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Number of compositions of a natural number | C ++ program to find the total number of compositions of a natural number ; Return 2 raised to power ( n - 1 ) ; Driver Code | #include <iostream> NEW_LINE using namespace std ; #define ull unsigned long long NEW_LINE ull countCompositions ( ull n ) { return ( 1L ) << ( n - 1 ) ; } int main ( ) { ull n = 4 ; cout << countCompositions ( n ) << " STRNEWLINE " ; return 0 ; } |
Program to count digits in an integer ( 4 Different Methods ) | Recursive C ++ program to count number of digits in a number ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int countDigit ( long long n ) { if ( n / 10 == 0 ) return 1 ; return 1 + countDigit ( n / 10 ) ; } int main ( void ) { long long n = 345289467 ; cout << " Number ▁ of ▁ digits ▁ : " << countDigit ( n ) ; return 0 ; } |
Tribonacci Numbers | A DP based CPP program to print first n Tribonacci numbers . ; Driver code | #include <iostream> NEW_LINE using namespace std ; int printTrib ( int n ) { int dp [ n ] ; dp [ 0 ] = dp [ 1 ] = 0 ; dp [ 2 ] = 1 ; for ( int i = 3 ; i < n ; i ++ ) dp [ i ] = dp [ i - 1 ] + dp [ i - 2 ] + dp [ i - 3 ] ; for ( int i = 0 ; i < n ; i ++ ) cout << dp [ i ] << " ▁ " ; } int main ( ) { int n = 10 ; printTrib ( n ) ; return 0 ; } |
Tribonacci Numbers | Program to print first n tribonacci numbers Matrix Multiplication function for 3 * 3 matrix ; Recursive function to raise the matrix T to the power n ; base condition . ; recursively call to square the matrix ; calculating square of the matrix T ; if n is odd multiply it one time with M ; base condition ; T [ 0 ] [ 0 ] contains the tribonacci number so return it ; Driver Code | #include <iostream> NEW_LINE using namespace std ; void multiply ( int T [ 3 ] [ 3 ] , int M [ 3 ] [ 3 ] ) { int a , b , c , d , e , f , g , h , i ; a = T [ 0 ] [ 0 ] * M [ 0 ] [ 0 ] + T [ 0 ] [ 1 ] * M [ 1 ] [ 0 ] + T [ 0 ] [ 2 ] * M [ 2 ] [ 0 ] ; b = T [ 0 ] [ 0 ] * M [ 0 ] [ 1 ] + T [ 0 ] [ 1 ] * M [ 1 ] [ 1 ] + T [ 0 ] [ 2 ] * M [ 2 ] [ 1 ] ; c = T [ 0 ] [ 0 ] * M [ 0 ] [ 2 ] + T [ 0 ] [ 1 ] * M [ 1 ] [ 2 ] + T [ 0 ] [ 2 ] * M [ 2 ] [ 2 ] ; d = T [ 1 ] [ 0 ] * M [ 0 ] [ 0 ] + T [ 1 ] [ 1 ] * M [ 1 ] [ 0 ] + T [ 1 ] [ 2 ] * M [ 2 ] [ 0 ] ; e = T [ 1 ] [ 0 ] * M [ 0 ] [ 1 ] + T [ 1 ] [ 1 ] * M [ 1 ] [ 1 ] + T [ 1 ] [ 2 ] * M [ 2 ] [ 1 ] ; f = T [ 1 ] [ 0 ] * M [ 0 ] [ 2 ] + T [ 1 ] [ 1 ] * M [ 1 ] [ 2 ] + T [ 1 ] [ 2 ] * M [ 2 ] [ 2 ] ; g = T [ 2 ] [ 0 ] * M [ 0 ] [ 0 ] + T [ 2 ] [ 1 ] * M [ 1 ] [ 0 ] + T [ 2 ] [ 2 ] * M [ 2 ] [ 0 ] ; h = T [ 2 ] [ 0 ] * M [ 0 ] [ 1 ] + T [ 2 ] [ 1 ] * M [ 1 ] [ 1 ] + T [ 2 ] [ 2 ] * M [ 2 ] [ 1 ] ; i = T [ 2 ] [ 0 ] * M [ 0 ] [ 2 ] + T [ 2 ] [ 1 ] * M [ 1 ] [ 2 ] + T [ 2 ] [ 2 ] * M [ 2 ] [ 2 ] ; T [ 0 ] [ 0 ] = a ; T [ 0 ] [ 1 ] = b ; T [ 0 ] [ 2 ] = c ; T [ 1 ] [ 0 ] = d ; T [ 1 ] [ 1 ] = e ; T [ 1 ] [ 2 ] = f ; T [ 2 ] [ 0 ] = g ; T [ 2 ] [ 1 ] = h ; T [ 2 ] [ 2 ] = i ; } void power ( int T [ 3 ] [ 3 ] , int n ) { if ( n == 0 n == 1 ) return ; int M [ 3 ] [ 3 ] = { { 1 , 1 , 1 } , { 1 , 0 , 0 } , { 0 , 1 , 0 } } ; power ( T , n / 2 ) ; multiply ( T , T ) ; if ( n % 2 ) multiply ( T , M ) ; } int tribonacci ( int n ) { int T [ 3 ] [ 3 ] = { { 1 , 1 , 1 } , { 1 , 0 , 0 } , { 0 , 1 , 0 } } ; if ( n == 0 n == 1 ) return 0 ; else power ( T , n - 2 ) ; return T [ 0 ] [ 0 ] ; } int main ( ) { int n = 10 ; for ( int i = 0 ; i < n ; i ++ ) cout << tribonacci ( i ) << " ▁ " ; cout << endl ; return 0 ; } |
Geometric mean ( Two Methods ) | Program to calculate the geometric mean of the given array elements . ; function to calculate geometric mean and return float value . ; declare sum variable and initialize it to 1. ; Compute the sum of all the elements in the array . ; compute geometric mean through formula antilog ( ( ( log ( 1 ) + log ( 2 ) + . . . + log ( n ) ) / n ) and return the value to main function . ; Driver function ; function call | #include <bits/stdc++.h> NEW_LINE using namespace std ; float geometricMean ( int arr [ ] , int n ) { float sum = 0 ; for ( int i = 0 ; i < n ; i ++ ) sum = sum + log ( arr [ i ] ) ; sum = sum / n ; return exp ( sum ) ; } int main ( ) { int arr [ ] = { 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 } ; int n = sizeof ( arr ) / sizeof ( arr [ 0 ] ) ; cout << geometricMean ( arr , n ) ; return 0 ; } |
Smallest number k such that the product of digits of k is equal to n | C ++ implementation to find smallest number k such that the product of digits of k is equal to n ; function to find smallest number k such that the product of digits of k is equal to n ; if ' n ' is a single digit number , then it is the required number ; stack the store the digits ; repeatedly divide ' n ' by the numbers from 9 to 2 until all the numbers are used or ' n ' > 1 ; save the digit ' i ' that divides ' n ' onto the stack ; if true , then no number ' k ' can be formed ; pop digits from the stack ' digits ' and add them to ' k ' ; required smallest number ; Driver program to test above | #include <bits/stdc++.h> NEW_LINE using namespace std ; long long int smallestNumber ( int n ) { if ( n >= 0 && n <= 9 ) return n ; stack < int > digits ; for ( int i = 9 ; i >= 2 && n > 1 ; i -- ) { while ( n % i == 0 ) { digits . push ( i ) ; n = n / i ; } } if ( n != 1 ) return -1 ; long long int k = 0 ; while ( ! digits . empty ( ) ) { k = k * 10 + digits . top ( ) ; digits . pop ( ) ; } return k ; } int main ( ) { int n = 100 ; cout << smallestNumber ( n ) ; return 0 ; } |
Check if a number is magic ( Recursive sum of digits is 1 ) | CPP program to check if a number is Magic number . ; Note that the loop continues if n is 0 and sum is non - zero . It stops when n becomes 0 and sum becomes single digit . ; Return true if sum becomes 1. ; Driver code | #include <iostream> NEW_LINE using namespace std ; bool isMagic ( int n ) { int sum = 0 ; while ( n > 0 sum > 9 ) { if ( n == 0 ) { n = sum ; sum = 0 ; } sum += n % 10 ; n /= 10 ; } return ( sum == 1 ) ; } int main ( ) { int n = 1234 ; if ( isMagic ( n ) ) cout << " Magic ▁ Number " ; else cout << " Not ▁ a ▁ magic ▁ Number " ; return 0 ; } |
Check if a number is magic ( Recursive sum of digits is 1 ) | C ++ program to check Whether the number is Magic or not . ; Accepting sample input ; Condition to check Magic number | #include <iostream> NEW_LINE using namespace std ; int main ( ) { int x = 1234 ; if ( x % 9 == 1 ) cout << ( " Magic ▁ Number " ) ; else cout << ( " Not ▁ a ▁ Magic ▁ Number " ) ; return 0 ; } |
Rearrangement of a number which is also divisible by it | CPP program for finding rearrangement of n that is divisible by n ; perform hashing for given n ; perform hashing ; check whether any arrangement exists ; Create a hash for given number n The hash is of size 10 and stores count of each digit in n . ; check for all possible multipliers ; check hash table for both . Please refer below link for help of equal ( ) https : www . geeksforgeeks . org / stdequal - in - cpp / ; driver program | #include <bits/stdc++.h> NEW_LINE using namespace std ; void storeDigitCounts ( int n , vector < int > & hash ) { while ( n ) { hash [ n % 10 ] ++ ; n /= 10 ; } } int rearrange ( int n ) { vector < int > hash_n ( 10 , 0 ) ; storeDigitCounts ( n , hash_n ) ; for ( int mult = 2 ; mult < 10 ; mult ++ ) { int curr = n * mult ; vector < int > hash_curr ( 10 , 0 ) ; storeDigitCounts ( curr , hash_curr ) ; if ( equal ( hash_n . begin ( ) , hash_n . end ( ) , hash_curr . begin ( ) ) ) return curr ; } return -1 ; } int main ( ) { int n = 10035 ; cout << rearrange ( n ) ; return 0 ; } |
Sylvester 's sequence | CPP program to print terms of Sylvester 's sequence ; To store the product . ; To store the current number . ; Loop till n . ; Driven Program | #include <bits/stdc++.h> NEW_LINE using namespace std ; #define N 1000000007 NEW_LINE void printSequence ( int n ) { int a = 1 ; int ans = 2 ; for ( int i = 1 ; i <= n ; i ++ ) { cout << ans << " ▁ " ; ans = ( ( a % N ) * ( ans % N ) ) % N ; a = ans ; ans = ( ans + 1 ) % N ; } } int main ( ) { int n = 6 ; printSequence ( n ) ; return 0 ; } |
Program to find sum of first n natural numbers | CPP program to find sum of first n natural numbers . ; Returns sum of first n natural numbers ; Driver code | #include <iostream> NEW_LINE using namespace std ; int findSum ( int n ) { int sum = 0 ; for ( int x = 1 ; x <= n ; x ++ ) sum = sum + x ; return sum ; } int main ( ) { int n = 5 ; cout << findSum ( n ) ; return 0 ; } |
Hailstone Numbers | C ++ program to generate hailstone numbers and calculate steps required to reduce them to 1 ; function to print hailstone numbers and to calculate the number of steps required ; N is initially 1. ; N is reduced to 1. ; If N is Even . ; N is Odd . ; Driver code ; Function to generate Hailstone Numbers ; Output : Number of Steps | #include <bits/stdc++.h> NEW_LINE using namespace std ; int HailstoneNumbers ( int N ) { static int c ; cout << N << " ▁ " ; if ( N == 1 && c == 0 ) { return c ; } else if ( N == 1 && c != 0 ) { c ++ ; return c ; } else if ( N % 2 == 0 ) { c ++ ; HailstoneNumbers ( N / 2 ) ; } else if ( N % 2 != 0 ) { c ++ ; HailstoneNumbers ( 3 * N + 1 ) ; } } int main ( ) { int N = 7 ; int x ; x = HailstoneNumbers ( N ) ; cout << endl ; cout << " Number ▁ of ▁ Steps : ▁ " << x ; return 0 ; } |
Find m | CPP program to find m - th summation ; Function to return mth summation ; base case ; driver program | #include <bits/stdc++.h> NEW_LINE using namespace std ; int SUM ( int n , int m ) { if ( m == 1 ) return ( n * ( n + 1 ) / 2 ) ; int sum = SUM ( n , m - 1 ) ; return ( sum * ( sum + 1 ) / 2 ) ; } int main ( ) { int n = 5 ; int m = 3 ; cout << " SUM ( " << n << " , ▁ " << m << " ) : ▁ " << SUM ( n , m ) ; return 0 ; } |
Count number of digits after decimal on dividing a number | CPP program to count digits after dot when a number is divided by another . ; int ans = 0 ; Initialize result ; calculating remainder ; if this remainder appeared before then the numbers are irrational and would not converge to a solution the digits after decimal will be infinite ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int count ( int x , int y ) { unordered_map < int , int > m ; while ( x % y != 0 ) { x = x % y ; ans ++ ; if ( m . find ( x ) != m . end ( ) ) return -1 ; m [ x ] = 1 ; x = x * 10 ; } return ans ; } int main ( ) { int res = count ( 1 , 2 ) ; ( res == -1 ) ? cout << " INF " : cout << res ; cout << endl ; res = count ( 5 , 3 ) ; ( res == -1 ) ? cout << " INF " : cout << res ; cout << endl ; res = count ( 3 , 5 ) ; ( res == -1 ) ? cout << " INF " : cout << res ; return 0 ; } |
Find smallest number n such that n XOR n + 1 equals to given k . | CPP to find n such that XOR of n and n + 1 is equals to given n ; function to return the required n ; if k is of form 2 ^ i - 1 ; driver program | #include <bits/stdc++.h> NEW_LINE using namespace std ; int xorCalc ( int k ) { if ( k == 1 ) return 2 ; if ( ( ( k + 1 ) & k ) == 0 ) return k / 2 ; return -1 ; } int main ( ) { int k = 31 ; cout << xorCalc ( k ) ; return 0 ; } |
Find n | C ++ program to find n - th number containing only 4 and 7. ; If n is odd , append 4 and move to parent ; If n is even , append 7 and move to parent ; Reverse res and return . ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; string findNthNo ( int n ) { string res = " " ; while ( n >= 1 ) { if ( n & 1 ) { res = res + "4" ; n = ( n - 1 ) / 2 ; } else { res = res + "7" ; n = ( n - 2 ) / 2 ; } } reverse ( res . begin ( ) , res . end ( ) ) ; return res ; } int main ( ) { int n = 13 ; cout << findNthNo ( n ) ; return 0 ; } |
Total number of divisors for a given number | CPP program for finding number of divisor ; program for finding no . of divisors ; sieve method for prime calculation ; Traversing through all prime numbers ; calculate number of divisor with formula total div = ( p1 + 1 ) * ( p2 + 1 ) * ... . . * ( pn + 1 ) where n = ( a1 ^ p1 ) * ( a2 ^ p2 ) . ... * ( an ^ pn ) ai being prime divisor for n and pi are their respective power in factorization ; driver program | #include <bits/stdc++.h> NEW_LINE using namespace std ; int divCount ( int n ) { bool hash [ n + 1 ] ; memset ( hash , true , sizeof ( hash ) ) ; for ( int p = 2 ; p * p < n ; p ++ ) if ( hash [ p ] == true ) for ( int i = p * 2 ; i < n ; i += p ) hash [ i ] = false ; int total = 1 ; for ( int p = 2 ; p <= n ; p ++ ) { if ( hash [ p ] ) { int count = 0 ; if ( n % p == 0 ) { while ( n % p == 0 ) { n = n / p ; count ++ ; } total = total * ( count + 1 ) ; } } } return total ; } int main ( ) { int n = 24 ; cout << divCount ( n ) ; return 0 ; } |
Maximum number of unique prime factors | C ++ program to find maximum number of prime factors for a number in range [ 1 , N ] ; Return smallest number having maximum prime factors . ; Sieve of eratosthenes method to count number of unique prime factors . ; Return maximum element in arr [ ] ; Driven Program | #include <bits/stdc++.h> NEW_LINE using namespace std ; int maxPrimefactorNum ( int N ) { int arr [ N + 1 ] ; memset ( arr , 0 , sizeof ( arr ) ) ; for ( int i = 2 ; i * i <= N ; i ++ ) { if ( ! arr [ i ] ) for ( int j = 2 * i ; j <= N ; j += i ) arr [ j ] ++ ; arr [ i ] = 1 ; } return * max_element ( arr , arr + N ) ; } int main ( ) { int N = 40 ; cout << maxPrimefactorNum ( N ) << endl ; return 0 ; } |
Decimal to binary conversion without using arithmetic operators | C ++ implementation of decimal to binary conversion without using arithmetic operators ; function for decimal to binary conversion without using arithmetic operators ; to store the binary equivalent of decimal ; to get the last binary digit of the number ' n ' and accumulate it at the beginning of ' bin ' ; right shift ' n ' by 1 ; required binary number ; Driver program to test above | #include <bits/stdc++.h> NEW_LINE using namespace std ; string decToBin ( int n ) { if ( n == 0 ) return "0" ; string bin = " " ; while ( n > 0 ) { bin = ( ( n & 1 ) == 0 ? '0' : '1' ) + bin ; n >>= 1 ; } return bin ; } int main ( ) { int n = 38 ; cout << decToBin ( n ) ; return 0 ; } |
Difference between two given times | C ++ program to find difference between two given times . ; remove ' : ' and convert it into an integer ; Main function which finds difference ; change string ( eg . 2 : 21 -- > 221 , 00 : 23 -- > 23 ) ; difference between hours ; difference between minutes ; convert answer again in string with ' : ' ; Driver program to test above functions | #include <bits/stdc++.h> NEW_LINE using namespace std ; int removeColon ( string s ) { if ( s . size ( ) == 4 ) s . replace ( 1 , 1 , " " ) ; if ( s . size ( ) == 5 ) s . replace ( 2 , 1 , " " ) ; return stoi ( s ) ; } string diff ( string s1 , string s2 ) { int time1 = removeColon ( s1 ) ; int time2 = removeColon ( s2 ) ; int hourDiff = time2 / 100 - time1 / 100 - 1 ; int minDiff = time2 % 100 + ( 60 - time1 % 100 ) ; if ( minDiff >= 60 ) { hourDiff ++ ; minDiff = minDiff - 60 ; } string res = to_string ( hourDiff ) + ' : ' + to_string ( minDiff ) ; return res ; } int main ( ) { string s1 = "14:00" ; string s2 = "16:45" ; cout << diff ( s1 , s2 ) << endl ; return 0 ; } |
Sum of array elements that is first continuously increasing then decreasing | Efficient C ++ method to find sum of the elements of array that is halfway increasing and then halfway decreassing ; Driver code | #include <iostream> NEW_LINE using namespace std ; int arraySum ( int arr [ ] , int n ) { int x = ( n + 1 ) / 2 ; return ( arr [ 0 ] - 1 ) * n + x * x ; } int main ( ) { int arr [ ] = { 10 , 11 , 12 , 13 , 12 , 11 , 10 } ; int n = sizeof ( arr ) / sizeof ( arr [ 0 ] ) ; cout << arraySum ( arr , n ) ; return 0 ; } |
Balance pans using given weights that are powers of a number | C ++ code to check whether scale can be balanced or not ; method returns true if balancing of scale is possible ; baseForm vector will store T 's representation on base a in reverse order ; convert T to representation on base a ; make first digit of representation as 0 ; loop over base representation of T ; if any digit is not 0 , 1 , ( a - 1 ) or a then balancing is not possible ; if digit is a or ( a - 1 ) then increase left index ' s ▁ count / ▁ ( case , ▁ when ▁ this ▁ weight ▁ is ▁ transferred ▁ to ▁ T ' s side ) ; if representation is processed then balancing is possible ; Driver code to test above methods | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool isBalancePossible ( int T , int a ) { vector < int > baseForm ; while ( T ) { baseForm . push_back ( T % a ) ; T /= a ; } baseForm . push_back ( 0 ) ; for ( int i = 0 ; i < baseForm . size ( ) ; i ++ ) { if ( baseForm [ i ] != 0 && baseForm [ i ] != 1 && baseForm [ i ] != ( a - 1 ) && baseForm [ i ] != a ) return false ; if ( baseForm [ i ] == a || baseForm [ i ] == ( a - 1 ) ) baseForm [ i + 1 ] += 1 ; } return true ; } int main ( ) { int T = 11 ; int a = 4 ; bool balancePossible = isBalancePossible ( T , a ) ; if ( balancePossible ) cout << " Balance ▁ is ▁ possible " << endl ; else cout << " Balance ▁ is ▁ not ▁ possible " << endl ; return 0 ; } |
Number of digits in the product of two numbers | C ++ implementation to count number of digits in the product of two numbers ; function to count number of digits in the product of two numbers ; if either of the number is 0 , then product will be 0 ; required count of digits ; Driver program to test above | #include <bits/stdc++.h> NEW_LINE using namespace std ; int countDigits ( int a , int b ) { if ( a == 0 b == 0 ) return 1 ; return floor ( log10 ( abs ( a ) ) + log10 ( abs ( b ) ) ) + 1 ; } int main ( ) { int a = 33 ; int b = -24 ; cout << countDigits ( a , b ) ; return 0 ; } |
Distributing M items in a circle of size N starting from K | C ++ program to find the position where last item is delivered . ; n == > Size of circle m == > Number of items k == > Initial position ; n - k + 1 is number of positions before we reach beginning of circle If m is less than this value , then we can simply return ( m - 1 ) th position ; Let us compute remaining items before we reach beginning . ; We compute m % n to skip all complete rounds . If we reach end , we return n else we return m % n ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int lastPosition ( int n , int m , int k ) { if ( m <= n - k + 1 ) return m + k - 1 ; m = m - ( n - k + 1 ) ; return ( m % n == 0 ) ? n : ( m % n ) ; } int main ( ) { int n = 5 ; int m = 8 ; int k = 2 ; cout << lastPosition ( n , m , k ) ; return 0 ; } |
An interesting solution to get all prime numbers smaller than n | C ++ program to Prints prime numbers smaller than n ; Compute factorials and apply Wilson 's theorem. ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; void primesInRange ( int n ) { int fact = 1 ; for ( int k = 2 ; k < n ; k ++ ) { fact = fact * ( k - 1 ) ; if ( ( fact + 1 ) % k == 0 ) cout << k << endl ; } } int main ( ) { int n = 15 ; primesInRange ( n ) ; } |
A product array puzzle | Set 2 ( O ( 1 ) Space ) | C ++ program for product array puzzle with O ( n ) time and O ( 1 ) space . ; epsilon value to maintain precision ; to hold sum of all values ; output product for each index antilog to find original product value ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; #define EPS 1e-9 NEW_LINE void productPuzzle ( int a [ ] , int n ) { long double sum = 0 ; for ( int i = 0 ; i < n ; i ++ ) sum += ( long double ) log10 ( a [ i ] ) ; for ( int i = 0 ; i < n ; i ++ ) cout << ( int ) ( EPS + pow ( ( long double ) 10.00 , sum - log10 ( a [ i ] ) ) ) << " ▁ " ; } int main ( ) { int a [ ] = { 10 , 3 , 5 , 6 , 2 } ; int n = sizeof ( a ) / sizeof ( a [ 0 ] ) ; cout << " The ▁ product ▁ array ▁ is : ▁ STRNEWLINE " ; productPuzzle ( a , n ) ; return 0 ; } |
Change all even bits in a number to 0 | C ++ program to change even bits to 0. ; Returns modified number with all even bits 0. ; To store sum of bits at even positions . ; To store bits to shift ; One by one put all even bits to end ; If current last bit is set , add it to ans ; Next shift position ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int changeEvenBits ( int n ) { int to_subtract = 0 ; int m = 0 ; for ( int x = n ; x ; x >>= 2 ) { if ( x & 1 ) to_subtract += ( 1 << m ) ; m += 2 ; } return n - to_subtract ; } int main ( ) { int n = 30 ; cout << changeEvenBits ( n ) << endl ; return 0 ; } |
Find the number closest to n and divisible by m | C ++ implementation to find the number closest to n and divisible by m ; function to find the number closest to n and divisible by m ; find the quotient ; 1 st possible closest number ; 2 nd possible closest number ; if true , then n1 is the required closest number ; else n2 is the required closest number ; Driver program to test above | #include <bits/stdc++.h> NEW_LINE using namespace std ; int closestNumber ( int n , int m ) { int q = n / m ; int n1 = m * q ; int n2 = ( n * m ) > 0 ? ( m * ( q + 1 ) ) : ( m * ( q - 1 ) ) ; if ( abs ( n - n1 ) < abs ( n - n2 ) ) return n1 ; return n2 ; } int main ( ) { int n = 13 , m = 4 ; cout << closestNumber ( n , m ) << endl ; n = -15 ; m = 6 ; cout << closestNumber ( n , m ) << endl ; n = 0 ; m = 8 ; cout << closestNumber ( n , m ) << endl ; n = 18 ; m = -7 ; cout << closestNumber ( n , m ) << endl ; return 0 ; } |
Check if a given number is Pronic | C / C ++ program to check if a number is pronic ; function to check Pronic Number ; Checking Pronic Number by multiplying consecutive numbers ; Driver Code ; Printing Pronic Numbers upto 200 | #include <iostream> NEW_LINE #include <math.h> NEW_LINE using namespace std ; bool checkPronic ( int x ) { for ( int i = 0 ; i <= ( int ) ( sqrt ( x ) ) ; i ++ ) if ( x == i * ( i + 1 ) ) return true ; return false ; } int main ( void ) { for ( int i = 0 ; i <= 200 ; i ++ ) if ( checkPronic ( i ) ) cout << i << " ▁ " ; return 0 ; } |
Find minimum sum of factors of number | CPP program to find minimum sum of product of number ; To find minimum sum of product of number ; Find factors of number and add to the sum ; Return sum of numbers having minimum product ; Driver program to test above function | #include <bits/stdc++.h> NEW_LINE using namespace std ; int findMinSum ( int num ) { int sum = 0 ; for ( int i = 2 ; i * i <= num ; i ++ ) { while ( num % i == 0 ) { sum += i ; num /= i ; } } sum += num ; return sum ; } int main ( ) { int num = 12 ; cout << findMinSum ( num ) ; return 0 ; } |
Minimum number with digits as 4 and 7 only and given sum | C ++ program to find smallest number with given sum of digits . ; Prints minimum number with given digit sum and only allowed digits as 4 and 7. ; Cases where all remaining digits are 4 or 7 ( Remaining sum of digits should be multiple of 4 or 7 ) ; If both 4 s and 7 s are there in digit sum , we subtract a 4. ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; void findMin ( int sum ) { int a = 0 , b = 0 ; while ( sum > 0 ) { if ( sum % 7 == 0 ) { b ++ ; sum -= 7 ; } else if ( sum % 4 == 0 ) { a ++ ; sum -= 4 ; } else { a ++ ; sum -= 4 ; } } if ( sum < 0 ) { printf ( " - 1n " ) ; return ; } for ( int i = 0 ; i < a ; i ++ ) printf ( "4" ) ; for ( int i = 0 ; i < b ; i ++ ) printf ( "7" ) ; printf ( " n " ) ; } int main ( ) { findMin ( 15 ) ; return 0 ; } |
Add minimum number to an array so that the sum becomes even | CPP program to add minimum number so that the sum of array becomes even ; Function to find out minimum number ; Count odd number of terms in array ; Driver code | #include <iostream> NEW_LINE using namespace std ; int minNum ( int arr [ ] , int n ) { int odd = 0 ; for ( int i = 0 ; i < n ; i ++ ) if ( arr [ i ] % 2 ) odd += 1 ; return ( odd % 2 ) ? 1 : 2 ; } int main ( ) { int arr [ ] = { 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 } ; int n = sizeof ( arr ) / sizeof ( arr [ 0 ] ) ; cout << minNum ( arr , n ) << " n " ; return 0 ; } |
Find maximum number that can be formed using digits of a given number | CPP program to print the maximum number from the set of digits of a given number ; Function to print the maximum number ; hashed array to store count of digits ; Converting given number to string ; Updating the count array ; result is to store the final number ; Traversing the count array to calculate the maximum number ; return the result ; Driver program to test above function | #include <bits/stdc++.h> NEW_LINE using namespace std ; int printMaxNum ( int num ) { int count [ 10 ] = { 0 } ; string str = to_string ( num ) ; for ( int i = 0 ; i < str . length ( ) ; i ++ ) count [ str [ i ] - '0' ] ++ ; int result = 0 , multiplier = 1 ; for ( int i = 0 ; i <= 9 ; i ++ ) { while ( count [ i ] > 0 ) { result = result + ( i * multiplier ) ; count [ i ] -- ; multiplier = multiplier * 10 ; } } return result ; } int main ( ) { int num = 38293367 ; cout << printMaxNum ( num ) ; return 0 ; } |
Form the largest number using at most one swap operation | C ++ implementation to form the largest number by applying atmost one swap operation ; function to form the largest number by applying atmost one swap operation ; for the rightmost digit , there will be no greater right digit ; index of the greatest right digit till the current index from the right direction ; traverse the array from second right element up to the left element ; if ' num [ i ] ' is less than the greatest digit encountered so far ; else ; there is no greater right digit for ' num [ i ] ' ; update ' right ' index ; traverse the ' rightMax [ ] ' array from left to right ; if for the current digit , greater right digit exists then swap it with its greater right digit and break ; performing the required swap operation ; required largest number ; Driver program to test above | #include <bits/stdc++.h> NEW_LINE using namespace std ; string largestNumber ( string num ) { int n = num . size ( ) ; int rightMax [ n ] , right ; rightMax [ n - 1 ] = -1 ; right = n - 1 ; for ( int i = n - 2 ; i >= 0 ; i -- ) { if ( num [ i ] < num [ right ] ) rightMax [ i ] = right ; else { rightMax [ i ] = -1 ; right = i ; } } for ( int i = 0 ; i < n ; i ++ ) { if ( rightMax [ i ] != -1 ) { swap ( num [ i ] , num [ rightMax [ i ] ] ) ; break ; } } return num ; } int main ( ) { string num = "8725634" ; cout << " Largest ▁ number : " << largestNumber ( num ) ; return 0 ; } |
Binomial Random Variables | C ++ program to compute Binomial Probability ; function to calculate nCr i . e . , number of ways to choose r out of n objects ; Since nCr is same as nC ( n - r ) To decrease number of iterations ; function to calculate binomial r . v . probability ; Driver code | #include <iostream> NEW_LINE #include <cmath> NEW_LINE using namespace std ; int nCr ( int n , int r ) { if ( r > n / 2 ) r = n - r ; int answer = 1 ; for ( int i = 1 ; i <= r ; i ++ ) { answer *= ( n - r + i ) ; answer /= i ; } return answer ; } float binomialProbability ( int n , int k , float p ) { return nCr ( n , k ) * pow ( p , k ) * pow ( 1 - p , n - k ) ; } int main ( ) { int n = 10 ; int k = 5 ; float p = 1.0 / 3 ; float probability = binomialProbability ( n , k , p ) ; cout << " Probability ▁ of ▁ " << k ; cout << " ▁ heads ▁ when ▁ a ▁ coin ▁ is ▁ tossed ▁ " << n ; cout << " ▁ times ▁ where ▁ probability ▁ of ▁ each ▁ head ▁ is ▁ " << p << endl ; cout << " ▁ is ▁ = ▁ " << probability << endl ; } |
Find pair with maximum GCD in an array | C ++ Code to find pair with maximum GCD in an array ; function to find GCD of pair with max GCD in the array ; Computing highest element ; Array to store the count of divisors i . e . Potential GCDs ; Iterating over every element ; Calculating all the divisors ; Divisor found ; Incrementing count for divisor ; Element / divisor is also a divisor Checking if both divisors are not same ; Checking the highest potential GCD ; If this divisor can divide at least 2 numbers , it is a GCD of at least 1 pair ; Driver code ; Array in which pair with max GCD is to be found ; Size of array | #include <bits/stdc++.h> NEW_LINE using namespace std ; int findMaxGCD ( int arr [ ] , int n ) { int high = 0 ; for ( int i = 0 ; i < n ; i ++ ) high = max ( high , arr [ i ] ) ; int divisors [ high + 1 ] = { 0 } ; for ( int i = 0 ; i < n ; i ++ ) { for ( int j = 1 ; j <= sqrt ( arr [ i ] ) ; j ++ ) { if ( arr [ i ] % j == 0 ) { divisors [ j ] ++ ; if ( j != arr [ i ] / j ) divisors [ arr [ i ] / j ] ++ ; } } } for ( int i = high ; i >= 1 ; i -- ) if ( divisors [ i ] > 1 ) return i ; } int main ( ) { int arr [ ] = { 1 , 2 , 4 , 8 , 8 , 12 } ; int n = sizeof ( arr ) / sizeof ( arr [ 0 ] ) ; cout << findMaxGCD ( arr , n ) ; return 0 ; } |
Find pair with maximum GCD in an array | C ++ Code to Find pair with maximum GCD in an array ; function to find GCD of pair with max GCD in the array ; Calculating MAX in array ; Maintaining count array ; Variable to store the multiples of a number ; Iterating from MAX to 1 GCD is always between MAX and 1. The first GCD found will be the highest as we are decrementing the potential GCD ; Iterating from current potential GCD till it is less than MAX ; A multiple found ; Incrementing potential GCD by itself To check i , 2 i , 3 i ... . ; 2 multiples found , max GCD found ; Driver code ; Array in which pair with max GCD is to be found ; Size of array | #include <bits/stdc++.h> NEW_LINE using namespace std ; int findMaxGCD ( int arr [ ] , int n ) { int high = 0 ; for ( int i = 0 ; i < n ; i ++ ) high = max ( high , arr [ i ] ) ; int count [ high + 1 ] = { 0 } ; for ( int i = 0 ; i < n ; i ++ ) count [ arr [ i ] ] ++ ; int counter = 0 ; for ( int i = high ; i >= 1 ; i -- ) { int j = i ; counter = 0 ; while ( j <= high ) { if ( count [ j ] >= 2 ) return j ; else if ( count [ j ] == 1 ) counter ++ ; j += i ; if ( counter == 2 ) return i ; } } } int main ( ) { int arr [ ] = { 1 , 2 , 4 , 8 , 8 , 12 } ; int n = sizeof ( arr ) / sizeof ( arr [ 0 ] ) ; cout << findMaxGCD ( arr , n ) ; return 0 ; } |
Evil Number | C / C ++ program to check if a number is Evil number or Odious Number ; returns number of 1 s from the binary number ; counting 1 s ; Check if number is evil or not ; converting n to binary form ; calculating remainder ; storing the remainders in binary form as a number ; Calling the count_one function to count and return number of 1 s in bin ; Driver Code | #include <iostream> NEW_LINE using namespace std ; #include <math.h> NEW_LINE int count_one ( int n ) { int c_one = 0 ; while ( n != 0 ) { int rem = n % 10 ; if ( rem == 1 ) c_one = c_one + 1 ; n = n / 10 ; } return c_one ; } int checkEvil ( int n ) { int i = 0 , bin = 0 , n_one = 0 ; while ( n != 0 ) { int r = n % 2 ; bin = bin + r * ( int ) ( pow ( 10 , i ) ) ; n = n / 2 ; } n_one = count_one ( bin ) ; if ( n_one % 2 == 0 ) return 1 ; else return 0 ; } int main ( void ) { int i , check , num ; num = 32 ; check = checkEvil ( num ) ; if ( check == 1 ) cout << num << " ▁ is ▁ Evil ▁ Number STRNEWLINE " ; else cout << num << " ▁ is ▁ Odious ▁ Number STRNEWLINE " ; return 0 ; } |
Count number of pairs ( A <= N , B <= N ) such that gcd ( A , B ) is B | C ++ implementation of counting pairs such that gcd ( a , b ) = b ; returns number of valid pairs ; initialize k ; loop till imin <= n ; Initialize result ; max i with given k floor ( n / k ) ; adding k * ( number of i with floor ( n / i ) = k to ans ; set imin = imax + 1 and k = n / imin ; Driver function | #include <bits/stdc++.h> NEW_LINE using namespace std ; int CountPairs ( int n ) { int k = n ; int imin = 1 ; int ans = 0 ; while ( imin <= n ) { int imax = n / k ; ans += k * ( imax - imin + 1 ) ; imin = imax + 1 ; k = n / imin ; } return ans ; } int main ( ) { cout << CountPairs ( 1 ) << endl ; cout << CountPairs ( 2 ) << endl ; cout << CountPairs ( 3 ) << endl ; return 0 ; } |
Find the last digit of given series | C ++ program to calculate to find last digit of above expression ; Iterative Function to calculate ( x ^ y ) % p in O ( log y ) ; 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 >> 1L L ; y = y / 2 ; Returns modulo inverse of a with respect to m using extended Euclid Algorithm ; q is quotient ; m is remainder now , process same as Euclid 's algo ; Make x1 positive ; Function to calculate the above expression ; Initialize the result ; Compute first part of expression ; Compute second part of expression i . e . , ( ( 4 ^ ( n + 1 ) - 1 ) / 3 ) mod 10 Since division of 3 in modulo can ' t ▁ ▁ be ▁ performed ▁ directly ▁ therefore ▁ we ▁ ▁ need ▁ to ▁ find ▁ it ' s modulo Inverse ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; long long powermod ( long long x , long long y , long long p ) { while ( y > 0 ) { if ( y & 1LL ) res = ( res * x ) % p ; x = ( x * x ) % p ; } return res ; } long long modInverse ( long long a , long long m ) { long long m0 = m , t , q ; long long 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 ; } long long evaluteExpression ( long long & n ) { long long firstsum = 0 , mod = 10 ; for ( long long i = 2 , j = 0 ; ( 1LL << j ) <= n ; i *= i , ++ j ) firstsum = ( firstsum + i ) % mod ; long long secondsum = ( powermod ( 4LL , n + 1 , mod ) - 1 ) * modInverse ( 3LL , mod ) ; return ( firstsum * secondsum ) % mod ; } int main ( ) { long long n = 3 ; cout << evaluteExpression ( n ) << endl ; n = 10 ; cout << evaluteExpression ( n ) ; return 0 ; } |
Sum of all proper divisors of natural numbers in an array | C ++ program to find sum of proper divisors for every element in an array . ; To store prime factors and their powers ; Fills factors such that factors [ i ] is a vector of pairs containing prime factors ( of i ) and their powers . Also sets values in isPrime [ ] ; To check if a number is prime ; If i is prime , then update its powers in all multiples of it . ; Returns sum of proper divisors of num using factors [ ] ; Applying above discussed formula for every array element ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; #define MAX 1000001 NEW_LINE #define pii pair<int, int> NEW_LINE #define F first NEW_LINE #define S second NEW_LINE vector < pii > factors [ MAX ] ; void sieveOfEratothenese ( ) { bool isPrime [ MAX ] ; memset ( isPrime , true , sizeof ( isPrime ) ) ; isPrime [ 0 ] = isPrime [ 1 ] = false ; for ( int i = 2 ; i < MAX ; i ++ ) { if ( isPrime [ i ] ) { for ( int j = i ; j < MAX ; j += i ) { int k , l ; isPrime [ j ] = false ; for ( k = j , l = 0 ; k % i == 0 ; l ++ , k /= i ) ; factors [ j ] . push_back ( make_pair ( i , l ) ) ; } } } } int sumOfProperDivisors ( int num ) { int mul = 1 ; for ( int i = 0 ; i < factors [ num ] . size ( ) ; i ++ ) mul *= ( ( pow ( factors [ num ] [ i ] . F , factors [ num ] [ i ] . S + 1 ) - 1 ) / ( factors [ num ] [ i ] . F - 1 ) ) ; return mul - num ; } int main ( ) { sieveOfEratothenese ( ) ; int arr [ ] = { 8 , 13 , 24 , 36 , 59 , 75 , 91 } ; for ( int i = 0 ; i < sizeof ( arr ) / sizeof ( int ) ; i ++ ) cout << sumOfProperDivisors ( arr [ i ] ) << " ▁ " ; cout << endl ; return 0 ; } |
Finding power of prime number p in n ! | C ++ implementation of finding power of p in n ! ; Returns power of p in n ! ; initializing answer ; initializing ; loop until temp <= n ; add number of numbers divisible by n ; each time multiply temp by p ; Driver function | #include <bits/stdc++.h> NEW_LINE using namespace std ; int PowerOFPINnfactorial ( int n , int p ) { int ans = 0 ; int temp = p ; while ( temp <= n ) { ans += n / temp ; temp = temp * p ; } return ans ; } int main ( ) { cout << PowerOFPINnfactorial ( 4 , 2 ) << endl ; return 0 ; } |
Program for Decimal to Binary Conversion | C ++ implementation of the approach ; Function to return the binary equivalent of decimal value N ; To store the binary number ; Count used to store exponent value ; Driver code | #include <cmath> NEW_LINE #include <iostream> NEW_LINE using namespace std ; #define ull unsigned long long int NEW_LINE int decimalToBinary ( int N ) { ull B_Number = 0 ; int cnt = 0 ; while ( N != 0 ) { int rem = N % 2 ; ull c = pow ( 10 , cnt ) ; B_Number += rem * c ; N /= 2 ; cnt ++ ; } return B_Number ; } int main ( ) { int N = 17 ; cout << decimalToBinary ( N ) ; return 0 ; } |
Program for Binary To Decimal Conversion | C ++ program to convert binary to decimal ; Function to convert binary to decimal ; Initializing base value to 1 , i . e 2 ^ 0 ; Driver program to test above function | #include <iostream> NEW_LINE using namespace std ; int binaryToDecimal ( int n ) { int num = n ; int dec_value = 0 ; int base = 1 ; int temp = num ; while ( temp ) { int last_digit = temp % 10 ; temp = temp / 10 ; dec_value += last_digit * base ; base = base * 2 ; } return dec_value ; } int main ( ) { int num = 10101001 ; cout << binaryToDecimal ( num ) << endl ; } |
Calculating Factorials using Stirling Approximation | CPP program for calculating factorial of a number using Stirling Approximation ; function for calculating factorial ; evaluating factorial using stirling approximation ; driver program | #include <bits/stdc++.h> NEW_LINE using namespace std ; long int stirlingFactorial ( int n ) { if ( n == 1 ) return 1 ; long int z ; z = sqrt ( 2 * 3.14 * n ) * pow ( ( n / e ) , n ) ; return z ; } int main ( ) { cout << stirlingFactorial ( 1 ) << endl ; cout << stirlingFactorial ( 2 ) << endl ; cout << stirlingFactorial ( 3 ) << endl ; cout << stirlingFactorial ( 4 ) << endl ; cout << stirlingFactorial ( 5 ) << endl ; cout << stirlingFactorial ( 6 ) << endl ; cout << stirlingFactorial ( 7 ) << endl ; return 0 ; } |
Count pairs with Odd XOR | C ++ program to count pairs in array whose XOR is odd ; A function will return number of pair whose XOR is odd ; To store count of odd and even numbers ; Increase even if number is even otherwise increase odd ; Return number of pairs ; Driver program to test countXorPair ( ) | #include <iostream> NEW_LINE using namespace std ; int countXorPair ( int arr [ ] , int n ) { int odd = 0 , even = 0 ; for ( int i = 0 ; i < n ; i ++ ) { if ( arr [ i ] % 2 == 0 ) even ++ ; else odd ++ ; } return odd * even ; } int main ( ) { int arr [ ] = { 1 , 2 , 3 } ; int n = sizeof ( arr ) / sizeof ( arr [ 0 ] ) ; cout << countXorPair ( arr , n ) ; return 0 ; } |
Lychrel Number Implementation | C ++ Program to check whether the given number is Lychrel Number or not with given limit on number of iterations . ; Max Iterations ; Function to check whether number is Lychrel Number ; Function to check whether the number is Palindrome ; Function to reverse the number ; Driver program | #include <iostream> NEW_LINE using namespace std ; long reverse ( long ) ; bool isPalindrome ( long ) ; static int MAX_ITERATIONS = 20 ; string isLychrel ( long number ) { for ( int i = 0 ; i < MAX_ITERATIONS ; i ++ ) { number = number + reverse ( number ) ; if ( isPalindrome ( number ) ) return " false " ; } return " true " ; } bool isPalindrome ( long number ) { return number == reverse ( number ) ; } long reverse ( long number ) { long reverse = 0 ; while ( number > 0 ) { long remainder = number % 10 ; reverse = ( reverse * 10 ) + remainder ; number = number / 10 ; } return reverse ; } int main ( ) { long number = 295 ; cout << number << " ▁ is ▁ lychrel ? ▁ " << isLychrel ( number ) ; } |
Rectangular ( or Pronic ) Numbers | CPP Program to find n - th rectangular number ; Returns n - th rectangular number ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int findRectNum ( int n ) { return n * ( n + 1 ) ; } int main ( ) { int n = 6 ; cout << findRectNum ( n ) ; return 0 ; } |
Program for Muller Method | C ++ Program to find root of a function , f ( x ) ; Function to calculate f ( x ) ; Taking f ( x ) = x ^ 3 + 2 x ^ 2 + 10 x - 20 ; Calculating various constants required to calculate x3 ; Taking the root which is closer to x2 ; checking for resemblance of x3 with x2 till two decimal places ; Driver main function | #include <bits/stdc++.h> NEW_LINE using namespace std ; const int MAX_ITERATIONS = 10000 ; float f ( float x ) { return 1 * pow ( x , 3 ) + 2 * x * x + 10 * x - 20 ; } void Muller ( float a , float b , float c ) { int i ; float res ; for ( i = 0 ; ; ++ i ) { float f1 = f ( a ) ; float f2 = f ( b ) ; float f3 = f ( c ) ; float d1 = f1 - f3 ; float d2 = f2 - f3 ; float h1 = a - c ; float h2 = b - c ; float a0 = f3 ; float a1 = ( ( ( d2 * pow ( h1 , 2 ) ) - ( d1 * pow ( h2 , 2 ) ) ) / ( ( h1 * h2 ) * ( h1 - h2 ) ) ) ; float a2 = ( ( ( d1 * h2 ) - ( d2 * h1 ) ) / ( ( h1 * h2 ) * ( h1 - h2 ) ) ) ; float x = ( ( -2 * a0 ) / ( a1 + abs ( sqrt ( a1 * a1 - 4 * a0 * a2 ) ) ) ) ; float y = ( ( -2 * a0 ) / ( a1 - abs ( sqrt ( a1 * a1 - 4 * a0 * a2 ) ) ) ) ; if ( x >= y ) res = x + c ; else res = y + c ; float m = res * 100 ; float n = c * 100 ; m = floor ( m ) ; n = floor ( n ) ; if ( m == n ) break ; a = b ; b = c ; c = res ; if ( i > MAX_ITERATIONS ) { cout << " Root ▁ cannot ▁ be ▁ found ▁ using " << " ▁ Muller ' s ▁ method " ; break ; } } if ( i <= MAX_ITERATIONS ) cout << " The ▁ value ▁ of ▁ the ▁ root ▁ is ▁ " << res ; } int main ( ) { float a = 0 , b = 1 , c = 2 ; Muller ( a , b , c ) ; return 0 ; } |
Optimized Euler Totient Function for Multiple Evaluations | C ++ program to efficiently compute values of euler totient function for multiple inputs . ; Stores prime numbers upto MAX - 1 values ; Finds prime numbers upto MAX - 1 and stores them in vector p ; if prime [ i ] is not marked before ; fill vector for every newly encountered prime ; run this loop till square root of MAX , mark the index i * j as not prime ; function to find totient of n ; this loop runs sqrt ( n / ln ( n ) ) times ; subtract multiples of p [ i ] from r ; Remove all occurrences of p [ i ] in n ; when n has prime factor greater than sqrt ( n ) ; Driver code ; preprocess all prime numbers upto 10 ^ 5 | #include <bits/stdc++.h> NEW_LINE using namespace std ; #define ll long long NEW_LINE const int MAX = 100001 ; vector < ll > p ; void sieve ( ) { ll isPrime [ MAX + 1 ] ; for ( ll i = 2 ; i <= MAX ; i ++ ) { if ( isPrime [ i ] == 0 ) { p . push_back ( i ) ; for ( ll j = 2 ; i * j <= MAX ; j ++ ) isPrime [ i * j ] = 1 ; } } } ll phi ( ll n ) { ll res = n ; for ( ll i = 0 ; p [ i ] * p [ i ] <= n ; i ++ ) { if ( n % p [ i ] == 0 ) { res -= ( res / p [ i ] ) ; while ( n % p [ i ] == 0 ) n /= p [ i ] ; } } if ( n > 1 ) res -= ( res / n ) ; return res ; } int main ( ) { sieve ( ) ; cout << phi ( 11 ) << " STRNEWLINE " ; cout << phi ( 21 ) << " STRNEWLINE " ; cout << phi ( 31 ) << " STRNEWLINE " ; cout << phi ( 41 ) << " STRNEWLINE " ; cout << phi ( 51 ) << " STRNEWLINE " ; cout << phi ( 61 ) << " STRNEWLINE " ; cout << phi ( 91 ) << " STRNEWLINE " ; cout << phi ( 101 ) << " STRNEWLINE " ; return 0 ; } |
Finding n | C ++ implementation for finding nth number made of prime digits only ; Prints n - th number where each digit is a prime number ; Finding the length of n - th number ; Count of numbers with len - 1 digits ; Count of numbers with i digits ; if i is the length of such number then n < 4 * ( 4 ^ ( i - 1 ) - 1 ) / 3 and n >= 4 * ( 4 ^ i - 1 ) / 3 if a valid i is found break the loop ; check for i + 1 ; Finding ith digit at ith place ; j = 1 means 2 j = 2 means ... j = 4 means 7 ; if prev_count + 4 ^ ( len - i ) is less than n , increase prev_count by 4 ^ ( x - i ) ; else print the ith digit and break ; Driver function | #include <bits/stdc++.h> NEW_LINE using namespace std ; void nthprimedigitsnumber ( long long n ) { long long len = 1 ; long long prev_count = 0 ; while ( true ) { long long curr_count = prev_count + pow ( 4 , len ) ; if ( prev_count < n && curr_count >= n ) break ; len ++ ; prev_count = curr_count ; } for ( int i = 1 ; i <= len ; i ++ ) { for ( long long j = 1 ; j <= 4 ; j ++ ) { if ( prev_count + pow ( 4 , len - i ) < n ) prev_count += pow ( 4 , len - i ) ; else { if ( j == 1 ) cout << "2" ; else if ( j == 2 ) cout << "3" ; else if ( j == 3 ) cout << "5" ; else if ( j == 4 ) cout << "7" ; break ; } } } cout << endl ; } int main ( ) { nthprimedigitsnumber ( 10 ) ; nthprimedigitsnumber ( 21 ) ; return 0 ; } |
Cassiniâ €™ s Identity | C ++ implementation to demonstrate working of CassiniaTMs Identity ; Returns ( - 1 ) ^ n ; Driver program | #include <bits/stdc++.h> NEW_LINE using namespace std ; int cassini ( int n ) { return ( n & 1 ) ? -1 : 1 ; } int main ( ) { int n = 5 ; cout << cassini ( n ) ; return 0 ; } |
Find if a number is divisible by every number in a list | C ++ program which check is a number divided with every element in list or not ; Function which check is a number divided with every element in list or not ; driver program | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool findNoIsDivisibleOrNot ( int a [ ] , int n , int l ) { for ( int i = 0 ; i < l ; i ++ ) { if ( a [ i ] % n != 0 ) return false ; } return true ; } int main ( ) { int a [ ] = { 14 , 12 , 4 , 18 } ; int n = 2 ; int l = ( sizeof ( a ) / sizeof ( a [ 0 ] ) ) ; if ( findNoIsDivisibleOrNot ( a , n , l ) ) cout << " Yes " ; else cout << " No " ; return 0 ; } |
Find a range of composite numbers of given length | C ++ program to find a range of composite numbers of given length ; method to find factorial of given number ; to print range of length n having all composite integers ; Driver method | #include <bits/stdc++.h> NEW_LINE using namespace std ; int factorial ( int n ) { if ( n == 0 ) return 1 ; return n * factorial ( n - 1 ) ; } int printRange ( int n ) { int a = factorial ( n + 2 ) + 2 ; int b = a + n - 1 ; cout << " [ " << a << " , ▁ " << b << " ] " ; return 0 ; } int main ( ) { int n = 3 ; printRange ( n ) ; return 0 ; } |
Find minimum value to assign all array elements so that array product becomes greater | C ++ program to find minimum value that can be assigned to all elements so that product becomes greater than current product . ; sort the array to apply Binary search ; using log property add every logarithmic value of element to val ld val = 0 ; where ld is long double ; set left and right extremities to find min value ; multiplying n to mid , to find the correct min value ; Driver code | #include <bits/stdc++.h> NEW_LINE #define ll long long NEW_LINE #define ld long double NEW_LINE using namespace std ; ll findMinValue ( ll arr [ ] , ll n ) { sort ( arr , arr + n ) ; for ( int i = 0 ; i < n ; i ++ ) val += ( ld ) ( log ( ( ld ) ( arr [ i ] ) ) ) ; ll left = arr [ 0 ] , right = arr [ n - 1 ] + 1 ; ll ans ; while ( left <= right ) { ll mid = ( left + right ) / 2 ; ld temp = ( ld ) n * ( ld ) ( log ( ( ld ) ( mid ) ) ) ; if ( val < temp ) { ans = mid ; right = mid - 1 ; } else left = mid + 1 ; } return ans ; } int main ( ) { ll arr [ ] = { 4 , 2 , 1 , 10 , 6 } ; ll n = sizeof ( arr ) / sizeof ( arr [ 0 ] ) ; cout << findMinValue ( arr , n ) << endl ; return 0 ; } |
Find the sum of all the terms in the n | C ++ implementation to find the sum of all the terms in the nth row of the given series ; function to find the required sum ; sum = n * ( 2 * n ^ 2 + 1 ) ; Driver program to test above | #include <bits/stdc++.h> NEW_LINE using namespace std ; int sumOfTermsInNthRow ( int n ) { int sum = n * ( 2 * pow ( n , 2 ) + 1 ) ; return sum ; } int main ( ) { int n = 4 ; cout << " Sum ▁ of ▁ all ▁ the ▁ terms ▁ in ▁ nth ▁ row ▁ = ▁ " << sumOfTermsInNthRow ( n ) ; return 0 ; } |
First digit in product of an array of numbers | C ++ implementation to find first digit of a single number ; Keep dividing by 10 until it is greater than equal to 10 ; driver function | #include <bits/stdc++.h> NEW_LINE using namespace std ; int firstDigit ( int x ) { while ( x >= 10 ) x = x / 10 ; return x ; } int main ( ) { cout << firstDigit ( 12345 ) << endl ; cout << firstDigit ( 5432 ) << endl ; } |
Find the occurrences of digit d in the range [ 0. . n ] | C ++ program to count appearances of a digit ' d ' in range from [ 0. . n ] ; Count appearances in numbers starting from d . ; When the last digit is equal to d ; When the first digit is equal to d then ; increment result as well as number ; In case of reverse of number such as 12 and 21 ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int getOccurence ( int n , int d ) { int itr = d ; while ( itr <= n ) { if ( itr % 10 == d ) result ++ ; if ( itr != 0 && itr / 10 == d ) { result ++ ; itr ++ ; } else if ( itr / 10 == d - 1 ) itr = itr + ( 10 - d ) ; else itr = itr + 10 ; } return result ; } int main ( void ) { int n = 11 , d = 1 ; cout << getOccurence ( n , d ) ; return 0 ; } |
Numbers having Unique ( or Distinct ) digits | C ++ code for the above approach ; Function to print unique numbers ; Iterate from l to r ; Convert the no . to string ; Convert string to set using stl ; Output if condition satisfies ; Driver Code ; Input of the lower and higher limits ; Function Call | #include <bits/stdc++.h> NEW_LINE using namespace std ; void printUnique ( int l , int r ) { for ( int i = l ; i <= r ; i ++ ) { string s = to_string ( i ) ; set < int > uniDigits ( s . begin ( ) , s . end ( ) ) ; if ( s . size ( ) == uniDigits . size ( ) ) { cout << i << " ▁ " ; } } } int main ( ) { int l = 1 , r = 20 ; printUnique ( l , r ) ; return 0 ; } |
Program to calculate the value of sin ( x ) and cos ( x ) using Expansion | CPP code for implementing cos function ; Function for calculation ; Converting degrees to radian ; maps the sum along the series ; holds the actual value of sin ( n ) ; Main function | #include <iostream> NEW_LINE #include <math.h> NEW_LINE using namespace std ; void cal_cos ( float n ) { float accuracy = 0.0001 , x1 , denominator , cosx , cosval ; n = n * ( 3.142 / 180.0 ) ; x1 = 1 ; cosx = x1 ; cosval = cos ( n ) ; int i = 1 ; do { denominator = 2 * i * ( 2 * i - 1 ) ; x1 = - x1 * n * n / denominator ; cosx = cosx + x1 ; i = i + 1 ; } while ( accuracy <= fabs ( cosval - cosx ) ) ; cout << cosx ; } int main ( ) { float n = 30 ; cal_cos ( n ) ; } |
Find sum of digits in factorial of a number | C ++ program to find sum of digits in factorial of a number ; Function to multiply x with large number stored in vector v . Result is stored in v . ; Calculate res + prev carry ; updation at ith position ; Returns sum of digits in n ! ; One by one multiply i to current vector and update the vector . ; Find sum of digits in vector v [ ] ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; void multiply ( vector < int > & v , int x ) { int carry = 0 , res ; int size = v . size ( ) ; for ( int i = 0 ; i < size ; i ++ ) { int res = carry + v [ i ] * x ; v [ i ] = res % 10 ; carry = res / 10 ; } while ( carry != 0 ) { v . push_back ( carry % 10 ) ; carry /= 10 ; } } int findSumOfDigits ( int n ) { for ( int i = 1 ; i <= n ; i ++ ) multiply ( v , i ) ; int sum = 0 ; int size = v . size ( ) ; for ( int i = 0 ; i < size ; i ++ ) sum += v [ i ] ; return sum ; } int main ( ) { int n = 1000 ; cout << findSumOfDigits ( n ) ; return 0 ; } |
Find other two sides of a right angle triangle | C ++ program to print other two sides of right angle triangle given one side ; Finds two sides of a right angle triangle if it exist . ; if n is odd ; case of n = 1 handled separately ; case of n = 2 handled separately ; Driver program to test above function | #include <bits/stdc++.h> NEW_LINE using namespace std ; void printOtherSides ( int n ) { if ( n & 1 ) { if ( n == 1 ) cout << -1 << endl ; else { int b = ( n * n - 1 ) / 2 ; int c = ( n * n + 1 ) / 2 ; cout << " b ▁ = ▁ " << b << " , ▁ c ▁ = ▁ " << c << endl ; } } else { if ( n == 2 ) cout << -1 << endl ; else { int b = n * n / 4 - 1 ; int c = n * n / 4 + 1 ; cout << " b ▁ = ▁ " << b << " , ▁ c ▁ = ▁ " << c << endl ; } } } int main ( ) { int a = 3 ; printOtherSides ( a ) ; return 0 ; } |
Minimum positive integer to divide a number such that the result is an odd | C ++ program to make a number odd ; Return 1 if already odd ; Check on dividing with a number when the result becomes odd Return that number ; If n is divided by i and n / i is odd then return i ; Driver code | #include <iostream> NEW_LINE using namespace std ; int makeOdd ( int n ) { if ( n % 2 != 0 ) return 1 ; for ( int i = 2 ; i <= n ; i ++ ) if ( ( n % i == 0 ) && ( ( n / i ) % 2 == 1 ) ) return i ; } int main ( ) { int n = 36 ; cout << makeOdd ( n ) ; return 0 ; } |
XOR of all subarray XORs | Set 2 | C ++ program to get total xor of all subarray xors ; Returns XOR of all subarray xors ; if even number of terms are there , all numbers will appear even number of times . So result is 0. ; else initialize result by 0 as ( a xor 0 = a ) ; Driver code to test above methods | #include <bits/stdc++.h> NEW_LINE using namespace std ; int getTotalXorOfSubarrayXors ( int arr [ ] , int N ) { if ( N % 2 == 0 ) return 0 ; int res = 0 ; for ( int i = 0 ; i < N ; i += 2 ) res ^= arr [ i ] ; return res ; } int main ( ) { int arr [ ] = { 3 , 5 , 2 , 4 , 6 } ; int N = sizeof ( arr ) / sizeof ( arr [ 0 ] ) ; cout << getTotalXorOfSubarrayXors ( arr , N ) ; return 0 ; } |
Fill array with 1 's using minimum iterations of filling neighbors | C ++ program to find number of iterations to fill with all 1 s ; Returns count of iterations to fill arr [ ] with 1 s . ; Start traversing the array ; Traverse until a 0 is found ; Count contiguous 0 s ; Condition for Case 3 ; Condition to check if Case 1 satisfies : ; If count_zero is even ; If count_zero is odd ; Reset count_zero ; Case 2 ; Update res ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int countIterations ( int arr [ ] , int n ) { bool oneFound = false ; int res = 0 ; for ( int i = 0 ; i < n ; ) { if ( arr [ i ] == 1 ) oneFound = true ; while ( i < n && arr [ i ] == 1 ) i ++ ; int count_zero = 0 ; while ( i < n && arr [ i ] == 0 ) { count_zero ++ ; i ++ ; } if ( oneFound == false && i == n ) return -1 ; int curr_count ; if ( i < n && oneFound == true ) { if ( count_zero & 1 == 0 ) curr_count = count_zero / 2 ; else curr_count = ( count_zero + 1 ) / 2 ; count_zero = 0 ; } else { curr_count = count_zero ; count_zero = 0 ; } res = max ( res , curr_count ) ; } return res ; } int main ( ) { int arr [ ] = { 0 , 1 , 0 , 0 , 1 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 1 , 0 } ; int n = sizeof ( arr ) / sizeof ( arr [ 0 ] ) ; cout << countIterations ( arr , n ) ; return 0 ; } |
Interesting facts about Fibonacci numbers | C ++ program to demonstrate that Fibonacci numbers that are divisible by their indexes have indexes as either power of 5 or multiple of 12. ; storing Fibonacci numbers | #include <iostream> NEW_LINE using namespace std ; #define MAX 100 NEW_LINE int main ( ) { long long int arr [ MAX ] ; arr [ 0 ] = 0 ; arr [ 1 ] = 1 ; for ( int i = 2 ; i < MAX ; i ++ ) arr [ i ] = arr [ i - 1 ] + arr [ i - 2 ] ; cout << " Fibonacci ▁ numbers ▁ divisible ▁ by ▁ " " their ▁ indexes ▁ are ▁ : STRNEWLINE " ; for ( int i = 1 ; i < MAX ; i ++ ) if ( arr [ i ] % i == 0 ) cout << " ▁ " << i ; } |
Express a number as sum of consecutive numbers | C ++ program to print a consecutive sequence to express N if possible . ; Print consecutive numbers from last to first ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; void printConsecutive ( int last , int first ) { cout << first ++ ; for ( int x = first ; x <= last ; x ++ ) cout << " ▁ + ▁ " << x ; } void findConsecutive ( int N ) { for ( int last = 1 ; last < N ; last ++ ) { for ( int first = 0 ; first < last ; first ++ ) { if ( 2 * N == ( last - first ) * ( last + first + 1 ) ) { cout << N << " ▁ = ▁ " ; printConsecutive ( last , first + 1 ) ; return ; } } } cout << " - 1" ; } int main ( ) { int n = 12 ; findConsecutive ( n ) ; return 0 ; } |
Find n | C ++ program to find n - th number in a series made of digits 4 and 7 ; Return n - th number in series made of 4 and 7 ; create an array of size ( n + 1 ) ; If i is odd ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int printNthElement ( int n ) { int arr [ n + 1 ] ; arr [ 1 ] = 4 ; arr [ 2 ] = 7 ; for ( int i = 3 ; i <= n ; i ++ ) { if ( i % 2 != 0 ) arr [ i ] = arr [ i / 2 ] * 10 + 4 ; else arr [ i ] = arr [ ( i / 2 ) - 1 ] * 10 + 7 ; } return arr [ n ] ; } int main ( ) { int n = 6 ; cout << printNthElement ( n ) ; return 0 ; } |
Maximum sum of distinct numbers such that LCM of these numbers is N | C ++ program to find the max sum of numbers whose lcm is n ; Returns maximum sum of numbers with LCM as N ; Finding a divisor of n and adding it to max_sum ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int maxSumLCM ( int n ) { for ( int i = 1 ; i * i <= n ; i ++ ) { if ( n % i == 0 ) { max_sum += i ; if ( n / i != i ) max_sum += ( n / i ) ; } } return max_sum ; } int main ( ) { int n = 2 ; cout << MaxSumLCM ( n ) << endl ; return 0 ; } |
Square root of a number using log | C ++ program to demonstrate finding square root of a number using sqrt ( ) | #include <bits/stdc++.h> NEW_LINE int main ( void ) { double n = 12 ; printf ( " % lf ▁ " , sqrt ( n ) ) ; return 0 ; } |
Check if LCM of array elements is divisible by a prime number or not | C ++ program to find LCM of array of number is divisible by a prime number k or not ; Function to check any number of array is divisible by k or not ; If any array element is divisible by k , then LCM of whole array should also be divisible . ; Driver Code | #include <iostream> NEW_LINE using namespace std ; bool func ( int a [ ] , int k , int n ) { for ( int i = 0 ; i < n ; i ++ ) if ( a [ i ] % k == 0 ) return true ; return false ; } int main ( ) { int a [ ] = { 14 , 27 , 38 , 76 , 84 } ; int k = 19 ; bool res = func ( a , k , 5 ) ; if ( res ) cout << " true " ; else cout << " false " ; return 0 ; } |
Find the closest and smaller tidy number | C ++ program to find closest tidy number smaller than the given number ; check whether string violates tidy property ; if string violates tidy property , then decrease the value stored at that index by 1 and replace all the value stored right to that index by 9 ; Driver code ; num will store closest tidy number | #include <bits/stdc++.h> NEW_LINE using namespace std ; char * tidyNum ( char str [ ] , int len ) { for ( int i = len - 2 ; i >= 0 ; i -- ) { if ( str [ i ] > str [ i + 1 ] ) { ( char ) str [ i ] -- ; for ( int j = i + 1 ; j < len ; j ++ ) str [ j ] = '9' ; } } return str ; } int main ( ) { char str [ ] = "11333445538" ; int len = strlen ( str ) ; char * num = tidyNum ( str , len ) ; printf ( " % s STRNEWLINE " , num ) ; return 0 ; } |
Convert a binary number to hexadecimal number | C ++ implementation to convert a binary number to hexadecimal number ; Function to create map between binary number and its equivalent hexadecimal ; function to find hexadecimal equivalent of binary ; length of string before ' . ' ; add min 0 's in the beginning to make left substring length divisible by 4 ; if decimal point exists ; length of string after ' . ' ; add min 0 's in the end to make right substring length divisible by 4 ; create map between binary and its equivalent hex code ; one by one extract from left , substring of size 4 and add its hex code ; if ' . ' is encountered add it to result ; required hexadecimal number ; Driver program to test above | #include <bits/stdc++.h> NEW_LINE using namespace std ; void createMap ( unordered_map < string , char > * um ) { ( * um ) [ "0000" ] = '0' ; ( * um ) [ "0001" ] = '1' ; ( * um ) [ "0010" ] = '2' ; ( * um ) [ "0011" ] = '3' ; ( * um ) [ "0100" ] = '4' ; ( * um ) [ "0101" ] = '5' ; ( * um ) [ "0110" ] = '6' ; ( * um ) [ "0111" ] = '7' ; ( * um ) [ "1000" ] = '8' ; ( * um ) [ "1001" ] = '9' ; ( * um ) [ "1010" ] = ' A ' ; ( * um ) [ "1011" ] = ' B ' ; ( * um ) [ "1100" ] = ' C ' ; ( * um ) [ "1101" ] = ' D ' ; ( * um ) [ "1110" ] = ' E ' ; ( * um ) [ "1111" ] = ' F ' ; } string convertBinToHex ( string bin ) { int l = bin . size ( ) ; int t = bin . find_first_of ( ' . ' ) ; int len_left = t != -1 ? t : l ; for ( int i = 1 ; i <= ( 4 - len_left % 4 ) % 4 ; i ++ ) bin = '0' + bin ; if ( t != -1 ) { int len_right = l - len_left - 1 ; for ( int i = 1 ; i <= ( 4 - len_right % 4 ) % 4 ; i ++ ) bin = bin + '0' ; } unordered_map < string , char > bin_hex_map ; createMap ( & bin_hex_map ) ; int i = 0 ; string hex = " " ; while ( 1 ) { hex += bin_hex_map [ bin . substr ( i , 4 ) ] ; i += 4 ; if ( i == bin . size ( ) ) break ; if ( bin . at ( i ) == ' . ' ) { hex += ' . ' ; i ++ ; } } return hex ; } int main ( ) { string bin = "1111001010010100001.010110110011011" ; cout << " Hexadecimal ▁ number ▁ = ▁ " << convertBinToHex ( bin ) ; return 0 ; } |
Count of m digit integers that are divisible by an integer n | C ++ program to count m digit numbers having n as divisor . ; Returns count of m digit numbers having n as divisor ; generating largest number of m digit ; generating largest number of m - 1 digit ; returning number of dividend ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int findCount ( int m , int n ) { int num1 = 0 ; for ( int i = 0 ; i < m ; i ++ ) num1 = ( num1 * 10 ) + 9 ; int num2 = 0 ; for ( int i = 0 ; i < ( m - 1 ) ; i ++ ) num2 = ( num2 * 10 ) + 9 ; return ( ( num1 / n ) - ( num2 / n ) ) ; } int main ( ) { int m = 2 , n = 6 ; printf ( " % d STRNEWLINE " , findCount ( m , n ) ) ; return 0 ; } |
Find the n | Simple C ++ program to find n - th number made of even digits only ; function to calculate nth number made of even digits only ; variable to note how many such numbers have been found till now ; bool variable to check if 1 , 3 , 5 , 7 , 9 is there or not ; checking each digit of the number ; If 1 , 3 , 5 , 7 , 9 is found temp is changed to false ; temp is true it means that it does not have 1 , 3 , 5 , 7 , 9 ; If nth such number is found return it ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int findNthEvenDigitNumber ( int n ) { int count = 0 ; for ( int i = 0 ; ; i ++ ) { int curr = i ; bool isCurrEvenDigit = true ; while ( curr != 0 ) { if ( curr % 10 == 1 curr % 10 == 3 curr % 10 == 5 curr % 10 == 7 curr % 10 == 9 ) isCurrEvenDigit = false ; curr = curr / 10 ; } if ( isCurrEvenDigit == true ) count ++ ; if ( count == n ) return i ; } } int main ( ) { cout << findNthEvenDigitNumber ( 2 ) << endl ; cout << findNthEvenDigitNumber ( 10 ) << endl ; return 0 ; } |
Find the n | Efficient C ++ program to find n - th number made of even digits only ; function to find nth number made of even digits only ; If n = 1 return 0 ; vector to store the digits when converted into base 5 ; Reduce n to n - 1 to exclude 0 ; Reduce n to base 5 number and store digits ; pushing the digits into vector ; variable to represent the number after converting it to base 5. Since the digits are be in reverse order , we traverse vector from back ; return 2 * result ( to convert digits 0 , 1 , 2 , 3 , 4 to 0 , 2 , 4 , 6 , 8. ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int findNthEvenDigitNumber ( int n ) { if ( n == 1 ) return 0 ; vector < int > v ; n = n - 1 ; while ( n > 0 ) { v . push_back ( n % 5 ) ; n = n / 5 ; } int result = 0 ; for ( int i = v . size ( ) - 1 ; i >= 0 ; i -- ) { result = result * 10 ; result = result + v [ i ] ; } return 2 * result ; } int main ( ) { cout << findNthEvenDigitNumber ( 2 ) << endl ; cout << findNthEvenDigitNumber ( 10 ) << endl ; return 0 ; } |
Check if a large number is divisible by 25 or not | C ++ program to find if a number is divisible by 25 or not ; Function to find that number divisible by 25 or not . ; If length of string is single digit then it 's not divisible by 25 ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool isDivisibleBy25 ( string str ) { int n = str . length ( ) ; if ( n == 1 ) return false ; return ( ( str [ n - 1 ] - '0' == 0 && str [ n - 2 ] - '0' == 0 ) || ( ( str [ n - 2 ] - '0' ) * 10 + ( str [ n - 1 ] - '0' ) ) % 25 == 0 ) ; } int main ( ) { string str = "76955" ; isDivisibleBy25 ( str ) ? cout << " Yes " : cout << " No ▁ " ; return 0 ; } |
Check a large number is divisible by 16 or not | C ++ program to find if a number is divisible by 16 or not ; Function to find that number divisible by 16 or not ; Empty string ; If there is double digit ; If there is triple digit ; If number formed by last four digits is divisible by 16. ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool check ( string str ) { int n = str . length ( ) ; if ( n == 0 && n == 1 ) return false ; if ( n == 2 ) return ( ( ( str [ n - 2 ] - '0' ) * 10 + ( str [ n - 1 ] - '0' ) ) % 16 == 0 ) ; if ( n == 3 ) return ( ( ( str [ n - 3 ] - '0' ) * 100 + ( str [ n - 2 ] - '0' ) * 10 + ( str [ n - 1 ] - '0' ) ) % 16 == 0 ) ; int last = str [ n - 1 ] - '0' ; int second_last = str [ n - 2 ] - '0' ; int third_last = str [ n - 3 ] - '0' ; int fourth_last = str [ n - 4 ] - '0' ; return ( ( fourth_last * 1000 + third_last * 100 + second_last * 10 + last ) % 16 == 0 ) ; } int main ( ) { string str = "769528" ; check ( str ) ? cout << " Yes " : cout << " No ▁ " ; return 0 ; } |
Find Index of given fibonacci number in constant time | A simple C ++ program to find index of given Fibonacci number . ; if Fibonacci number is less than 2 , its index will be same as number ; iterate until generated fibonacci number is less than given fibonacci number ; res keeps track of number of generated fibonacci number ; Driver program to test above function | #include <bits/stdc++.h> NEW_LINE int findIndex ( int n ) { if ( n <= 1 ) return n ; int a = 0 , b = 1 , c = 1 ; int res = 1 ; while ( c < n ) { c = a + b ; res ++ ; a = b ; b = c ; } return res ; } int main ( ) { int result = findIndex ( 21 ) ; printf ( " % d STRNEWLINE " , result ) ; } |
Date after adding given number of days to the given date | C ++ program to find date after adding given number of days . ; Return if year is leap year or not . ; Given a date , returns number of days elapsed from the beginning of the current year ( 1 stjan ) . ; Given a year and days elapsed in it , finds date by storing results in d and m . ; Add x days to the given date . ; y2 is going to store result year and offset2 is going to store offset days in result year . ; x may store thousands of days . We find correct year and offset in the year . ; Find values of day and month from offset of result year . ; Driven Program | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool isLeap ( int y ) { if ( y % 100 != 0 && y % 4 == 0 y % 400 == 0 ) return true ; return false ; } int offsetDays ( int d , int m , int y ) { int offset = d ; switch ( m - 1 ) { case 11 : offset += 30 ; case 10 : offset += 31 ; case 9 : offset += 30 ; case 8 : offset += 31 ; case 7 : offset += 31 ; case 6 : offset += 30 ; case 5 : offset += 31 ; case 4 : offset += 30 ; case 3 : offset += 31 ; case 2 : offset += 28 ; case 1 : offset += 31 ; } if ( isLeap ( y ) && m > 2 ) offset += 1 ; return offset ; } void revoffsetDays ( int offset , int y , int * d , int * m ) { int month [ 13 ] = { 0 , 31 , 28 , 31 , 30 , 31 , 30 , 31 , 31 , 30 , 31 , 30 , 31 } ; if ( isLeap ( y ) ) month [ 2 ] = 29 ; int i ; for ( i = 1 ; i <= 12 ; i ++ ) { if ( offset <= month [ i ] ) break ; offset = offset - month [ i ] ; } * d = offset ; * m = i ; } void addDays ( int d1 , int m1 , int y1 , int x ) { int offset1 = offsetDays ( d1 , m1 , y1 ) ; int remDays = isLeap ( y1 ) ? ( 366 - offset1 ) : ( 365 - offset1 ) ; int y2 , offset2 ; if ( x <= remDays ) { y2 = y1 ; offset2 = offset1 + x ; } else { x -= remDays ; y2 = y1 + 1 ; int y2days = isLeap ( y2 ) ? 366 : 365 ; while ( x >= y2days ) { x -= y2days ; y2 ++ ; y2days = isLeap ( y2 ) ? 366 : 365 ; } offset2 = x ; } int m2 , d2 ; revoffsetDays ( offset2 , y2 , & d2 , & m2 ) ; cout << " d2 ▁ = ▁ " << d2 << " , ▁ m2 ▁ = ▁ " << m2 << " , ▁ y2 ▁ = ▁ " << y2 ; } int main ( ) { int d = 14 , m = 3 , y = 2015 ; int x = 366 ; addDays ( d , m , y , x ) ; return 0 ; } |
Determine whether a given number is a Hyperperfect Number | C ++ 4.3 . 2 program to check whether a given number is k - hyperperfect ; function to find the sum of all proper divisors ( excluding 1 and N ) ; Iterate only until sqrt N as we are going to generate pairs to produce divisors ; As divisors occur in pairs , we can take the values i and N / i as long as i divides N ; Function to check whether the given number is prime ; base and corner cases ; Since integers can be represented as some 6 * k + y where y >= 0 , we can eliminate all integers that can be expressed in this form ; start from 5 as this is the next prime number ; Returns true if N is a K - Hyperperfect number Else returns false . ; Condition from the definition of hyperperfect ; Driver function to test for hyperperfect numbers ; First two statements test against the condition N = 1 + K * ( sum ( proper divisors ) ) | #include <bits/stdc++.h> NEW_LINE using namespace std ; int divisorSum ( int N , int K ) { int sum = 0 ; for ( int i = 2 ; i <= ceil ( sqrt ( N ) ) ; i ++ ) if ( N % i == 0 ) sum += ( i + N / i ) ; return sum ; } bool isPrime ( int n ) { if ( n == 1 n == 0 ) 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 ; } bool isHyperPerfect ( int N , int K ) { int sum = divisorSum ( N , K ) ; if ( ( 1 + K * ( sum ) ) == N ) return true ; else return false ; } int main ( ) { int N1 = 1570153 , K1 = 12 ; int N2 = 321 , K2 = 3 ; if ( isHyperPerfect ( N1 , K1 ) ) cout << N1 << " ▁ is ▁ " << K1 << " - HyperPerfect " << " STRNEWLINE " ; else cout << N1 << " ▁ is ▁ not ▁ " << K1 << " - HyperPerfect " << " STRNEWLINE " ; if ( isHyperPerfect ( N2 , K2 ) ) cout << N2 << " ▁ is ▁ " << K2 << " - HyperPerfect " << " STRNEWLINE " ; else cout << N2 << " ▁ is ▁ not ▁ " << K2 << " - HyperPerfect " << " STRNEWLINE " ; return 0 ; } |
Generate k digit numbers with digits in strictly increasing order | C ++ program to generate well ordered numbers with k digits . ; number -- > Current value of number . x -- > Current digit to be considered k -- > Remaining number of digits ; Try all possible greater digits ; Generates all well ordered numbers of length k . ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; void printWellOrdered ( int number , int x , int k ) { if ( k == 0 ) { cout << number << " ▁ " ; return ; } for ( int i = ( x + 1 ) ; i < 10 ; i ++ ) printWellOrdered ( number * 10 + i , i , k - 1 ) ; } void generateWellOrdered ( int k ) { printWellOrdered ( 0 , 0 , k ) ; } int main ( ) { int k = 3 ; generateWellOrdered ( k ) ; return 0 ; } |
Multiply large integers under large modulo | C ++ program of finding modulo multiplication ; Returns ( a * b ) % mod ; Update a if it is more than or equal to mod ; If b is odd , add a with result ; Here we assume that doing 2 * a doesn 't cause overflow ; b >>= 1 ; b = b / 2 ; Driver program | #include <bits/stdc++.h> NEW_LINE using namespace std ; long long moduloMultiplication ( long long a , long long b , long long mod ) { a %= mod ; while ( b ) { if ( b & 1 ) res = ( res + a ) % mod ; a = ( 2 * a ) % mod ; } return res ; } int main ( ) { long long a = 426 ; long long b = 964 ; long long m = 235 ; cout << moduloMultiplication ( a , b , m ) ; return 0 ; } |
Number of occurrences of 2 as a digit in numbers from 0 to n | Write C ++ code here ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int numberOf2sinRange ( int n ) { string s = " " ; for ( int i = 0 ; i < n + 1 ; i ++ ) s += to_string ( i ) ; int count = 0 ; for ( int i = 0 ; i < s . length ( ) ; i ++ ) { if ( s [ i ] == '2' ) { count ++ ; } } return count ; } int main ( ) { int n = 30 ; cout << numberOf2sinRange ( n ) ; return 0 ; } |
Number of occurrences of 2 as a digit in numbers from 0 to n | C ++ program to count 2 s from 0 to n ; Counts the number of 2 s in a number at d - th digit ; if the digit in spot digit is ; Counts the number of '2' digits between 0 and n ; Convert integer to String to find its length ; Traverse every digit and count for every digit ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int count2sinRangeAtDigit ( int number , int d ) { int powerOf10 = ( int ) pow ( 10 , d ) ; int nextPowerOf10 = powerOf10 * 10 ; int right = number % powerOf10 ; int roundDown = number - number % nextPowerOf10 ; int roundup = roundDown + nextPowerOf10 ; int digit = ( number / powerOf10 ) % 10 ; if ( digit < 2 ) return roundDown / 10 ; if ( digit == 2 ) return roundDown / 10 + right + 1 ; return roundup / 10 ; } int numberOf2sinRange ( int number ) { stringstream convert ; convert << number ; string s = convert . str ( ) ; int len = s . length ( ) ; int count = 0 ; for ( int digit = 0 ; digit < len ; digit ++ ) count += count2sinRangeAtDigit ( number , digit ) ; return count ; } int main ( ) { cout << numberOf2sinRange ( 22 ) << endl ; cout << numberOf2sinRange ( 100 ) ; return 0 ; } |
Modulo 10 ^ 9 + 7 ( 1000000007 ) | ; f = f * i ; WRONG APPROACH as f may exceed ( 2 ^ 64 - 1 ) | unsigned long long factorial ( int n ) { const unsigned int M = 1000000007 ; unsigned long long f = 1 ; for ( int i = 1 ; i <= n ; i ++ ) return f % M ; } |
Modulo 10 ^ 9 + 7 ( 1000000007 ) | ; f = ( f * i ) % M ; Now f never can exceed 10 ^ 9 + 7 | unsigned long long factorial ( int n ) { const unsigned int M = 1000000007 ; unsigned long long f = 1 ; for ( int i = 1 ; i <= n ; i ++ ) return f ; } |
Modulo 10 ^ 9 + 7 ( 1000000007 ) | | int mod ( int a , int m ) { return ( a % m + m ) % m ; } |
Program to find Star number | C ++ program to find star number ; Returns n - th star number ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int findStarNum ( int n ) { return ( 6 * n * ( n - 1 ) + 1 ) ; } int main ( ) { int n = 3 ; cout << findStarNum ( n ) ; return 0 ; } |
Check if a large number is divisible by 5 or not | C ++ program to find if a number is divisible by 5 or not ; Function to find that number divisible by 5 or not . The function assumes that string length is at least one . ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool isDivisibleBy5 ( string str ) { int n = str . length ( ) ; return ( ( ( str [ n - 1 ] - '0' ) == 0 ) || ( ( str [ n - 1 ] - '0' ) == 5 ) ) ; } int main ( ) { string str = "76955" ; isDivisibleBy5 ( str ) ? cout << " Yes " : cout << " No ▁ " ; return 0 ; } |
Tidy Number ( Digits in non | C ++ program to check if a number is Tidy or not . ; Returns true if num is Tidy ; To store previous digit ( Assigning initial value which is more than any digit ) ; Traverse all digits from right to left and check if any digit is smaller than previous . ; Driver code | #include <iostream> NEW_LINE using namespace std ; bool isTidy ( int num ) { int prev = 10 ; while ( num ) { int rem = num % 10 ; num /= 10 ; if ( rem > prev ) return false ; prev = rem ; } return true ; } int main ( ) { int num = 1556 ; isTidy ( num ) ? cout << " Yes " : cout << " No " ; return 0 ; } |
Nth Square free number | Program to find the nth square free number ; Function to find nth square free number ; To maintain count of square free number ; Loop for square free numbers ; Checking whether square of a number is divisible by any number which is a perfect square ; If number is square free ; If the cnt becomes n , return the number ; Driver Program | #include <bits/stdc++.h> NEW_LINE using namespace std ; int squareFree ( int n ) { int cnt = 0 ; for ( int i = 1 ; ; i ++ ) { bool isSqFree = true ; for ( int j = 2 ; j * j <= i ; j ++ ) { if ( i % ( j * j ) == 0 ) { isSqFree = false ; break ; } } if ( isSqFree == true ) { cnt ++ ; if ( cnt == n ) return i ; } } return 0 ; } int main ( ) { int n = 10 ; cout << squareFree ( n ) << endl ; return 0 ; } |
Nth Square free number | Program to find the nth square free number ; Maximum prime number to be considered for square divisibility ; Maximum value of result . We do binary search from 1 to MAX_RES ; 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 ; Store all prime numbers in a [ ] ; Function to count integers upto k which are having perfect squares as factors . i is index of next prime number whose square needs to be checked . curr is current number whos square to be checked . ; variable to store square of prime ; If value of greatest integer becomes zero ; Counting integers with squares as factor ; Inclusion ( Recur for next prime number ) ; Exclusion ( Recur for next prime number ) ; Final count ; Function to return nth square free number ; Computing primes and storing it in an array a [ ] ; Applying binary search ; ' c ' contains Number of square free numbers less than or equal to ' mid ' ; If c < n , then search right side of mid else search left side of mid ; nth square free number ; Driver Program | #include <bits/stdc++.h> NEW_LINE using namespace std ; const int MAX_PRIME = 100000 ; const int MAX_RES = 2000000000l ; void SieveOfEratosthenes ( vector < long long > & a ) { bool prime [ MAX_PRIME + 1 ] ; memset ( prime , true , sizeof ( prime ) ) ; for ( long long p = 2 ; p * p <= MAX_PRIME ; p ++ ) { if ( prime [ p ] == true ) { for ( long long i = p * 2 ; i <= MAX_PRIME ; i += p ) prime [ i ] = false ; } } for ( long long p = 2 ; p <= MAX_PRIME ; p ++ ) if ( prime [ p ] ) a . push_back ( p ) ; } long long countSquares ( long long i , long long cur , long long k , vector < long long > & a ) { long long square = a [ i ] * a [ i ] ; long long newCur = square * cur ; if ( newCur > k ) return 0 ; long long cnt = k / ( newCur ) ; cnt += countSquares ( i + 1 , cur , k , a ) ; cnt -= countSquares ( i + 1 , newCur , k , a ) ; return cnt ; } long long squareFree ( long long n ) { vector < long long > a ; SieveOfEratosthenes ( a ) ; long long low = 1 ; long long high = MAX_RES ; while ( low < high ) { long long mid = low + ( high - low ) / 2 ; long long c = mid - countSquares ( 0 , 1 , mid , a ) ; if ( c < n ) low = mid + 1 ; else high = mid ; } return low ; } int main ( ) { int n = 10 ; cout << squareFree ( n ) << endl ; return 0 ; } |
Find if n can be written as product of k numbers | C ++ program to find if it is possible to write a number n as product of exactly k positive numbers greater than 1. ; Prints k factors of n if n can be written as multiple of k numbers . Else prints - 1. ; A vector to store all prime factors of n ; Insert all 2 's in vector ; n must be odd at this point So we skip one element ( i = i + 2 ) ; This is to handle when n > 2 and n is prime ; If size ( P ) < k , k factors are not possible ; printing first k - 1 factors ; calculating and printing product of rest of numbers ; Driver program to test above function | #include <bits/stdc++.h> NEW_LINE using namespace std ; void kFactors ( int n , int k ) { vector < int > P ; while ( n % 2 == 0 ) { P . push_back ( 2 ) ; n /= 2 ; } for ( int i = 3 ; i * i <= n ; i = i + 2 ) { while ( n % i == 0 ) { n = n / i ; P . push_back ( i ) ; } } if ( n > 2 ) P . push_back ( n ) ; if ( P . size ( ) < k ) { cout << " - 1" << endl ; return ; } for ( int i = 0 ; i < k - 1 ; i ++ ) cout << P [ i ] << " , ▁ " ; int product = 1 ; for ( int i = k - 1 ; i < P . size ( ) ; i ++ ) product = product * P [ i ] ; cout << product << endl ; } int main ( ) { int n = 54 , k = 3 ; kFactors ( n , k ) ; return 0 ; } |
Largest number smaller than or equal to n and digits in non | C ++ program for brute force approach to find largest number having digits in non - decreasing order . ; Returns the required number ; loop to recursively check the numbers less than or equal to given number ; Keep traversing digits from right to left . For every digit check if it is smaller than prev_dig ; We found the required number ; Driver program | #include <bits/stdc++.h> NEW_LINE using namespace std ; long long nondecdigits ( long long n ) { long long int x = 0 ; for ( x = n ; x >= 1 ; x -- ) { int no = x ; int prev_dig = 11 ; bool flag = true ; while ( no != 0 ) { if ( prev_dig < no % 10 ) { flag = false ; break ; } prev_dig = no % 10 ; no /= 10 ; } if ( flag == true ) break ; } return x ; } int main ( ) { long long n = 200 ; cout << nondecdigits ( n ) ; return 0 ; } |
Largest number smaller than or equal to n and digits in non | C ++ program for efficient approach to find largest number having digits in non - decreasing order . ; Prints the largest number smaller than s and digits in non - decreasing order . ; array to store digits of number ; conversion of characters of string int number ; variable holds the value of index after which all digits are set 9 ; Checking the condition if the digit is less than its left digit ; If first digit is 0 no need to print it ; Driver function | #include <bits/stdc++.h> NEW_LINE using namespace std ; void nondecdigits ( string s ) { long long m = s . size ( ) ; long long a [ m ] ; for ( long long i = 0 ; i < m ; i ++ ) a [ i ] = s [ i ] - '0' ; long long level = m - 1 ; for ( long long i = m - 1 ; i > 0 ; i -- ) { if ( a [ i ] < a [ i - 1 ] ) { a [ i - 1 ] -- ; level = i - 1 ; } } if ( a [ 0 ] != 0 ) { for ( long long i = 0 ; i <= level ; i ++ ) cout << a [ i ] ; for ( long long i = level + 1 ; i < m ; i ++ ) cout << "9" ; } else { for ( long long i = 1 ; i < level ; i ++ ) cout << a [ i ] ; for ( long long i = level + 1 ; i < m ; i ++ ) cout << "9" ; } } int main ( ) { string n = "200" ; nondecdigits ( n ) ; return 0 ; } |
Count Divisors of n in O ( n ^ 1 / 3 ) | C implementation of Naive method to count all divisors ; function to count the divisors ; If divisors are equal , count only one ; else Otherwise count both ; Driver program to test above function | #include <bits/stdc++.h> NEW_LINE using namespace std ; int countDivisors ( int n ) { int cnt = 0 ; for ( int i = 1 ; i <= sqrt ( n ) ; i ++ ) { if ( n % i == 0 ) { if ( n / i == i ) cnt ++ ; cnt = cnt + 2 ; } } return cnt ; } int main ( ) { printf ( " Total ▁ distinct ▁ divisors ▁ of ▁ 100 ▁ are ▁ : ▁ % d " , countDivisors ( 100 ) ) ; return 0 ; } |
Check if the door is open or closed | C ++ implementation of doors open or closed ; Function to check whether ' n ' has even number of factors or not ; if ' n ' is a perfect square it has odd number of factors ; else ' n ' has even number of factors ; Function to find and print status of each door ; If even number of factors final status is closed ; else odd number of factors final status is open ; Driver program | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool hasEvenNumberOfFactors ( int n ) { int root_n = sqrt ( n ) ; if ( ( root_n * root_n ) == n ) return false ; return true ; } void printStatusOfDoors ( int n ) { for ( int i = 1 ; i <= n ; i ++ ) { if ( hasEvenNumberOfFactors ( i ) ) cout << " closed " << " ▁ " ; else cout << " open " << " ▁ " ; } } int main ( ) { int n = 5 ; printStatusOfDoors ( n ) ; return 0 ; } |
Check if frequency of each digit is less than the digit | A C ++ program to validate a number ; Function to validate number ( Check iffrequency of a digit is less than thedigit itself or not ) ; If current digit of temp is same as i ; if frequency is greater than digit value , return false ; driver program | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool validate ( long long int n ) { for ( int i = 0 ; i < 10 ; i ++ ) { long long int temp = n ; int count = 0 ; while ( temp ) { if ( temp % 10 == i ) count ++ ; if ( count > i ) return false ; temp /= 10 ; } } return true ; } int main ( ) { long long int n = 1552793 ; if ( validate ( n ) ) cout << " True " ; else cout << " False " ; return 0 ; } |
Check if a larger number divisible by 36 | C ++ implementation to check divisibility by 36 ; Function to check whether a number is divisible by 36 or not ; null number cannot be divisible by 36 ; single digit number other than 0 is not divisible by 36 ; number formed by the last 2 digits ; if number is not divisible by 4 ; number is divisible by 4 calculate sum of digits ; sum of digits is not divisible by 9 ; number is divisible by 4 and 9 hence , number is divisible by 36 ; Driver program | #include <bits/stdc++.h> NEW_LINE using namespace std ; string divisibleBy36 ( string num ) { int l = num . length ( ) ; if ( l == 0 ) return " No " ; if ( l == 1 && num [ 0 ] != '0' ) return " No " ; int two_digit_num = ( num [ l - 2 ] - '0' ) * 10 + ( num [ l - 1 ] - '0' ) ; if ( two_digit_num % 4 != 0 ) return " No " ; int sum = 0 ; for ( int i = 0 ; i < l ; i ++ ) sum += ( num [ i ] - '0' ) ; if ( sum % 9 != 0 ) return " No " ; return " Yes " ; } int main ( ) { string num = "92567812197966231384" ; cout << divisibleBy36 ( num ) ; return 0 ; } |
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