ) { return ( getSum ( n ) > n ) ; } int main ( ) { checkAbundant ( 12 ) ? cout << " YES STRNEWLINE " : cout << " NO STRNEWLINE " ; checkAbundant ( 15 ) ? cout << " YES STRNEWLINE " : cout << " NO STRNEWLINE " ; return 0 ; } |
Powerful Number | C ++ program to find if a number is powerful or not . ; function to check if the number is powerful ; First divide the number repeatedly by 2 ; If only 2 ^ 1 divides n ( not higher powers ) , then return false ; if n is not a power of 2 then this loop will execute repeat above process ; Find highest power of " factor " that divides n ; If only factor ^ 1 divides n ( not higher powers ) , then return false ; n must be 1 now if it is not a prime numenr . Since prime numbers are not powerful , we return false if n is not 1. ; Driver program to test above function | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool isPowerful ( int n ) { while ( n % 2 == 0 ) { int power = 0 ; while ( n % 2 == 0 ) { n /= 2 ; power ++ ; } if ( power == 1 ) return false ; } for ( int factor = 3 ; factor <= sqrt ( n ) ; factor += 2 ) { int power = 0 ; while ( n % factor == 0 ) { n = n / factor ; power ++ ; } if ( power == 1 ) return false ; } return ( n == 1 ) ; } int main ( ) { isPowerful ( 20 ) ? cout << " YES STRNEWLINE " : cout << " NO STRNEWLINE " ; isPowerful ( 27 ) ? cout << " YES STRNEWLINE " : cout << " NO STRNEWLINE " ; return 0 ; } |
Deficient Number | C ++ program to implement an Optimized Solution to check Deficient Number ; Function to calculate sum of divisors ; Note that this loop runs till square root of n ; If divisors are equal , take only one of them ; else Otherwise take both ; Function to check Deficient Number ; Check if sum ( n ) < 2 * n ; Driver program to test above function | #include <bits/stdc++.h> NEW_LINE using namespace std ; int divisorsSum ( int n ) { for ( int i = 1 ; i <= sqrt ( n ) ; i ++ ) { if ( n % i == 0 ) { if ( n / i == i ) { sum = sum + i ; } { sum = sum + i ; sum = sum + ( n / i ) ; } } } return sum ; } bool isDeficient ( int n ) { return ( divisorsSum ( n ) < ( 2 * n ) ) ; } int main ( ) { isDeficient ( 12 ) ? cout << " YES STRNEWLINE " : cout << " NO STRNEWLINE " ; isDeficient ( 15 ) ? cout << " YES STRNEWLINE " : cout << " NO STRNEWLINE " ; return 0 ; } |
Harshad ( Or Niven ) Number | C ++ implementation of above approach ; Converting integer to string ; Initialising sum to 0 ; Traversing through the string ; Converting character to int ; Comparing number and sum ; Driver Code ; Passing this number to get result function | #include <bits/stdc++.h> NEW_LINE using namespace std ; string checkHarshad ( int n ) { string st = to_string ( n ) ; int sum = 0 ; int length = st . length ( ) ; for ( char i : st ) { sum = sum + ( i - '0' ) ; } if ( n % sum == 0 ) { return " Yes " ; } else { return " No " ; } } int main ( ) { int number = 18 ; cout << checkHarshad ( number ) << endl ; } |
Smith Number | C ++ program to check whether a number is Smith Number or not . ; array to store all prime less than and equal to 10 ^ 6 ; utility function for sieve of sundaram ; In general Sieve of Sundaram , produces primes smaller than ( 2 * x + 2 ) for a number given number x . Since we want primes smaller than MAX , we reduce MAX to half This array is used to separate numbers of the form i + j + 2 ij from others where 1 <= i <= j ; Main logic of Sundaram . Mark all numbers which do not generate prime number by doing 2 * i + 1 ; Since 2 is a prime number ; Print other primes . Remaining primes are of the form 2 * i + 1 such that marked [ i ] is false . ; Returns true if n is a Smith number , else false . ; Find sum the digits of prime factors of n ; If primes [ i ] is a prime factor , add its digits to pDigitSum . ; If n != 1 then one prime factor still to be summed up ; ; All prime factors digits summed up Now sum the original number digits ; If sum of digits in prime factors and sum of digits in original number are same , then return true . Else return false . ; Driver code ; Finding all prime numbers before limit . These numbers are used to find prime factors . | #include <bits/stdc++.h> NEW_LINE using namespace std ; const int MAX = 10000 ; vector < int > primes ; void sieveSundaram ( ) { bool marked [ MAX / 2 + 100 ] = { 0 } ; for ( int i = 1 ; i <= ( sqrt ( MAX ) - 1 ) / 2 ; i ++ ) for ( int j = ( i * ( i + 1 ) ) << 1 ; j <= MAX / 2 ; j = j + 2 * i + 1 ) marked [ j ] = true ; primes . push_back ( 2 ) ; for ( int i = 1 ; i <= MAX / 2 ; i ++ ) if ( marked [ i ] == false ) primes . push_back ( 2 * i + 1 ) ; } bool isSmith ( int n ) { int original_no = n ; int pDigitSum = 0 ; for ( int i = 0 ; primes [ i ] <= n / 2 ; i ++ ) { while ( n % primes [ i ] == 0 ) { int p = primes [ i ] ; n = n / p ; while ( p > 0 ) { pDigitSum += ( p % 10 ) ; p = p / 10 ; } } } if ( n != 1 && n != original_no ) { while ( n > 0 ) { pDigitSum = pDigitSum + n % 10 ; n = n / 10 ; } } int sumDigits = 0 ; while ( original_no > 0 ) { sumDigits = sumDigits + original_no % 10 ; original_no = original_no / 10 ; } return ( pDigitSum == sumDigits ) ; } int main ( ) { sieveSundaram ( ) ; cout << " Printing ▁ first ▁ few ▁ Smith ▁ Numbers " " ▁ using ▁ isSmith ( ) n " ; for ( int i = 1 ; i < 500 ; i ++ ) if ( isSmith ( i ) ) cout << i << " ▁ " ; return 0 ; } |
Kaprekar Number | C ++ program to check if a number is Kaprekar number or not ; Returns true if n is a Kaprekar number , else false ; Count number of digits in square ; Split the square at different poitns and see if sum of any pair of splitted numbers is equal to n . ; To avoid numbers like 10 , 100 , 1000 ( These are not Karprekar numbers ; Find sum of current parts and compare with n ; compare with original number ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool iskaprekar ( int n ) { if ( n == 1 ) return true ; int sq_n = n * n ; int count_digits = 0 ; while ( sq_n ) { count_digits ++ ; sq_n /= 10 ; } for ( int r_digits = 1 ; r_digits < count_digits ; r_digits ++ ) { int eq_parts = pow ( 10 , r_digits ) ; if ( eq_parts == n ) continue ; int sum = sq_n / eq_parts + sq_n % eq_parts ; if ( sum == n ) return true ; } return false ; } int main ( ) { cout << " Printing ▁ first ▁ few ▁ Kaprekar ▁ Numbers " " ▁ using ▁ iskaprekar ( ) STRNEWLINE " ; for ( int i = 1 ; i < 10000 ; i ++ ) if ( iskaprekar ( i ) ) cout << i << " ▁ " ; return 0 ; } |
Keith Number | C ++ program to check if a number is Keith or not ; Returns true if x is Keith , else false . ; Store all digits of x in a vector " terms " Also find number of digits and store in " n " . ; int temp = x , n = 0 ; n is number of digits in x ; To get digits in right order ( from MSB to LSB ) ; Keep finding next trms of a sequence generated using digits of x until we either reach x or a number greate than x ; Next term is sum of previous n terms ; When the control comes out of the while loop , either the next_term is equal to the number or greater than it . If next_term is equal to x , then x is a Keith number , else not ; Driver program | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool isKeith ( int x ) { vector < int > terms ; while ( temp > 0 ) { terms . push_back ( temp % 10 ) ; temp = temp / 10 ; n ++ ; } reverse ( terms . begin ( ) , terms . end ( ) ) ; int next_term = 0 , i = n ; while ( next_term < x ) { next_term = 0 ; for ( int j = 1 ; j <= n ; j ++ ) next_term += terms [ i - j ] ; terms . push_back ( next_term ) ; i ++ ; } return ( next_term == x ) ; } int main ( ) { isKeith ( 14 ) ? cout << " Yes STRNEWLINE " : cout << " No STRNEWLINE " ; isKeith ( 12 ) ? cout << " Yes STRNEWLINE " : cout << " No STRNEWLINE " ; isKeith ( 197 ) ? cout << " Yes STRNEWLINE " : cout << " No STRNEWLINE " ; return 0 ; } |
Check if a number can be expressed as a sum of consecutive numbers | C ++ program to check if a number can be expressed as sum of consecutive numbers ; This function returns true if n can be expressed sum of consecutive . ; We basically return true if n is a power of two ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool canBeSumofConsec ( unsigned int n ) { return ( ( n & ( n - 1 ) ) && n ) ; } int main ( ) { unsigned int n = 15 ; canBeSumofConsec ( n ) ? cout << " true " : cout << " false " ; return 0 ; } |
Check if a number can be expressed as a sum of consecutive numbers | ; Updating n with 2 n ; ( n & ( n - 1 ) ) = > Checking whether we can write 2 n as 2 ^ k if yes ( can 't represent 2n as 2^k) then answer 1 if no (can represent 2n as 2^k) then answer 0 | #include <bits/stdc++.h> NEW_LINE using namespace std ; long long int canBeSumofConsec ( long long int n ) { n = 2 * n ; return ( ( n & ( n - 1 ) ) != 0 ) ; } int main ( ) { long long int n = 10 ; cout << canBeSumofConsec ( n ) << " STRNEWLINE " ; } |
Even Fibonacci Numbers Sum | Find the sum of all the even - valued terms in the Fibonacci sequence which do not exceed given limit . ; Returns sum of even Fibonacci numbers which are less than or equal to given limit . ; Initialize first two even prime numbers and their sum ; calculating sum of even Fibonacci value ; get next even value of Fibonacci sequence ; If we go beyond limit , we break loop ; Move to next even number and update sum ; Driver code | #include <iostream> NEW_LINE using namespace std ; int evenFibSum ( int limit ) { if ( limit < 2 ) return 0 ; long long int ef1 = 0 , ef2 = 2 ; long long int sum = ef1 + ef2 ; while ( ef2 <= limit ) { long long int ef3 = 4 * ef2 + ef1 ; if ( ef3 > limit ) break ; ef1 = ef2 ; ef2 = ef3 ; sum += ef2 ; } return sum ; } int main ( ) { int limit = 400 ; cout << evenFibSum ( limit ) ; return 0 ; } |
Find numbers with K odd divisors in a given range | C ++ program to count numbers with k odd divisors in a range . ; Utility function to check if number is perfect square or not ; Utility Function to return count of divisors of a number ; Note that this loop runs till square root ; If divisors are equal , count it only once ; Otherwise print both ; Function to calculate all divisors having exactly k divisors between a and b ; calculate only for perfect square numbers ; check if number is perfect square or not ; total divisors of number equals to k or not ; Driver program to run the case | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool isPerfect ( int n ) { int s = sqrt ( n ) ; return ( s * s == n ) ; } int divisorsCount ( int n ) { int count = 0 ; for ( int i = 1 ; i <= sqrt ( n ) + 1 ; i ++ ) { if ( n % i == 0 ) { if ( n / i == i ) count += 1 ; else count += 2 ; } } return count ; } int kDivisors ( int a , int b , int k ) { for ( int i = a ; i <= b ; i ++ ) { if ( isPerfect ( i ) ) if ( divisors ( i ) == k ) count ++ ; } return count ; } int main ( ) { int a = 2 , b = 49 , k = 3 ; cout << kDivisors ( a , b , k ) ; return 0 ; } |
Nth Even Fibonacci Number | C ++ code to find Even Fibonacci Series using normal Recursion ; Function which return nth even fibonnaci number ; calculation of Fn = 4 * ( Fn - 1 ) + Fn - 2 ; Driver Code | #include <iostream> NEW_LINE using namespace std ; long int evenFib ( int n ) { if ( n < 1 ) return n ; if ( n == 1 ) return 2 ; return ( ( 4 * evenFib ( n - 1 ) ) + evenFib ( n - 2 ) ) ; } int main ( ) { int n = 7 ; cout << evenFib ( n ) ; return 0 ; } |
Querying maximum number of divisors that a number in a given range has | A C ++ implementation of the above idea to process queries of finding a number with maximum divisors . ; Finds smallest prime factor of all numbers inrange [ 1 , maxn ) and stores them in smallest_prime [ ] , smallest_prime [ i ] should contain the smallest prime that divides i ; Initialize the smallest_prime factors of all to infinity ; to be built like eratosthenes sieve ; prime number will have its smallest_prime equal to itself ; if ' i ' is the first prime number reaching ' j ' ; number of divisors of n = ( p1 ^ k1 ) * ( p2 ^ k2 ) . . . ( pn ^ kn ) are equal to ( k1 + 1 ) * ( k2 + 1 ) . . . ( kn + 1 ) this function finds the number of divisors of all numbersin range [ 1 , maxn ) and stores it in divisors [ ] divisors [ i ] stores the number of divisors i has ; we can obtain the prime factorization of the number n n = ( p1 ^ k1 ) * ( p2 ^ k2 ) . . . ( pn ^ kn ) using the smallest_prime [ ] array , we keep dividing n by its smallest_prime until it becomes 1 , whilst we check if we have need to set k zero ; use p ^ k , initialize k to 0 ; builds segment tree for divisors [ ] array ; leaf node ; build left and right subtree ; combine the information from left and right subtree at current node ; returns the maximum number of divisors in [ l , r ] ; If current node 's range is disjoint with query range ; If the current node stores information for the range that is completely inside the query range ; Returns maximum number of divisors from left or right subtree ; driver code ; First find smallest prime divisors for all the numbers ; Then build the divisors [ ] array to store the number of divisors ; Build segment tree for the divisors [ ] array | #include <bits/stdc++.h> NEW_LINE using namespace std ; #define maxn 1000005 NEW_LINE #define INF 99999999 NEW_LINE int smallest_prime [ maxn ] ; int divisors [ maxn ] ; int segmentTree [ 4 * maxn ] ; void findSmallestPrimeFactors ( ) { for ( int i = 0 ; i < maxn ; i ++ ) smallest_prime [ i ] = INF ; for ( long long i = 2 ; i < maxn ; i ++ ) { if ( smallest_prime [ i ] == INF ) { smallest_prime [ i ] = i ; for ( long long j = i * i ; j < maxn ; j += i ) if ( smallest_prime [ j ] > i ) smallest_prime [ j ] = i ; } } } void buildDivisorsArray ( ) { for ( int i = 1 ; i < maxn ; i ++ ) { divisors [ i ] = 1 ; int n = i , p = smallest_prime [ i ] , k = 0 ; while ( n > 1 ) { n = n / p ; k ++ ; if ( smallest_prime [ n ] != p ) { divisors [ i ] = divisors [ i ] * ( k + 1 ) ; k = 0 ; } p = smallest_prime [ n ] ; } } } void buildSegtmentTree ( int node , int a , int b ) { if ( a == b ) { segmentTree [ node ] = divisors [ a ] ; return ; } buildSegtmentTree ( 2 * node , a , ( a + b ) / 2 ) ; buildSegtmentTree ( 2 * node + 1 , ( ( a + b ) / 2 ) + 1 , b ) ; segmentTree [ node ] = max ( segmentTree [ 2 * node ] , segmentTree [ 2 * node + 1 ] ) ; } int query ( int node , int a , int b , int l , int r ) { if ( l > b a > r ) return -1 ; if ( a >= l && b <= r ) return segmentTree [ node ] ; return max ( query ( 2 * node , a , ( a + b ) / 2 , l , r ) , query ( 2 * node + 1 , ( ( a + b ) / 2 ) + 1 , b , l , r ) ) ; } int main ( ) { findSmallestPrimeFactors ( ) ; buildDivisorsArray ( ) ; buildSegtmentTree ( 1 , 1 , maxn - 1 ) ; cout << " Maximum ▁ divisors ▁ that ▁ a ▁ number ▁ has ▁ " << " ▁ in ▁ [ 1 , ▁ 100 ] ▁ are ▁ " << query ( 1 , 1 , maxn - 1 , 1 , 100 ) << endl ; cout << " Maximum ▁ divisors ▁ that ▁ a ▁ number ▁ has " << " ▁ in ▁ [ 10 , ▁ 48 ] ▁ are ▁ " << query ( 1 , 1 , maxn - 1 , 10 , 48 ) << endl ; cout << " Maximum ▁ divisors ▁ that ▁ a ▁ number ▁ has " << " ▁ in ▁ [ 1 , ▁ 10 ] ▁ are ▁ " << query ( 1 , 1 , maxn - 1 , 1 , 10 ) << endl ; return 0 ; } |
N 'th Smart Number | C ++ implementation to find n 'th smart number ; Limit on result ; Function to calculate n 'th smart number ; Initialize all numbers as not prime ; iterate to mark all primes and smart number ; Traverse all numbers till maximum limit ; ' i ' is maked as prime number because it is not multiple of any other prime ; mark all multiples of ' i ' as non prime ; If i is the third prime factor of j then add it to result as it has at least three prime factors . ; Sort all smart numbers ; return n 'th smart number ; Driver program to run the case | #include <bits/stdc++.h> NEW_LINE using namespace std ; const int MAX = 3000 ; int smartNumber ( int n ) { int primes [ MAX ] = { 0 } ; vector < int > result ; for ( int i = 2 ; i < MAX ; i ++ ) { if ( primes [ i ] == 0 ) { primes [ i ] = 1 ; for ( int j = i * 2 ; j < MAX ; j = j + i ) { primes [ j ] -= 1 ; if ( ( primes [ j ] + 3 ) == 0 ) result . push_back ( j ) ; } } } sort ( result . begin ( ) , result . end ( ) ) ; return result [ n - 1 ] ; } int main ( ) { int n = 50 ; cout << smartNumber ( n ) ; return 0 ; } |
Repeated subtraction among two numbers | C ++ program to count of steps until one of the two numbers become 0. ; Returns count of steps before one of the numbers become 0 after repeated subtractions . ; If y divides x , then simply return x / y . ; Else recur . Note that this function works even if x is smaller than y because in that case first recursive call exchanges roles of x and y . ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int countSteps ( int x , int y ) { if ( x % y == 0 ) return x / y ; return x / y + countSteps ( y , x % y ) ; } int main ( ) { int x = 100 , y = 19 ; cout << countSteps ( x , y ) ; return 0 ; } |
Common Divisors of Two Numbers | C ++ implementation of program ; Map to store the count of each prime factor of a ; Function that calculate the count of each prime factor of a number ; Function to calculate all common divisors of two given numbers a , b -- > input integer numbers ; Find count of each prime factor of a ; stores number of common divisors ; Find the count of prime factors of b using distinct prime factors of a ; Prime factor of common divisor has minimum cnt of both a and b ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; map < int , int > ma ; void primeFactorize ( int a ) { for ( int i = 2 ; i * i <= a ; i += 2 ) { int cnt = 0 ; while ( a % i == 0 ) { cnt ++ ; a /= i ; } ma [ i ] = cnt ; } if ( a > 1 ) { ma [ a ] = 1 ; } } int commDiv ( int a , int b ) { primeFactorize ( a ) ; int res = 1 ; for ( auto m = ma . begin ( ) ; m != ma . end ( ) ; m ++ ) { int cnt = 0 ; int key = m -> first ; int value = m -> second ; while ( b % key == 0 ) { b /= key ; cnt ++ ; } res *= ( min ( cnt , value ) + 1 ) ; } return res ; } int main ( ) { int a = 12 , b = 24 ; cout << commDiv ( a , b ) << endl ; return 0 ; } |
Count number of solutions of x ^ 2 = 1 ( mod p ) in given range | C ++ program to count number of values that satisfy x ^ 2 = 1 mod p where x lies in range [ 1 , n ] ; Initialize result ; Traverse all numbers smaller than given number p . Note that we don 't traverse from 1 to n, but 1 to p ; If x is a solution , then count all numbers of the form x + i * p such that x + i * p is in range [ 1 , n ] ; The largest number in the form of x + p * i in range [ 1 , n ] ; Add count of numbers of the form x + p * i . 1 is added for x itself . ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; typedef long long ll ; int findCountOfSolutions ( int n , int p ) { ll ans = 0 ; for ( ll x = 1 ; x < p ; x ++ ) { if ( ( x * x ) % p == 1 ) { ll last = x + p * ( n / p ) ; if ( last > n ) last -= p ; ans += ( ( last - x ) / p + 1 ) ; } } return ans ; } int main ( ) { ll n = 10 , p = 5 ; printf ( " % lld STRNEWLINE " , findCountOfSolutions ( n , p ) ) ; return 0 ; } |
Happy Number | method return true if n is Happy Number numSquareSum method is given in below detailed code snippet | int isHappyNumber ( int n ) { set < int > st ; while ( 1 ) { n = numSquareSum ( n ) ; if ( n == 1 ) return true ; if ( st . find ( n ) != st . end ( ) ) return false ; st . insert ( n ) ; } } |
Kaprekar Constant | C ++ program to demonstrate working of Kaprekar constant ; This function checks validity of kaprekar ' s ▁ constant . ▁ It ▁ returns ▁ kaprekar ' s constant for any four digit number " n " such that all digits of n are not same . ; Store current n as previous number ; Get four digits of given number ; Sort all four digits in ascending order And giet in the form of number " asc " ; Get all four dgits in descending order in the form of number " desc " ; Get the difference of two numbers ; If difference is same as previous , we have reached kaprekar 's constant ; Else recur ; A wrapper over kaprekarRec ( ) ; Driver code ; Trying few four digit numbers , we always get 6174 | #include <bits/stdc++.h> NEW_LINE using namespace std ; int kaprekarRec ( int n , int & prev ) { if ( n == 0 ) return 0 ; prev = n ; int digits [ 4 ] ; for ( int i = 0 ; i < 4 ; i ++ ) { digits [ i ] = n % 10 ; n = n / 10 ; } sort ( digits , digits + 4 ) ; int asc = 0 ; for ( int i = 0 ; i < 4 ; i ++ ) asc = asc * 10 + digits [ i ] ; sort ( digits , digits + 4 , std :: greater < int > ( ) ) ; int desc = 0 ; for ( int i = 0 ; i < 4 ; i ++ ) desc = desc * 10 + digits [ i ] ; int diff = abs ( asc - desc ) ; if ( diff == prev ) return diff ; return kaprekarRec ( diff , prev ) ; } int kaprekar ( int n ) { int prev = 0 ; return kaprekarRec ( n , prev ) ; } int main ( ) { cout << kaprekar ( 1000 ) << endl ; cout << kaprekar ( 1112 ) << endl ; cout << kaprekar ( 9812 ) << endl ; return 0 ; } |
Bakhshali Approximation for computing square roots | This program gives result approximated to 5 decimal places . ; This will be the nearest perfect square to s ; This is the sqrt of pSq ; Find the nearest perfect square to s ; calculate d ; calculate P ; calculate A ; calculate sqrt ( S ) . ; Driver program to test above function | #include <iostream> NEW_LINE float sqroot ( float s ) { int pSq = 0 ; int N = 0 ; for ( int i = static_cast < int > ( s ) ; i > 0 ; i -- ) { for ( int j = 1 ; j < i ; j ++ ) { if ( j * j == i ) { pSq = i ; N = j ; break ; } } if ( pSq > 0 ) break ; } float d = s - pSq ; float P = d / ( 2.0 * N ) ; float A = N + P ; float sqrt_of_s = A - ( ( P * P ) / ( 2.0 * A ) ) ; return sqrt_of_s ; } int main ( ) { float num = 9.2345 ; float sqroot_of_num = sqroot ( num ) ; std :: cout << " Square ▁ root ▁ of ▁ " << num << " ▁ = ▁ " << sqroot_of_num ; return 0 ; } |
Breaking an Integer to get Maximum Product | C / C ++ program to find maximum product by breaking the Integer ; method return x ^ a in log ( a ) time ; Method returns maximum product obtained by breaking N ; base case 2 = 1 + 1 ; base case 3 = 2 + 1 ; breaking based on mod with 3 ; If divides evenly , then break into all 3 ; If division gives mod as 1 , then break as 4 + power of 3 for remaining part ; If division gives mod as 2 , then break as 2 + power of 3 for remaining part ; Driver code to test above methods | #include <bits/stdc++.h> NEW_LINE using namespace std ; int power ( int x , int a ) { int res = 1 ; while ( a ) { if ( a & 1 ) res = res * x ; x = x * x ; a >>= 1 ; } return res ; } int breakInteger ( int N ) { if ( N == 2 ) return 1 ; if ( N == 3 ) return 2 ; int maxProduct ; switch ( N % 3 ) { case 0 : maxProduct = power ( 3 , N / 3 ) ; break ; case 1 : maxProduct = 2 * 2 * power ( 3 , ( N / 3 ) - 1 ) ; break ; case 2 : maxProduct = 2 * power ( 3 , N / 3 ) ; break ; } return maxProduct ; } int main ( ) { int maxProduct = breakInteger ( 10 ) ; cout << maxProduct << endl ; return 0 ; } |
Finding sum of digits of a number until sum becomes single digit | C ++ program to find sum of digits of a number until sum becomes single digit . ; Loop to do sum while sum is not less than or equal to 9 ; Driver program to test the above function | #include <bits/stdc++.h> NEW_LINE using namespace std ; int digSum ( int n ) { int sum = 0 ; while ( n > 0 sum > 9 ) { if ( n == 0 ) { n = sum ; sum = 0 ; } sum += n % 10 ; n /= 10 ; } return sum ; } int main ( ) { int n = 1234 ; cout << digSum ( n ) ; return 0 ; } |
Finding sum of digits of a number until sum becomes single digit | Driver program to test the above function | #include <bits/stdc++.h> NEW_LINE using namespace std ; int digSum ( int n ) { if ( n == 0 ) return 0 ; return ( n % 9 == 0 ) ? 9 : ( n % 9 ) ; } int main ( ) { int n = 9999 ; cout << digSum ( n ) ; return 0 ; } |
Multiples of 3 or 7 | A better C ++ program to find count of all numbers that multiples ; Returns count of all numbers smaller than or equal to n and multiples of 3 or 7 or both ; Driver code | #include <iostream> NEW_LINE using namespace std ; int countMultiples ( int n ) { return n / 3 + n / 7 - n / 21 ; } int main ( ) { cout << " Count ▁ = ▁ " << countMultiples ( 25 ) ; } |
Find Last Digit of a ^ b for Large Numbers | C ++ code to find last digit of a ^ b ; Function to find b % a ; Initialize result ; calculating mod of b with a to make b like 0 <= b < a ; return mod ; return modulo ; function to find last digit of a ^ b ; if a and b both are 0 ; if exponent is 0 ; if base is 0 ; if exponent is divisible by 4 that means last digit will be pow ( a , 4 ) % 10. if exponent is not divisible by 4 that means last digit will be pow ( a , b % 4 ) % 10 ; Find last digit in ' a ' and compute its exponent ; Return last digit of result ; Driver program to run test case | #include <bits/stdc++.h> NEW_LINE using namespace std ; int Modulo ( int a , char b [ ] ) { int mod = 0 ; for ( int i = 0 ; i < strlen ( b ) ; i ++ ) mod = ( mod * 10 + b [ i ] - '0' ) % a ; } int LastDigit ( char a [ ] , char b [ ] ) { int len_a = strlen ( a ) , len_b = strlen ( b ) ; if ( len_a == 1 && len_b == 1 && b [ 0 ] == '0' && a [ 0 ] == '0' ) return 1 ; if ( len_b == 1 && b [ 0 ] == '0' ) return 1 ; if ( len_a == 1 && a [ 0 ] == '0' ) return 0 ; int exp = ( Modulo ( 4 , b ) == 0 ) ? 4 : Modulo ( 4 , b ) ; int res = pow ( a [ len_a - 1 ] - '0' , exp ) ; return res % 10 ; } int main ( ) { char a [ ] = "117" , b [ ] = "3" ; cout << LastDigit ( a , b ) ; return 0 ; } |
Reverse and Add Function | C ++ Program to implement reverse and add function ; Iterative function to reverse digits of num ; Function to check whether he number is palindrome or not ; Reverse and Add Function ; Reversing the digits of the number ; Adding the reversed number with the original ; Checking whether the number is palindrome or not ; Driver Program | #include <bits/stdc++.h> NEW_LINE using namespace std ; long long reversDigits ( long long num ) { long long rev_num = 0 ; while ( num > 0 ) { rev_num = rev_num * 10 + num % 10 ; num = num / 10 ; } return rev_num ; } bool isPalindrome ( long long num ) { return ( reversDigits ( num ) == num ) ; } void ReverseandAdd ( long long num ) { long long rev_num = 0 ; while ( num <= 4294967295 ) { rev_num = reversDigits ( num ) ; num = num + rev_num ; if ( isPalindrome ( num ) ) { printf ( " % lld STRNEWLINE " , num ) ; break ; } else if ( num > 4294967295 ) { printf ( " No ▁ palindrome ▁ exist " ) ; } } } int main ( ) { ReverseandAdd ( 195 ) ; ReverseandAdd ( 265 ) ; return 0 ; } |
Stein 's Algorithm for finding GCD | Recursive C ++ program to implement Stein 's Algorithm ; Function to implement Stein 's Algorithm ; GCD ( 0 , b ) == b ; GCD ( a , 0 ) == a , GCD ( 0 , 0 ) == 0 ; look for factors of 2 if ( ~ a & 1 ) a is even ; if ( b & 1 ) b is odd ; else both a and b are even ; if ( ~ b & 1 ) a is odd , b is even ; reduce larger number ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int gcd ( int a , int b ) { if ( a == b ) return a ; if ( a == 0 ) return b ; if ( b == 0 ) return a ; { return gcd ( a >> 1 , b ) ; return gcd ( a >> 1 , b >> 1 ) << 1 ; } return gcd ( a , b >> 1 ) ; if ( a > b ) return gcd ( ( a - b ) >> 1 , b ) ; return gcd ( ( b - a ) >> 1 , a ) ; } int main ( ) { int a = 34 , b = 17 ; printf ( " Gcd ▁ of ▁ given ▁ numbers ▁ is ▁ % d STRNEWLINE " , gcd ( a , b ) ) ; return 0 ; } |
Sub | C ++ program to answer multiple queries of divisibility by 3 in substrings of a number ; Array to store the sum of digits ; Utility function to evaluate a character 's integer value ; This function receives the string representation of the number and precomputes the sum array ; This function receives l and r representing the indices and prints the required output ; Driver function to check the program | #include <iostream> NEW_LINE using namespace std ; int sum [ 1000005 ] ; int toInt ( char x ) { return int ( x ) - '0' ; } void prepareSum ( string s ) { sum [ 0 ] = 0 ; for ( int i = 0 ; i < s . length ( ) ; i ++ ) sum [ i + 1 ] = sum [ i ] + toInt ( s [ i ] ) ; } void query ( int l , int r ) { if ( ( sum [ r + 1 ] - sum [ l ] ) % 3 == 0 ) cout << " Divisible ▁ by ▁ 3 STRNEWLINE " ; else cout << " Not ▁ divisible ▁ by ▁ 3 STRNEWLINE " ; } int main ( ) { string n = "12468236544" ; prepareSum ( n ) ; query ( 0 , 1 ) ; query ( 1 , 2 ) ; query ( 3 , 6 ) ; query ( 0 , 10 ) ; return 0 ; } |
Print all n | A C ++ recursive program to print all n - digit numbers whose sum of digits equals to given sum ; n , sum -- > value of inputs out -- > output array index -- > index of next digit to be filled in output array ; Base case ; If number becomes N - digit ; if sum of its digits is equal to given sum , print it ; Traverse through every digit . Note that here we ' re ▁ considering ▁ leading ▁ 0' s as digits ; append current digit to number ; recurse for next digit with reduced sum ; This is mainly a wrapper over findNDigitNumsUtil . It explicitly handles leading digit ; output array to store N - digit numbers ; fill 1 st position by every digit from 1 to 9 and calls findNDigitNumsUtil ( ) for remaining positions ; Driver program | #include <bits/stdc++.h> NEW_LINE using namespace std ; void findNDigitNumsUtil ( int n , int sum , char * out , int index ) { if ( index > n sum < 0 ) return ; if ( index == n ) { if ( sum == 0 ) { out [ index ] = ' \0' ; cout << out << " ▁ " ; } return ; } for ( int i = 0 ; i <= 9 ; i ++ ) { out [ index ] = i + '0' ; findNDigitNumsUtil ( n , sum - i , out , index + 1 ) ; } } void findNDigitNums ( int n , int sum ) { char out [ n + 1 ] ; for ( int i = 1 ; i <= 9 ; i ++ ) { out [ 0 ] = i + '0' ; findNDigitNumsUtil ( n , sum - i , out , 1 ) ; } } int main ( ) { int n = 2 , sum = 3 ; findNDigitNums ( n , sum ) ; return 0 ; } |
Palindromic Primes | C ++ Program to print all palindromic primes smaller than or equal to n . ; A function that returns true only if num contains one digit ; comparison operation is faster than division operation . So using following instead of " return ▁ num ▁ / ▁ 10 ▁ = = ▁ 0 ; " ; A recursive function to find out whether num is palindrome or not . Initially , dupNum contains address of a copy of num . ; Base case ( needed for recursion termination ) : This statement / mainly compares the first digit with the last digit ; This is the key line in this method . Note that all recursive / calls have a separate copy of num , but they all share same copy of * dupNum . We divide num while moving up the recursion tree ; The following statements are executed when we move up the recursion call tree ; At this point , if num % 10 contains i ' th ▁ ▁ digit ▁ from ▁ beginning , ▁ then ▁ ( * dupNum ) %10 ▁ ▁ contains ▁ i ' th digit from end ; The main function that uses recursive function isPalUtil ( ) to find out whether num is palindrome or not ; If num is negative , make it positive ; Create a separate copy of num , so that modifications made to address dupNum don 't change the input number. int *dupNum = new int(num); *dupNum = num ; Function to generate all primes and checking whether number is palindromic or not ; Create a boolean array " prime [ 0 . . n ] " and initialize all entries it as true . A value in prime [ i ] will finally be false if i is Not a prime , else true . ; If prime [ p ] is not changed , then it is a prime ; Update all multiples of p ; Print all palindromic prime numbers ; checking whether the given number is prime palindromic or not ; Driver Program | #include <bits/stdc++.h> NEW_LINE using namespace std ; int oneDigit ( int num ) { return ( num >= 0 && num < 10 ) ; } bool isPalUtil ( int num , int * dupNum ) { if ( oneDigit ( num ) ) return ( num == ( * dupNum ) % 10 ) ; if ( ! isPalUtil ( num / 10 , dupNum ) ) return false ; * dupNum /= 10 ; return ( num % 10 == ( * dupNum ) % 10 ) ; } int isPal ( int num ) { if ( num < 0 ) num = - num ; return isPalUtil ( num , dupNum ) ; } void printPalPrimesLessThanN ( int n ) { bool prime [ n + 1 ] ; memset ( prime , true , sizeof ( prime ) ) ; for ( int p = 2 ; p * p <= n ; p ++ ) { if ( prime [ p ] == true ) { for ( int i = p * 2 ; i <= n ; i += p ) prime [ i ] = false ; } } for ( int p = 2 ; p <= n ; p ++ ) if ( prime [ p ] && isPal ( p ) ) cout << p << " ▁ " ; } int main ( ) { int n = 100 ; printf ( " Palindromic ▁ primes ▁ smaller ▁ than ▁ or ▁ " " equal ▁ to ▁ % d ▁ are ▁ : STRNEWLINE " , n ) ; printPalPrimesLessThanN ( n ) ; } |
Almost Prime Numbers | Program to print first n numbers that are k - primes ; A function to count all prime factors of a given number ; Count the number of 2 s that divide n ; n must be odd at this point . So we can skip one element ( Note i = i + 2 ) ; While i divides n , count i and divide n ; This condition is to handle the case when n is a prime number greater than 2 ; A function to print the first n numbers that are k - almost primes . ; Print this number if it is k - prime ; Increment count of k - primes printed so far ; Driver program to test above function | #include <bits/stdc++.h> NEW_LINE using namespace std ; int countPrimeFactors ( int n ) { int count = 0 ; while ( n % 2 == 0 ) { n = n / 2 ; count ++ ; } for ( int i = 3 ; i <= sqrt ( n ) ; i = i + 2 ) { while ( n % i == 0 ) { n = n / i ; count ++ ; } } if ( n > 2 ) count ++ ; return ( count ) ; } void printKAlmostPrimes ( int k , int n ) { for ( int i = 1 , num = 2 ; i <= n ; num ++ ) { if ( countPrimeFactors ( num ) == k ) { printf ( " % d ▁ " , num ) ; i ++ ; } } return ; } int main ( ) { int n = 10 , k = 2 ; printf ( " First ▁ % d ▁ % d - almost ▁ prime ▁ numbers ▁ : ▁ STRNEWLINE " , n , k ) ; printKAlmostPrimes ( k , n ) ; return 0 ; } |
Program to add two fractions | C ++ program to add 2 fractions ; Function to return gcd of a and b ; Function to convert the obtained fraction into it 's simplest form ; Finding gcd of both terms ; Converting both terms into simpler terms by dividing them by common factor ; Function to add two fractions ; Finding gcd of den1 and den2 ; Denominator of final fraction obtained finding LCM of den1 and den2 LCM * GCD = a * b ; Changing the fractions to have same denominator Numerator of the final fraction obtained ; Calling function to convert final fraction into it 's simplest form ; Driver program | #include <bits/stdc++.h> NEW_LINE using namespace std ; int gcd ( int a , int b ) { if ( a == 0 ) return b ; return gcd ( b % a , a ) ; } void lowest ( int & den3 , int & num3 ) { int common_factor = gcd ( num3 , den3 ) ; den3 = den3 / common_factor ; num3 = num3 / common_factor ; } void addFraction ( int num1 , int den1 , int num2 , int den2 , int & num3 , int & den3 ) { den3 = gcd ( den1 , den2 ) ; den3 = ( den1 * den2 ) / den3 ; num3 = ( num1 ) * ( den3 / den1 ) + ( num2 ) * ( den3 / den2 ) ; lowest ( den3 , num3 ) ; } int main ( ) { int num1 = 1 , den1 = 500 , num2 = 2 , den2 = 1500 , den3 , num3 ; addFraction ( num1 , den1 , num2 , den2 , num3 , den3 ) ; printf ( " % d / % d ▁ + ▁ % d / % d ▁ is ▁ equal ▁ to ▁ % d / % d STRNEWLINE " , num1 , den1 , num2 , den2 , num3 , den3 ) ; return 0 ; } |
The Lazy Caterer 's Problem | A C ++ program to find the solution to The Lazy Caterer 's Problem ; This function receives an integer n and returns the maximum number of pieces that can be made form pancake using n cuts ; Use the formula ; Driver Code | #include <iostream> NEW_LINE using namespace std ; int findPieces ( int n ) { return ( n * ( n + 1 ) ) / 2 + 1 ; } int main ( ) { cout << findPieces ( 1 ) << endl ; cout << findPieces ( 2 ) << endl ; cout << findPieces ( 3 ) << endl ; cout << findPieces ( 50 ) << endl ; return 0 ; } |
Count digits in a factorial | Set 2 | A optimised program to find the number of digits in a factorial ; Returns the number of digits present in n ! Since the result can be large long long is used as return type ; factorial of - ve number doesn 't exists ; base case ; Use Kamenetsky formula to calculate the number of digits ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; long long findDigits ( int n ) { if ( n < 0 ) return 0 ; if ( n <= 1 ) return 1 ; double x = ( ( n * log10 ( n / M_E ) + log10 ( 2 * M_PI * n ) / 2.0 ) ) ; return floor ( x ) + 1 ; } int main ( ) { cout << findDigits ( 1 ) << endl ; cout << findDigits ( 50000000 ) << endl ; cout << findDigits ( 1000000000 ) << endl ; cout << findDigits ( 120 ) << endl ; return 0 ; } |
Count digits in a factorial | Set 1 | A C ++ program to find the number of digits in a factorial ; This function receives an integer n , and returns the number of digits present in n ! ; factorial exists only for n >= 0 ; base case ; else iterate through n and calculate the value ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int findDigits ( int n ) { if ( n < 0 ) return 0 ; if ( n <= 1 ) return 1 ; double digits = 0 ; for ( int i = 2 ; i <= n ; i ++ ) digits += log10 ( i ) ; return floor ( digits ) + 1 ; } int main ( ) { cout << findDigits ( 1 ) << endl ; cout << findDigits ( 5 ) << endl ; cout << findDigits ( 10 ) << endl ; cout << findDigits ( 120 ) << endl ; return 0 ; } |
Find number of subarrays with even sum | C ++ program to count number of sub - arrays whose sum is even using brute force Time Complexity - O ( N ^ 2 ) Space Complexity - O ( 1 ) ; Find sum of all subarrays and increment result if sum is even ; Driver code | #include <iostream> NEW_LINE using namespace std ; int countEvenSum ( int arr [ ] , int n ) { int result = 0 ; for ( int i = 0 ; i <= n - 1 ; i ++ ) { int sum = 0 ; for ( int j = i ; j <= n - 1 ; j ++ ) { sum = sum + arr [ j ] ; if ( sum % 2 == 0 ) result ++ ; } } return ( result ) ; } int main ( ) { int arr [ ] = { 1 , 2 , 2 , 3 , 4 , 1 } ; int n = sizeof ( arr ) / sizeof ( arr [ 0 ] ) ; cout << " The ▁ Number ▁ of ▁ Subarrays ▁ with ▁ even " " ▁ sum ▁ is ▁ " << countEvenSum ( arr , n ) ; return ( 0 ) ; } |
Find number of subarrays with even sum | C ++ program to count number of sub - arrays with even sum using an efficient algorithm Time Complexity - O ( N ) Space Complexity - O ( 1 ) ; A temporary array of size 2. temp [ 0 ] is going to store count of even subarrays and temp [ 1 ] count of odd . temp [ 0 ] is initialized as 1 because there a single even element is also counted as a subarray ; Initialize count . sum is sum of elements under modulo 2 and ending with arr [ i ] . ; i ' th ▁ iteration ▁ computes ▁ sum ▁ of ▁ arr [ 0 . . i ] ▁ ▁ under ▁ modulo ▁ 2 ▁ and ▁ increments ▁ even / odd ▁ count ▁ ▁ according ▁ to ▁ sum ' s value ; 2 is added to handle negative numbers ; Increment even / odd count ; Use handshake lemma to count even subarrays ( Note that an even cam be formed by two even or two odd ) ; Driver code | #include <iostream> NEW_LINE using namespace std ; int countEvenSum ( int arr [ ] , int n ) { int temp [ 2 ] = { 1 , 0 } ; int result = 0 , sum = 0 ; for ( int i = 0 ; i <= n - 1 ; i ++ ) { sum = ( ( sum + arr [ i ] ) % 2 + 2 ) % 2 ; temp [ sum ] ++ ; } result = result + ( temp [ 0 ] * ( temp [ 0 ] - 1 ) / 2 ) ; result = result + ( temp [ 1 ] * ( temp [ 1 ] - 1 ) / 2 ) ; return ( result ) ; } int main ( ) { int arr [ ] = { 1 , 2 , 2 , 3 , 4 , 1 } ; int n = sizeof ( arr ) / sizeof ( arr [ 0 ] ) ; cout << " The ▁ Number ▁ of ▁ Subarrays ▁ with ▁ even " " ▁ sum ▁ is ▁ " << countEvenSum ( arr , n ) ; return ( 0 ) ; } |
Sum of Fibonacci Numbers | C ++ Program to find sum of Fibonacci numbers ; Computes value of first fibonacci numbers ; Initialize result ; Add remaining terms ; Driver program to test above function | #include <bits/stdc++.h> NEW_LINE using namespace std ; int calculateSum ( int n ) { if ( n <= 0 ) return 0 ; int fibo [ n + 1 ] ; fibo [ 0 ] = 0 , fibo [ 1 ] = 1 ; int sum = fibo [ 0 ] + fibo [ 1 ] ; for ( int i = 2 ; i <= n ; i ++ ) { fibo [ i ] = fibo [ i - 1 ] + fibo [ i - 2 ] ; sum += fibo [ i ] ; } return sum ; } int main ( ) { int n = 4 ; cout << " Sum ▁ of ▁ Fibonacci ▁ numbers ▁ is ▁ : ▁ " << calculateSum ( n ) << endl ; return 0 ; } |
Sum of Fibonacci Numbers | C ++ Program to find sum of Fibonacci numbers in O ( Log n ) time . ; Create an array for memoization ; Returns n 'th Fibonacci number using table f[] ; Base cases ; If fib ( n ) is already computed ; Applying above formula [ Note value n & 1 is 1 if n is odd , else 0 ] . ; Computes value of first Fibonacci numbers ; Driver program to test above function | #include <bits/stdc++.h> NEW_LINE using namespace std ; const int MAX = 1000 ; int f [ MAX ] = { 0 } ; int fib ( int n ) { if ( n == 0 ) return 0 ; if ( n == 1 n == 2 ) return ( f [ n ] = 1 ) ; if ( f [ n ] ) return f [ n ] ; int k = ( n & 1 ) ? ( n + 1 ) / 2 : n / 2 ; f [ n ] = ( n & 1 ) ? ( fib ( k ) * fib ( k ) + fib ( k - 1 ) * fib ( k - 1 ) ) : ( 2 * fib ( k - 1 ) + fib ( k ) ) * fib ( k ) ; return f [ n ] ; } int calculateSum ( int n ) { return fib ( n + 2 ) - 1 ; } int main ( ) { int n = 4 ; cout << " Sum ▁ of ▁ Fibonacci ▁ numbers ▁ is ▁ : ▁ " << calculateSum ( n ) << endl ; return 0 ; } |
Find all combinations that add upto given number | C ++ program to find out all combinations of positive numbers that add upto given number ; arr - array to store the combination index - next location in array num - given number reducedNum - reduced number ; Base condition ; If combination is found , print it ; Find the previous number stored in arr [ ] It helps in maintaining increasing order ; note loop starts from previous number i . e . at array location index - 1 ; next element of array is k ; call recursively with reduced number ; Function to find out all combinations of positive numbers that add upto given number . It uses findCombinationsUtil ( ) ; array to store the combinations It can contain max n elements ; find all combinations ; Driver code | #include <iostream> NEW_LINE using namespace std ; void findCombinationsUtil ( int arr [ ] , int index , int num , int reducedNum ) { if ( reducedNum < 0 ) return ; if ( reducedNum == 0 ) { for ( int i = 0 ; i < index ; i ++ ) cout << arr [ i ] << " ▁ " ; cout << endl ; return ; } int prev = ( index == 0 ) ? 1 : arr [ index - 1 ] ; for ( int k = prev ; k <= num ; k ++ ) { arr [ index ] = k ; findCombinationsUtil ( arr , index + 1 , num , reducedNum - k ) ; } } void findCombinations ( int n ) { int arr [ n ] ; findCombinationsUtil ( arr , 0 , n , n ) ; } int main ( ) { int n = 5 ; findCombinations ( n ) ; return 0 ; } |
Combinatorial Game Theory | Set 2 ( Game of Nim ) | A C ++ program to implement Game of Nim . The program assumes that both players are playing optimally ; A Structure to hold the two parameters of a moveA move has two parameters - 1 ) pile_index = The index of pile from which stone is going to be removed2 ) stones_removed = Number of stones removed from the pile indexed = pile_index ; A C function to output the current game state . ; A C function that returns True if game has ended and False if game is not yet over ; A C function to declare the winner of the game ; A C function to calculate the Nim - Sum at any point of the game . ; A C function to make moves of the Nim Game ; The player having the current turn is on a winning position . So he / she / it play optimally and tries to make Nim - Sum as 0 ; If this is not an illegal move then make this move . ; If you want to input yourself then remove the rand ( ) functions and modify the code to take inputs . But remember , you still won 't be able to change your fate/prediction. ; Create an array to hold indices of non - empty piles ; A C function to play the Game of Nim ; Driver program to test above functions ; Test Case 1 ; We will predict the results before playing The COMPUTER starts first ; Let us play the game with COMPUTER starting first and check whether our prediction was right or not ; Test Case 2 int piles [ ] = { 3 , 4 , 7 } ; int n = sizeof ( piles ) / sizeof ( piles [ 0 ] ) ; We will predict the results before playing The HUMAN ( You ) starts first ; Let us play the game with COMPUTER starting first and check whether our prediction was right or not playGame ( piles , n , HUMAN ) ; | #include <iostream> NEW_LINE #include <math.h> NEW_LINE using namespace std ; #define COMPUTER 1 NEW_LINE #define HUMAN 2 NEW_LINE struct move { int pile_index ; int stones_removed ; } ; void showPiles ( int piles [ ] , int n ) { int i ; cout << " Current ▁ Game ▁ Status ▁ - > ▁ " ; for ( i = 0 ; i < n ; i ++ ) cout << " ▁ " << piles [ i ] ; cout << " STRNEWLINE " ; return ; } bool gameOver ( int piles [ ] , int n ) { int i ; for ( i = 0 ; i < n ; i ++ ) if ( piles [ i ] != 0 ) return ( false ) ; return ( true ) ; } void declareWinner ( int whoseTurn ) { if ( whoseTurn == COMPUTER ) cout << " HUMAN won " else cout << " COMPUTER won " return ; } int calculateNimSum ( int piles [ ] , int n ) { int i , nimsum = piles [ 0 ] ; for ( i = 1 ; i < n ; i ++ ) nimsum = nimsum ^ piles [ i ] ; return ( nimsum ) ; } void makeMove ( int piles [ ] , int n , struct move * moves ) { int i , nim_sum = calculateNimSum ( piles , n ) ; if ( nim_sum != 0 ) { for ( i = 0 ; i < n ; i ++ ) { if ( ( piles [ i ] ^ nim_sum ) < piles [ i ] ) { ( * moves ) . pile_index = i ; ( * moves ) . stones_removed = piles [ i ] - ( piles [ i ] ^ nim_sum ) ; piles [ i ] = ( piles [ i ] ^ nim_sum ) ; break ; } } } else { int non_zero_indices [ n ] , count ; for ( i = 0 , count = 0 ; i < n ; i ++ ) if ( piles [ i ] > 0 ) non_zero_indices [ count ++ ] = i ; ( * moves ) . pile_index = ( rand ( ) % ( count ) ) ; ( * moves ) . stones_removed = 1 + ( rand ( ) % ( piles [ ( * moves ) . pile_index ] ) ) ; piles [ ( * moves ) . pile_index ] = piles [ ( * moves ) . pile_index ] - ( * moves ) . stones_removed ; if ( piles [ ( * moves ) . pile_index ] < 0 ) piles [ ( * moves ) . pile_index ] = 0 ; } return ; } void playGame ( int piles [ ] , int n , int whoseTurn ) { cout << " GAME STARTS " struct move moves ; while ( gameOver ( piles , n ) == false ) { showPiles ( piles , n ) ; makeMove ( piles , n , & moves ) ; if ( whoseTurn == COMPUTER ) { cout << " COMPUTER ▁ removes " << moves . stones_removed << " stones ▁ from ▁ pile ▁ at ▁ index ▁ " << moves . pile_index << endl ; whoseTurn = HUMAN ; } else { cout << " HUMAN ▁ removes " << moves . stones_removed << " stones ▁ from ▁ pile ▁ at ▁ index ▁ " << moves . pile_index << endl ; whoseTurn = COMPUTER ; } } showPiles ( piles , n ) ; declareWinner ( whoseTurn ) ; return ; } void knowWinnerBeforePlaying ( int piles [ ] , int n , int whoseTurn ) { cout << " Prediction ▁ before ▁ playing ▁ the ▁ game ▁ - > ▁ " ; if ( calculateNimSum ( piles , n ) != 0 ) { if ( whoseTurn == COMPUTER ) cout << " COMPUTER ▁ will ▁ win STRNEWLINE " ; else cout << " HUMAN ▁ will ▁ win STRNEWLINE " ; } else { if ( whoseTurn == COMPUTER ) cout << " HUMAN ▁ will ▁ win STRNEWLINE " ; else cout << " COMPUTER ▁ will ▁ win STRNEWLINE " ; } return ; } int main ( ) { int piles [ ] = { 3 , 4 , 5 } ; int n = sizeof ( piles ) / sizeof ( piles [ 0 ] ) ; knowWinnerBeforePlaying ( piles , n , COMPUTER ) ; playGame ( piles , n , COMPUTER ) ; knowWinnerBeforePlaying ( piles , n , COMPUTER ) ; return ( 0 ) ; } |
Find Square Root under Modulo p | Set 2 ( Shanks Tonelli algorithm ) | C ++ program to implement Shanks Tonelli algorithm for finding Modular Square Roots ; utility function to find pow ( base , exponent ) % modulus ; utility function to find gcd ; Returns k such that b ^ k = 1 ( mod p ) ; Initializing k with first odd prime number ; function return p - 1 ( = x argument ) as x * 2 ^ e , where x will be odd sending e as reference because updation is needed in actual e ; Main function for finding the modular square root ; a and p should be coprime for finding the modular square root ; If below expression return ( p - 1 ) then modular square root is not possible ; expressing p - 1 , in terms of s * 2 ^ e , where s is odd number ; finding smallest q such that q ^ ( ( p - 1 ) / 2 ) ( mod p ) = p - 1 ; q - 1 is in place of ( - 1 % p ) ; Initializing variable x , b and g ; keep looping until b become 1 or m becomes 0 ; finding m such that b ^ ( 2 ^ m ) = 1 ; updating value of x , g and b according to algorithm ; driver program to test above function ; p should be prime | #include <bits/stdc++.h> NEW_LINE using namespace std ; int pow ( int base , int exponent , int modulus ) { int result = 1 ; base = base % modulus ; while ( exponent > 0 ) { if ( exponent % 2 == 1 ) result = ( result * base ) % modulus ; exponent = exponent >> 1 ; base = ( base * base ) % modulus ; } return result ; } int gcd ( int a , int b ) { if ( b == 0 ) return a ; else return gcd ( b , a % b ) ; } int order ( int p , int b ) { if ( gcd ( p , b ) != 1 ) { printf ( " p ▁ and ▁ b ▁ are ▁ not ▁ co - prime . STRNEWLINE " ) ; return -1 ; } int k = 3 ; while ( 1 ) { if ( pow ( b , k , p ) == 1 ) return k ; k ++ ; } } int convertx2e ( int x , int & e ) { e = 0 ; while ( x % 2 == 0 ) { x /= 2 ; e ++ ; } return x ; } int STonelli ( int n , int p ) { if ( gcd ( n , p ) != 1 ) { printf ( " a ▁ and ▁ p ▁ are ▁ not ▁ coprime STRNEWLINE " ) ; return -1 ; } if ( pow ( n , ( p - 1 ) / 2 , p ) == ( p - 1 ) ) { printf ( " no ▁ sqrt ▁ possible STRNEWLINE " ) ; return -1 ; } int s , e ; s = convertx2e ( p - 1 , e ) ; int q ; for ( q = 2 ; ; q ++ ) { if ( pow ( q , ( p - 1 ) / 2 , p ) == ( p - 1 ) ) break ; } int x = pow ( n , ( s + 1 ) / 2 , p ) ; int b = pow ( n , s , p ) ; int g = pow ( q , s , p ) ; int r = e ; while ( 1 ) { int m ; for ( m = 0 ; m < r ; m ++ ) { if ( order ( p , b ) == -1 ) return -1 ; if ( order ( p , b ) == pow ( 2 , m ) ) break ; } if ( m == 0 ) return x ; x = ( x * pow ( g , pow ( 2 , r - m - 1 ) , p ) ) % p ; g = pow ( g , pow ( 2 , r - m ) , p ) ; b = ( b * g ) % p ; if ( b == 1 ) return x ; r = m ; } } int main ( ) { int n = 2 ; int p = 113 ; int x = STonelli ( n , p ) ; if ( x == -1 ) printf ( " Modular ▁ square ▁ root ▁ is ▁ not ▁ exist STRNEWLINE " ) ; else printf ( " Modular ▁ square ▁ root ▁ of ▁ % d ▁ and ▁ % d ▁ is ▁ % d STRNEWLINE " , n , p , x ) ; } |
Check if a number is a power of another number | CPP program to check given number number y is power of x ; logarithm function to calculate value ; compare to the result1 or result2 both are equal ; Driven program | #include <iostream> NEW_LINE #include <math.h> NEW_LINE using namespace std ; bool isPower ( int x , int y ) { int res1 = log ( y ) / log ( x ) ; return ( res1 == res2 ) ; } int main ( ) { cout << isPower ( 27 , 729 ) << endl ; return 0 ; } |
Program to find the Roots of Quadratic equation | C ++ program to find roots of a quadratic equation ; Prints roots of quadratic equation ax * 2 + bx + x ; If a is 0 , then equation is not quadratic , but linear ; else d < 0 ; Driver code ; Function call | #include <bits/stdc++.h> NEW_LINE using namespace std ; void findRoots ( int a , int b , int c ) { if ( a == 0 ) { cout << " Invalid " ; return ; } int d = b * b - 4 * a * c ; double sqrt_val = sqrt ( abs ( d ) ) ; if ( d > 0 ) { cout << " Roots ▁ are ▁ real ▁ and ▁ different ▁ STRNEWLINE " ; cout << ( double ) ( - b + sqrt_val ) / ( 2 * a ) << " STRNEWLINE " << ( double ) ( - b - sqrt_val ) / ( 2 * a ) ; } else if ( d == 0 ) { cout << " Roots ▁ are ▁ real ▁ and ▁ same ▁ STRNEWLINE " ; cout << - ( double ) b / ( 2 * a ) ; } { cout << " Roots ▁ are ▁ complex ▁ STRNEWLINE " ; cout << - ( double ) b / ( 2 * a ) << " ▁ + ▁ i " << sqrt_val << " STRNEWLINE " << - ( double ) b / ( 2 * a ) << " ▁ - ▁ i " << sqrt_val ; } } int main ( ) { int a = 1 , b = -7 , c = 12 ; findRoots ( a , b , c ) ; return 0 ; } |
Check perfect square using addition / subtraction | C ++ program to check if n is perfect square or not ; This function returns true if n is perfect square , else false ; sum is sum of all odd numbers . i is used one by one hold odd numbers ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool isPerfectSquare ( int n ) { for ( int sum = 0 , i = 1 ; sum < n ; i += 2 ) { sum += i ; if ( sum == n ) return true ; } return false ; } int main ( ) { isPerfectSquare ( 35 ) ? cout << " Yes STRNEWLINE " : cout << " No STRNEWLINE " ; isPerfectSquare ( 49 ) ? cout << " Yes STRNEWLINE " : cout << " No STRNEWLINE " ; return 0 ; } |
Count ' d ' digit positive integers with 0 as a digit | C ++ program to find the count of positive integer of a given number of digits that contain atleast one zero ; Returns count of ' d ' digit integers have 0 as a digit ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int findCount ( int d ) { return 9 * ( pow ( 10 , d - 1 ) - pow ( 9 , d - 1 ) ) ; } int main ( ) { int d = 1 ; cout << findCount ( d ) << endl ; d = 2 ; cout << findCount ( d ) << endl ; d = 4 ; cout << findCount ( d ) << endl ; return 0 ; } |
Dyck path | C ++ program to count number of Dyck Paths ; Returns count Dyck paths in n x n grid ; Compute value of 2 nCn ; return 2 nCn / ( n + 1 ) ; Driver Code | #include <iostream> NEW_LINE using namespace std ; int countDyckPaths ( unsigned int n ) { int res = 1 ; for ( int i = 0 ; i < n ; ++ i ) { res *= ( 2 * n - i ) ; res /= ( i + 1 ) ; } return res / ( n + 1 ) ; } int main ( ) { int n = 4 ; cout << " Number ▁ of ▁ Dyck ▁ Paths ▁ is ▁ " << countDyckPaths ( n ) ; return 0 ; } |
Triangular Numbers | C ++ program to check if a number is a triangular number using simple approach . ; Returns true if ' num ' is triangular , else false ; Base case ; A Triangular number must be sum of first n natural numbers ; Driver code | #include <iostream> NEW_LINE using namespace std ; bool isTriangular ( int num ) { if ( num < 0 ) return false ; int sum = 0 ; for ( int n = 1 ; sum <= num ; n ++ ) { sum = sum + n ; if ( sum == num ) return true ; } return false ; } int main ( ) { int n = 55 ; if ( isTriangular ( n ) ) cout << " The ▁ number ▁ is ▁ a ▁ triangular ▁ number " ; else cout << " The ▁ number ▁ is ▁ NOT ▁ a ▁ triangular ▁ number " ; return 0 ; } |
Triangular Numbers | C ++ program to check if a number is a triangular number using quadratic equation . ; Returns true if num is triangular ; Considering the equation n * ( n + 1 ) / 2 = num The equation is : a ( n ^ 2 ) + bn + c = 0 "; ; Find roots of equation ; checking if root1 is natural ; checking if root2 is natural ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool isTriangular ( int num ) { if ( num < 0 ) return false ; int c = ( -2 * num ) ; int b = 1 , a = 1 ; int d = ( b * b ) - ( 4 * a * c ) ; if ( d < 0 ) return false ; float root1 = ( - b + sqrt ( d ) ) / ( 2 * a ) ; float root2 = ( - b - sqrt ( d ) ) / ( 2 * a ) ; if ( root1 > 0 && floor ( root1 ) == root1 ) return true ; if ( root2 > 0 && floor ( root2 ) == root2 ) return true ; return false ; } int main ( ) { int num = 55 ; if ( isTriangular ( num ) ) cout << " The ▁ number ▁ is ▁ a ▁ triangular ▁ number " ; else cout << " The ▁ number ▁ is ▁ NOT ▁ a ▁ triangular ▁ number " ; return 0 ; } |
Frobenius coin problem | C ++ program to find the largest number that cannot be formed from given two coins ; Utility function to find gcd ; Function to print the desired output ; Solution doesn 't exist if GCD is not 1 ; Else apply the formula ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int gcd ( int a , int b ) { int c ; while ( a != 0 ) { c = a ; a = b % a ; b = c ; } return b ; } void forbenius ( int X , int Y ) { if ( gcd ( X , Y ) != 1 ) { cout << " NA STRNEWLINE " ; return ; } int A = ( X * Y ) - ( X + Y ) ; int N = ( X - 1 ) * ( Y - 1 ) / 2 ; cout << " Largest ▁ Amount ▁ = ▁ " << A << endl ; cout << " Total ▁ Count ▁ = ▁ " << N << endl ; } int main ( ) { int X = 2 , Y = 5 ; forbenius ( X , Y ) ; X = 5 , Y = 10 ; cout << endl ; forbenius ( X , Y ) ; return 0 ; } |
Gray to Binary and Binary to Gray conversion | C ++ program for Binary To Gray and Gray to Binary conversion ; Helper function to xor two characters ; Helper function to flip the bit ; function to convert binary string to gray string ; MSB of gray code is same as binary code ; Compute remaining bits , next bit is computed by doing XOR of previous and current in Binary ; Concatenate XOR of previous bit with current bit ; function to convert gray code string to binary string ; MSB of binary code is same as gray code ; Compute remaining bits ; If current bit is 0 , concatenate previous bit ; Else , concatenate invert of previous bit ; Driver program to test above functions | #include <iostream> NEW_LINE using namespace std ; char xor_c ( char a , char b ) { return ( a == b ) ? '0' : '1' ; } char flip ( char c ) { return ( c == '0' ) ? '1' : '0' ; } string binarytoGray ( string binary ) { string gray = " " ; gray += binary [ 0 ] ; for ( int i = 1 ; i < binary . length ( ) ; i ++ ) { gray += xor_c ( binary [ i - 1 ] , binary [ i ] ) ; } return gray ; } string graytoBinary ( string gray ) { string binary = " " ; binary += gray [ 0 ] ; for ( int i = 1 ; i < gray . length ( ) ; i ++ ) { if ( gray [ i ] == '0' ) binary += binary [ i - 1 ] ; else binary += flip ( binary [ i - 1 ] ) ; } return binary ; } int main ( ) { string binary = "01001" ; cout << " Gray ▁ code ▁ of ▁ " << binary << " ▁ is ▁ " << binarytoGray ( binary ) << endl ; string gray = "01101" ; cout << " Binary ▁ code ▁ of ▁ " << gray << " ▁ is ▁ " << graytoBinary ( gray ) << endl ; return 0 ; } |
Solving f ( n ) = ( 1 ) + ( 2 * 3 ) + ( 4 * 5 * 6 ) . . . n using Recursion | CPP Program to print the solution of the series f ( n ) = ( 1 ) + ( 2 * 3 ) + ( 4 * 5 * 6 ) . . . n using recursion ; Recursive function for finding sum of series calculated - number of terms till which sum of terms has been calculated current - number of terms for which sum has to becalculated N - Number of terms in the function to be calculated ; checking termination condition ; product of terms till current ; recursive call for adding terms next in the series ; Driver Code ; input number of terms in the series ; invoking the function to calculate the sum | #include <bits/stdc++.h> NEW_LINE using namespace std ; int seriesSum ( int calculated , int current , int N ) { int i , cur = 1 ; if ( current == N + 1 ) return 0 ; for ( i = calculated ; i < calculated + current ; i ++ ) cur *= i ; return cur + seriesSum ( i , current + 1 , N ) ; } int main ( ) { int N = 5 ; cout << seriesSum ( 1 , 1 , N ) << endl ; return 0 ; } |
How to avoid overflow in modular multiplication ? | C ++ program for modular multiplication without any overflow ; To compute ( a * b ) % mod ; ll res = 0 ; Initialize result ; If b is odd , add ' a ' to result ; Multiply ' a ' with 2 ; Divide b by 2 ; Return result ; Driver program | #include <iostream> NEW_LINE using namespace std ; typedef long long int ll ; ll mulmod ( ll a , ll b , ll mod ) { a = a % mod ; while ( b > 0 ) { if ( b % 2 == 1 ) res = ( res + a ) % mod ; a = ( a * 2 ) % mod ; b /= 2 ; } return res % mod ; } int main ( ) { ll a = 9223372036854775807 , b = 9223372036854775807 ; cout << mulmod ( a , b , 100000000000 ) ; return 0 ; } |
Count inversions in an array | Set 3 ( Using BIT ) | C ++ program to count inversions using Binary Indexed Tree ; Returns sum of arr [ 0. . index ] . This function assumes that the array is preprocessed and partial sums of array elements are stored in BITree [ ] . ; Traverse ancestors of BITree [ index ] ; Add current element of BITree to sum ; Move index to parent node in getSum View ; Updates a node in Binary Index Tree ( BITree ) at given index in BITree . The given value ' val ' is added to BITree [ i ] and all of its ancestors in tree . ; Traverse all ancestors and add ' val ' ; Add ' val ' to current node of BI Tree ; Update index to that of parent in update View ; Converts an array to an array with values from 1 to n and relative order of smaller and greater elements remains same . For example , { 7 , - 90 , 100 , 1 } is converted to { 3 , 1 , 4 , 2 } ; Create a copy of arrp [ ] in temp and sort the temp array in increasing order ; Traverse all array elements ; lower_bound ( ) Returns pointer to the first element greater than or equal to arr [ i ] ; Returns inversion count arr [ 0. . n - 1 ] ; Convert arr [ ] to an array with values from 1 to n and relative order of smaller and greater elements remains same . For example , { 7 , - 90 , 100 , 1 } is converted to { 3 , 1 , 4 , 2 } ; Create a BIT with size equal to maxElement + 1 ( Extra one is used so that elements can be directly be used as index ) ; Traverse all elements from right . ; Get count of elements smaller than arr [ i ] ; Add current element to BIT ; Driver program | #include <bits/stdc++.h> NEW_LINE using namespace std ; int getSum ( int BITree [ ] , int index ) { while ( index > 0 ) { sum += BITree [ index ] ; index -= index & ( - index ) ; } return sum ; } void updateBIT ( int BITree [ ] , int n , int index , int val ) { while ( index <= n ) { BITree [ index ] += val ; index += index & ( - index ) ; } } void convert ( int arr [ ] , int n ) { int temp [ n ] ; for ( int i = 0 ; i < n ; i ++ ) temp [ i ] = arr [ i ] ; sort ( temp , temp + n ) ; for ( int i = 0 ; i < n ; i ++ ) { arr [ i ] = lower_bound ( temp , temp + n , arr [ i ] ) - temp + 1 ; } } int getInvCount ( int arr [ ] , int n ) { convert ( arr , n ) ; int BIT [ n + 1 ] ; for ( int i = 1 ; i <= n ; i ++ ) BIT [ i ] = 0 ; for ( int i = n - 1 ; i >= 0 ; i -- ) { invcount += getSum ( BIT , arr [ i ] - 1 ) ; updateBIT ( BIT , n , arr [ i ] , 1 ) ; } return invcount ; } int main ( ) { int arr [ ] = { 8 , 4 , 2 , 1 } ; int n = sizeof ( arr ) / sizeof ( int ) ; cout << " Number ▁ of ▁ inversions ▁ are ▁ : ▁ " << getInvCount ( arr , n ) ; return 0 ; } |
Compute n ! under modulo p | Simple method to compute n ! % p ; Returns value of n ! % p ; Driver program | #include <bits/stdc++.h> NEW_LINE using namespace std ; int modFact ( int n , int p ) { if ( n >= p ) return 0 ; int result = 1 ; for ( int i = 1 ; i <= n ; i ++ ) result = ( result * i ) % p ; return result ; } int main ( ) { int n = 25 , p = 29 ; cout << modFact ( n , p ) ; return 0 ; } |
Chinese Remainder Theorem | Set 2 ( Inverse Modulo based Implementation ) | A C ++ program to demonstrate working of Chinise remainder Theorem ; Returns modulo inverse of a with respect to m using extended Euclid Algorithm . Refer below post for details : https : www . geeksforgeeks . org / multiplicative - inverse - under - modulo - m / ; Apply extended Euclid Algorithm ; q is quotient ; m is remainder now , process same as euclid 's algo ; Make x1 positive ; k is size of num [ ] and rem [ ] . Returns the smallest number x such that : x % num [ 0 ] = rem [ 0 ] , x % num [ 1 ] = rem [ 1 ] , ... ... ... ... ... ... x % num [ k - 2 ] = rem [ k - 1 ] Assumption : Numbers in num [ ] are pairwise coprime ( gcd for every pair is 1 ) ; Compute product of all numbers ; Initialize result ; Apply above formula ; Driver method | #include <bits/stdc++.h> NEW_LINE using namespace std ; int inv ( int a , int m ) { int m0 = m , t , q ; int x0 = 0 , x1 = 1 ; if ( m == 1 ) return 0 ; while ( a > 1 ) { q = a / m ; t = m ; m = a % m , a = t ; t = x0 ; x0 = x1 - q * x0 ; x1 = t ; } if ( x1 < 0 ) x1 += m0 ; return x1 ; } int findMinX ( int num [ ] , int rem [ ] , int k ) { int prod = 1 ; for ( int i = 0 ; i < k ; i ++ ) prod *= num [ i ] ; int result = 0 ; for ( int i = 0 ; i < k ; i ++ ) { int pp = prod / num [ i ] ; result += rem [ i ] * inv ( pp , num [ i ] ) * pp ; } return result % prod ; } int main ( void ) { int num [ ] = { 3 , 4 , 5 } ; int rem [ ] = { 2 , 3 , 1 } ; int k = sizeof ( num ) / sizeof ( num [ 0 ] ) ; cout << " x ▁ is ▁ " << findMinX ( num , rem , k ) ; return 0 ; } |
Chinese Remainder Theorem | Set 1 ( Introduction ) | A C ++ program to demonstrate working of Chinise remainder Theorem ; k is size of num [ ] and rem [ ] . Returns the smallest number x such that : x % num [ 0 ] = rem [ 0 ] , x % num [ 1 ] = rem [ 1 ] , ... ... ... ... ... ... x % num [ k - 2 ] = rem [ k - 1 ] Assumption : Numbers in num [ ] are pairwise coprime ( gcd for every pair is 1 ) ; As per the Chinese remainder theorem , this loop will always break . ; Check if remainder of x % num [ j ] is rem [ j ] or not ( for all j from 0 to k - 1 ) ; If all remainders matched , we found x ; Else try next number ; Driver method | #include <bits/stdc++.h> NEW_LINE using namespace std ; int findMinX ( int num [ ] , int rem [ ] , int k ) { while ( true ) { int j ; for ( j = 0 ; j < k ; j ++ ) if ( x % num [ j ] != rem [ j ] ) break ; if ( j == k ) return x ; x ++ ; } return x ; } int main ( void ) { int num [ ] = { 3 , 4 , 5 } ; int rem [ ] = { 2 , 3 , 1 } ; int k = sizeof ( num ) / sizeof ( num [ 0 ] ) ; cout << " x ▁ is ▁ " << findMinX ( num , rem , k ) ; return 0 ; } |
Compute nCr % p | Set 2 ( Lucas Theorem ) | A Lucas Theorem based solution to compute nCr % p ; Returns nCr % p . In this Lucas Theorem based program , this function is only called for n < p and r < p . ; The array C is going to store last row of pascal triangle at the end . And last entry of last row is nCr ; One by constructs remaining rows of Pascal Triangle from top to bottom ; Fill entries of current row using previous row values ; nCj = ( n - 1 ) Cj + ( n - 1 ) C ( j - 1 ) ; ; Lucas Theorem based function that returns nCr % p This function works like decimal to binary conversion recursive function . First we compute last digits of n and r in base p , then recur for remaining digits ; Base case ; Compute last digits of n and r in base p ; Compute result for last digits computed above , and for remaining digits . Multiply the two results and compute the result of multiplication in modulo p . return ( nCrModpLucas ( n / p , r / p , p ) * Last digits of n and r nCrModpDP ( ni , ri , p ) ) % p ; Remaining digits ; Driver program | #include <bits/stdc++.h> NEW_LINE using namespace std ; int nCrModpDP ( int n , int r , int p ) { int C [ r + 1 ] ; memset ( C , 0 , sizeof ( C ) ) ; for ( int i = 1 ; i <= n ; i ++ ) { for ( int j = min ( i , r ) ; j > 0 ; j -- ) C [ j ] = ( C [ j ] + C [ j - 1 ] ) % p ; } return C [ r ] ; } int nCrModpLucas ( int n , int r , int p ) { if ( r == 0 ) return 1 ; int ni = n % p , ri = r % p ; } int main ( ) { int n = 1000 , r = 900 , p = 13 ; cout << " Value ▁ of ▁ nCr ▁ % ▁ p ▁ is ▁ " << nCrModpLucas ( n , r , p ) ; return 0 ; } |
Fibonacci Coding | C ++ program for Fibonacci Encoding of a positive integer n ; To limit on the largest Fibonacci number to be used ; Array to store fibonacci numbers . fib [ i ] is going to store ( i + 2 ) 'th Fibonacci number ; Stores values in fib and returns index of the largest fibonacci number smaller than n . ; Fib [ 0 ] stores 2 nd Fibonacci No . ; Fib [ 1 ] stores 3 rd Fibonacci No . ; Keep Generating remaining numbers while previously generated number is smaller ; Return index of the largest fibonacci number smaller than or equal to n . Note that the above loop stopped when fib [ i - 1 ] became larger . ; Returns pointer to the char string which corresponds to code for n ; allocate memory for codeword ; index of the largest Fibonacci f <= n ; Mark usage of Fibonacci f ( 1 bit ) ; Subtract f from n ; Move to Fibonacci just smaller than f ; Mark all Fibonacci > n as not used ( 0 bit ) , progress backwards ; additional '1' bit ; return pointer to codeword ; driver function | #include <bits/stdc++.h> NEW_LINE using namespace std ; #define N 30 NEW_LINE int fib [ N ] ; int largestFiboLessOrEqual ( int n ) { fib [ 0 ] = 1 ; fib [ 1 ] = 2 ; int i ; for ( i = 2 ; fib [ i - 1 ] <= n ; i ++ ) fib [ i ] = fib [ i - 1 ] + fib [ i - 2 ] ; return ( i - 2 ) ; } char * fibonacciEncoding ( int n ) { int index = largestFiboLessOrEqual ( n ) ; char * codeword = ( char * ) malloc ( sizeof ( char ) * ( index + 3 ) ) ; int i = index ; while ( n ) { codeword [ i ] = '1' ; n = n - fib [ i ] ; i = i - 1 ; while ( i >= 0 && fib [ i ] > n ) { codeword [ i ] = '0' ; i = i - 1 ; } } codeword [ index + 1 ] = '1' ; codeword [ index + 2 ] = ' \0' ; return codeword ; } int main ( ) { int n = 143 ; cout << " Fibonacci ▁ code ▁ word ▁ for ▁ " << n << " ▁ is ▁ " << fibonacciEncoding ( n ) ; return 0 ; } |
Print all Good numbers in given range | C ++ program to print good numbers in a given range [ L , R ] ; To check whether n is a good number and doesn ' t ▁ contain ▁ digit ▁ ' d '. ; Get last digit and initialize sum from right side ; If last digit is d , return ; Traverse remaining digits ; Current digit ; If digit is d or digit is less than or equal to sum of digits on right side ; Update sum and n ; Print Good numbers in range [ L , R ] ; Traverse all numbers in given range ; If current numbers is good , print it . ; Driver program ; Print good numbers in [ L , R ] | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool isValid ( int n , int d ) { int digit = n % 10 ; int sum = digit ; if ( digit == d ) return false ; n /= 10 ; while ( n ) { digit = n % 10 ; if ( digit == d digit <= sum ) return false ; else { sum += digit ; n /= 10 ; } } return 1 ; } void printGoodNumbers ( int L , int R , int d ) { for ( int i = L ; i <= R ; i ++ ) { if ( isValid ( i , d ) ) cout << i << " ▁ " ; } } int main ( ) { int L = 410 , R = 520 , d = 3 ; printGoodNumbers ( L , R , d ) ; return 0 ; } |
Count number of squares in a rectangle | C ++ program to count squares in a rectangle of size m x n ; Returns count of all squares in a rectangle of size m x n ; If n is smaller , swap m and n ; Now n is greater dimension , apply formula ; Driver Code | #include <iostream> NEW_LINE using namespace std ; int countSquares ( int m , int n ) { if ( n < m ) { int temp = m ; m = n ; n = temp ; } return n * ( n + 1 ) * ( 3 * m - n + 1 ) / 6 ; } int main ( ) { int m = 4 , n = 3 ; cout << " Count ▁ of ▁ squares ▁ is ▁ " << countSquares ( m , n ) ; } |
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