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Hilbert Number | CPP program to find nth hilbert Number ; Utility function to return Nth Hilbert Number ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; long nthHilbertNumber ( int n ) { return 4 * ( n - 1 ) + 1 ; } int main ( ) { int n = 5 ; cout << nthHilbertNumber ( n ) ; return 0 ; } |
Program to find the nth Kynea number | CPP code to find nth Kynea number ; Function to calculate nth kynea number ; Firstly calculate 2 ^ n + 1 ; Now calculate ( 2 ^ n + 1 ) ^ 2 ; Now calculate ( ( 2 ^ n + 1 ) ^ 2 ) - 2 ; return nth Kynea number ; Driver Program ; print nth kynea number | #include <bits/stdc++.h> NEW_LINE using namespace std ; long nthKyneaNumber ( int n ) { n = ( 1 << n ) + 1 ; n = n * n ; n = n - 2 ; return n ; } int main ( ) { int n = 8 ; cout << nthKyneaNumber ( n ) ; return 0 ; } |
Program to find the nth Kynea number | CPP code to find nth Kynea number ; Function to calculate nth kynea number ; Calculate nth kynea number ; Driver Program ; print nth kynea number | #include <bits/stdc++.h> NEW_LINE using namespace std ; long nthKyneaNumber ( int n ) { return ( ( 1 << ( 2 * n ) ) + ( 1 << ( n + 1 ) ) - 1 ) ; } int main ( ) { int n = 2 ; cout << nthKyneaNumber ( n ) ; return 0 ; } |
Program to check whether a number is Proth number or not | CPP program to check Proth number ; Utility function to check power of two ; Function to check if the Given number is Proth number or not ; check if k divides n or not ; Check if n / k is power of 2 or not ; update k to next odd number ; If we reach here means there exists no value of K Such that k is odd number and n / k is a power of 2 greater than k ; Driver code ; Get n ; Check n for Proth Number | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool isPowerOfTwo ( int n ) { return ( n && ! ( n & ( n - 1 ) ) ) ; } bool isProthNumber ( int n ) { int k = 1 ; while ( k < ( n / k ) ) { if ( n % k == 0 ) { if ( isPowerOfTwo ( n / k ) ) return true ; } k = k + 2 ; } return false ; } int main ( ) { int n = 25 ; if ( isProthNumber ( n - 1 ) ) cout << " YES " ; else cout << " NO " ; return 0 ; } |
Find last two digits of sum of N factorials | C ++ program to find the unit place digit of the first N natural numbers factorials ; Function to find the unit ' s ▁ and ▁ ten ' s place digit ; Let us write for cases when N is smaller than or equal to 10. ; We know following ( 1 ! + 2 ! + 3 ! + 4 ! ... + 10 ! ) % 100 = 13 else ( N >= 10 ) ; Driver code | #include <iostream> NEW_LINE using namespace std ; #define ll long int NEW_LINE int get_last_two_digit ( long long int N ) { if ( N <= 10 ) { ll ans = 0 , fac = 1 ; for ( int i = 1 ; i <= N ; i ++ ) { fac = fac * i ; ans += fac ; } return ans % 100 ; } return 13 ; } int main ( ) { long long int N = 1 ; for ( N = 1 ; N <= 10 ; N ++ ) cout << " For ▁ N ▁ = ▁ " << N << " ▁ : ▁ " << get_last_two_digit ( N ) << endl ; return 0 ; } |
Check whether product of ' n ' numbers is even or odd | C ++ implementation to check whether product of ' n ' numbers is even or odd ; function to check whether product of ' n ' numbers is even or odd ; if a single even number is found , then final product will be an even number ; product is an odd number ; Driver program to test above | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool isProductEven ( int arr [ ] , int n ) { for ( int i = 0 ; i < n ; i ++ ) if ( ( arr [ i ] & 1 ) == 0 ) return true ; return false ; } int main ( ) { int arr [ ] = { 2 , 4 , 3 , 5 } ; int n = sizeof ( arr ) / sizeof ( arr [ 0 ] ) ; if ( isProductEven ( arr , n ) ) cout << " Even " ; else cout << " Odd " ; return 0 ; } |
Sum of squares of Fibonacci numbers | C ++ Program to find sum of squares of Fibonacci numbers ; Function to calculate sum of squares of Fibonacci numbers ; Initialize result ; Add remaining terms ; Driver program to test above function | #include <bits/stdc++.h> NEW_LINE using namespace std ; int calculateSquareSum ( int n ) { if ( n <= 0 ) return 0 ; int fibo [ n + 1 ] ; fibo [ 0 ] = 0 , fibo [ 1 ] = 1 ; int sum = ( fibo [ 0 ] * fibo [ 0 ] ) + ( fibo [ 1 ] * fibo [ 1 ] ) ; for ( int i = 2 ; i <= n ; i ++ ) { fibo [ i ] = fibo [ i - 1 ] + fibo [ i - 2 ] ; sum += ( fibo [ i ] * fibo [ i ] ) ; } return sum ; } int main ( ) { int n = 6 ; cout << " Sum ▁ of ▁ squares ▁ of ▁ Fibonacci ▁ numbers ▁ is ▁ : ▁ " << calculateSquareSum ( n ) << endl ; return 0 ; } |
Value of the series ( 1 ^ 3 + 2 ^ 3 + 3 ^ 3 + ... + n ^ 3 ) mod 4 for a given n | C ++ implementation of the approach ; function for obtaining the value of f ( n ) mod 4 ; Find the remainder of n when divided by 4 ; If n is of the form 4 k or 4 k + 3 ; If n is of the form 4 k + 1 or 4 k + 2 ; Driver code | #include <iostream> NEW_LINE using namespace std ; int fnMod ( int n ) { int rem = n % 4 ; if ( rem == 0 rem == 3 ) return 0 ; else if ( rem == 1 rem == 2 ) return 1 ; } int main ( ) { int n = 6 ; cout << fnMod ( n ) ; return 0 ; } |
Minimum increment operations to make the array in increasing order | C ++ program to find minimum moves required to make the array in increasing order ; function to find minimum moves required to make the array in increasing order ; to store answer ; iterate over an array ; non - increasing order ; add moves to answer ; increase the element ; return required answer ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int MinimumMoves ( int a [ ] , int n , int x ) { int ans = 0 ; for ( int i = 1 ; i < n ; i ++ ) { if ( a [ i ] <= a [ i - 1 ] ) { int p = ( a [ i - 1 ] - a [ i ] ) / x + 1 ; ans += p ; a [ i ] += p * x ; } } return ans ; } int main ( ) { int arr [ ] = { 1 , 3 , 3 , 2 } ; int x = 2 ; int n = sizeof ( arr ) / sizeof ( arr [ 0 ] ) ; cout << MinimumMoves ( arr , n , x ) ; return 0 ; } |
Check if a large number is divisible by 2 , 3 and 5 or not | CPP program to Check if a large number is divisible by 2 , 3 and 5 or not . ; function to return sum of digits of a number ; function to Check if a large number is divisible by 2 , 3 and 5 or not ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int SumOfDigits ( string str , int n ) { int sum = 0 ; for ( int i = 0 ; i < n ; i ++ ) sum += ( int ) ( str [ i ] - '0' ) ; return sum ; } bool Divisible ( string str , int n ) { if ( SumOfDigits ( str , n ) % 3 == 0 and str [ n - 1 ] == '0' ) return true ; return false ; } int main ( ) { string str = "263730746028908374890" ; int n = str . size ( ) ; if ( Divisible ( str , n ) ) cout << " YES " ; else cout << " NO " ; return 0 ; } |
Count all the numbers in a range with smallest factor as K | C ++ program to find the count of numbers in a range whose smallest factor is K ; Function to check if k is a prime number or not ; Corner case ; Check from 2 to n - 1 ; Function to check if a number is not divisible by any number between 2 and K - 1 ; to check if the num is divisible by any numbers between 2 and k - 1 ; if not divisible by any number between 2 and k - 1 but divisible by k ; Function to find count of numbers in range [ a , b ] with smallest factor as K ; a number can be divisible only by k and not by any number less than k only if k is a prime ; to check if a number has smallest factor as K ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool isPrime ( int k ) { if ( k <= 1 ) return false ; for ( int i = 2 ; i < k ; i ++ ) if ( k % i == 0 ) return false ; return true ; } int check ( int num , int k ) { int flag = 1 ; for ( int i = 2 ; i < k ; i ++ ) { if ( num % i == 0 ) flag = 0 ; } if ( flag == 1 ) { if ( num % k == 0 ) return 1 ; else return 0 ; } else return 0 ; } int findCount ( int a , int b , int k ) { int count = 0 ; if ( ! isPrime ( k ) ) return 0 ; else { int ans ; for ( int i = a ; i <= b ; i ++ ) { ans = check ( i , k ) ; if ( ans == 1 ) count ++ ; else continue ; } } return count ; } int main ( ) { int a = 2020 , b = 6300 , k = 29 ; cout << findCount ( a , b , k ) ; return 0 ; } |
Largest number with the given set of N digits that is divisible by 2 , 3 and 5 | C ++ implementation of above approach ; Function to find the largest integer with the given set ; find sum of all the digits look if any 0 is present or not ; if 0 is not present , the resultant number won 't be divisible by 5 ; sort all the elements in a non - decreasing manner ; if there is just one element 0 ; find the remainder of the sum of digits when divided by 3 ; there can a remainder as 1 or 2 ; traverse from the end of the digits ; first element which has the same remainder remove it ; if there is no element which has a same remainder as y ; subtract it by 3 ( could be one or two ) ; delete two minimal digits which has a remainder as y ; print all the digits as a single integer ; Driver code ; initialize the number of set of digits ; initialize all the set of digits in a vector | #include <bits/stdc++.h> NEW_LINE using namespace std ; #define ll long long NEW_LINE int findLargest ( int n , vector < int > & v ) { int flag = 0 ; ll sum = 0 ; for ( int i = 0 ; i < n ; i ++ ) { if ( v [ i ] == 0 ) flag = 1 ; sum += v [ i ] ; } if ( ! flag ) cout << " Not ▁ possible " << endl ; else { sort ( v . begin ( ) , v . end ( ) , greater < int > ( ) ) ; if ( v [ 0 ] == 0 ) { cout << "0" << endl ; return 0 ; } else { int flag = 0 ; int y = sum % 3 ; if ( y != 0 ) { for ( int i = n - 1 ; i >= 0 ; i -- ) { if ( v [ i ] % 3 == y ) { v . erase ( v . begin ( ) + i ) ; flag = 1 ; break ; } } if ( flag == 0 ) { y = 3 - y ; int cnt = 0 ; for ( int i = n - 1 ; i >= 0 ; i -- ) { if ( v [ i ] % 3 == y ) { v . erase ( v . begin ( ) + i ) ; cnt ++ ; if ( cnt >= 2 ) break ; } } } } if ( * v . begin ( ) == 0 ) cout << "0" << endl ; else for ( int i : v ) { cout << i ; } } } } int main ( ) { int n = 11 ; vector < int > v { 3 , 9 , 9 , 6 , 4 , 3 , 6 , 4 , 9 , 6 , 0 } ; findLargest ( n , v ) ; return 0 ; } |
Find the k | C ++ program to find the K - th smallest factor ; Function to find the k 'th divisor ; initialize vectors v1 and v2 ; store all the divisors in the two vectors accordingly ; reverse the vector v2 to sort it in increasing order ; if k is greater than the size of vectors then no divisor can be possible ; else print the ( k - 1 ) th value of vector ; If K is lying in first vector ; If K is lying in second vector ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; void findkth ( int n , int k ) { vector < int > v1 ; vector < int > v2 ; for ( int i = 1 ; i <= sqrt ( n ) ; i ++ ) { if ( n % i == 0 ) { v1 . push_back ( i ) ; if ( i != sqrt ( n ) ) v2 . push_back ( n / i ) ; } } reverse ( v2 . begin ( ) , v2 . end ( ) ) ; if ( k > ( v1 . size ( ) + v2 . size ( ) ) ) cout << " Doesn ' t ▁ Exist " ; else { if ( k <= v1 . size ( ) ) cout << v1 [ k - 1 ] ; else cout << v2 [ k - v1 . size ( ) - 1 ] ; } } int main ( ) { int n = 15 , k = 2 ; findkth ( n , k ) ; return 0 ; } |
Number of solutions for x < y , where a <= x <= b and c <= y <= d and x , y are integers | C ++ implementation of above approach ; function to Find the number of solutions for x < y , where a <= x <= b and c <= y <= d and x , y integers . ; to store answer ; iterate explicitly over all possible values of x ; return answer ; Driver code ; function call | #include <bits/stdc++.h> NEW_LINE using namespace std ; int NumberOfSolutions ( int a , int b , int c , int d ) { int ans = 0 ; for ( int i = a ; i <= b ; i ++ ) if ( d >= max ( c , i + 1 ) ) ans += d - max ( c , i + 1 ) + 1 ; return ans ; } int main ( ) { int a = 2 , b = 3 , c = 3 , d = 4 ; cout << NumberOfSolutions ( a , b , c , d ) ; return 0 ; } |
Minimum value possible of a given function from the given set | C ++ implementation of above approach ; Function to find the value of F ( n ) ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; #define ll long long NEW_LINE ll findF_N ( ll n ) { ll ans = 0 ; for ( ll i = 0 ; i < n ; ++ i ) ans += ( i + 1 ) * ( n - i - 1 ) ; return ans ; } int main ( ) { ll n = 3 ; cout << findF_N ( n ) ; return 0 ; } |
Find N digits number which is divisible by D | CPP program to Find N digits number which is divisible by D ; Function to return N digits number which is divisible by D ; to store answer ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; string findNumber ( int n , int d ) { string ans = " " ; if ( d != 10 ) { ans += to_string ( d ) ; for ( int i = 1 ; i < n ; i ++ ) ans += '0' ; } else { if ( n == 1 ) ans += " Impossible " ; else { ans += '1' ; for ( int i = 1 ; i < n ; i ++ ) ans += '0' ; } } return ans ; } int main ( ) { int n = 12 , d = 3 ; cout << findNumber ( n , d ) ; return 0 ; } |
Count all the numbers less than 10 ^ 6 whose minimum prime factor is N | C ++ implementation of above approach ; the sieve of prime number and count of minimum prime factor ; form the prime sieve ; 1 is not a prime number ; form the sieve ; if i is prime ; if i is the least prime factor ; mark the number j as non prime ; count the numbers whose least prime factor is i ; Driver code ; form the sieve ; display ; display | #include <bits/stdc++.h> NEW_LINE using namespace std ; #define MAX 1000000 NEW_LINE int sieve_Prime [ MAX + 4 ] = { 0 } , sieve_count [ MAX + 4 ] = { 0 } ; void form_sieve ( ) { sieve_Prime [ 1 ] = 1 ; for ( int i = 2 ; i <= MAX ; i ++ ) { if ( sieve_Prime [ i ] == 0 ) { for ( int j = i * 2 ; j <= MAX ; j += i ) { if ( sieve_Prime [ j ] == 0 ) { sieve_Prime [ j ] = 1 ; sieve_count [ i ] ++ ; } } } } } int main ( ) { form_sieve ( ) ; int n = 2 ; cout << " Count ▁ = ▁ " << ( sieve_count [ n ] + 1 ) << endl ; n = 3 ; cout << " Count ▁ = ▁ " << ( sieve_count [ n ] + 1 ) << endl ; return 0 ; } |
How to access elements of a Square Matrix | C ++ Program to read a square matrix and print the elements on secondary diagonal ; Get the square matrix ; Display the matrix ; Print the elements on secondary diagonal ; check for elements on secondary diagonal | #include <bits/stdc++.h> NEW_LINE using namespace std ; int main ( ) { int matrix [ 5 ] [ 5 ] , row_index , column_index , x = 0 , size = 5 ; for ( row_index = 0 ; row_index < size ; row_index ++ ) { for ( column_index = 0 ; column_index < size ; column_index ++ ) { matrix [ row_index ] [ column_index ] = ++ x ; } } cout << " The ▁ matrix ▁ is STRNEWLINE " ; for ( row_index = 0 ; row_index < size ; row_index ++ ) { for ( column_index = 0 ; column_index < size ; column_index ++ ) { cout << matrix [ row_index ] [ column_index ] << " ▁ " ; } cout << endl ; } cout << " Elements on Secondary diagonal : " for ( row_index = 0 ; row_index < size ; row_index ++ ) { for ( column_index = 0 ; column_index < size ; column_index ++ ) { if ( ( row_index + column_index ) == size - 1 ) cout << matrix [ row_index ] [ column_index ] << " , ▁ " ; } } return 0 ; } |
How to access elements of a Square Matrix | C ++ Program to read a square matrix and print the elements above secondary diagonal ; Get the square matrix ; Display the matrix ; Print the elements above secondary diagonal ; check for elements above secondary diagonal | #include <bits/stdc++.h> NEW_LINE using namespace std ; int main ( ) { int matrix [ 5 ] [ 5 ] , row_index , column_index , x = 0 , size = 5 ; for ( row_index = 0 ; row_index < size ; row_index ++ ) { for ( column_index = 0 ; column_index < size ; column_index ++ ) { matrix [ row_index ] [ column_index ] = ++ x ; } } cout << " The ▁ matrix ▁ is STRNEWLINE " ; for ( row_index = 0 ; row_index < size ; row_index ++ ) { for ( column_index = 0 ; column_index < size ; column_index ++ ) { cout << matrix [ row_index ] [ column_index ] << " ▁ " ; } cout << endl ; } cout << " Elements above Secondary diagonal are : " ; for ( row_index = 0 ; row_index < size ; row_index ++ ) { for ( column_index = 0 ; column_index < size ; column_index ++ ) { if ( ( row_index + column_index ) < size - 1 ) cout << matrix [ row_index ] [ column_index ] << " , ▁ " ; } } return 0 ; } |
How to access elements of a Square Matrix | C ++ Program to read a square matrix and print the Corner Elements ; Get the square matrix ; Display the matrix ; Print the Corner elements ; check for corner elements | #include <bits/stdc++.h> NEW_LINE using namespace std ; int main ( ) { int matrix [ 5 ] [ 5 ] , row_index , column_index , x = 0 , size = 5 ; for ( row_index = 0 ; row_index < size ; row_index ++ ) { for ( column_index = 0 ; column_index < size ; column_index ++ ) { matrix [ row_index ] [ column_index ] = ++ x ; } } cout << " The ▁ matrix ▁ is " << endl ; for ( row_index = 0 ; row_index < size ; row_index ++ ) { for ( column_index = 0 ; column_index < size ; column_index ++ ) { cout << " TABSYMBOL " << matrix [ row_index ] [ column_index ] ; } cout << endl ; } cout << " Corner Elements are : " ; for ( row_index = 0 ; row_index < size ; row_index ++ ) { for ( column_index = 0 ; column_index < size ; column_index ++ ) { if ( ( row_index == 0 row_index == size - 1 ) && ( column_index == 0 column_index == size - 1 ) ) cout << matrix [ row_index ] [ column_index ] << " , " ; } } return 0 ; } |
Find the largest good number in the divisors of given number N | CPP program to find the largest , good number in the divisors of given number N . ; function to return distinct prime factors ; to store distinct prime factors ; run a loop upto sqrt ( n ) ; place this prime factor in vector ; This condition is to handle the case when n is a prime number greater than 1 ; function that returns good number ; distinct prime factors ; to store answer ; product of all distinct prime factors is required answer ; Driver code ; function call | #include <bits/stdc++.h> NEW_LINE using namespace std ; vector < int > PrimeFactors ( int n ) { vector < int > v ; int x = n ; for ( int i = 2 ; i * i <= n ; i ++ ) { if ( x % i == 0 ) { v . push_back ( i ) ; while ( x % i == 0 ) x /= i ; } } if ( x > 1 ) v . push_back ( x ) ; return v ; } int GoodNumber ( int n ) { vector < int > v = PrimeFactors ( n ) ; int ans = 1 ; for ( int i = 0 ; i < v . size ( ) ; i ++ ) ans *= v [ i ] ; return ans ; } int main ( ) { int n = 12 ; cout << GoodNumber ( n ) ; return 0 ; } |
Find Largest Special Prime which is less than or equal to a given number | CPP program to find the Largest Special Prime which is less than or equal to a given number ; Function to check whether the number is a special prime or not ; While number is not equal to zero ; If the number is not prime return false . ; Else remove the last digit by dividing the number by 10. ; If the number has become zero then the number is special prime , hence return true ; Function to find the Largest Special Prime which is less than or equal to a given number ; Initially all numbers are considered Primes . ; There is always an answer possible ; Checking if the number is a special prime or not ; If yes print the number and break the loop . ; Else decrement the number . ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool checkSpecialPrime ( bool * sieve , int num ) { while ( num ) { if ( ! sieve [ num ] ) { return false ; } num /= 10 ; } return true ; } void findSpecialPrime ( int N ) { bool sieve [ N + 10 ] ; memset ( sieve , true , sizeof ( sieve ) ) ; sieve [ 0 ] = sieve [ 1 ] = false ; for ( long long i = 2 ; i <= N ; i ++ ) { if ( sieve [ i ] ) { for ( long long j = i * i ; j <= N ; j += i ) { sieve [ j ] = false ; } } } while ( true ) { if ( checkSpecialPrime ( sieve , N ) ) { cout << N << ' ' ; break ; } else N -- ; } } int main ( ) { findSpecialPrime ( 379 ) ; findSpecialPrime ( 100 ) ; return 0 ; } |
Check if an integer can be expressed as a sum of two semi | CPP Code to check if an integer can be expressed as sum of two semi - primes ; Utility function to compute semi - primes in a range ; num /= j , ++ cnt ; Increment count of prime numbers ; If number is greater than 1 , add it to the count variable as it indicates the number remain is prime number ; if count is equal to '2' then number is semi - prime ; Utility function to check if a number sum of two semi - primes ; arr [ i ] is already a semi - prime if n - arr [ i ] is also a semi - prime then we a number can be expressed as sum of two semi - primes ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; #define MAX 1000000 NEW_LINE vector < int > arr ; bool sprime [ MAX ] ; void computeSemiPrime ( ) { memset ( sprime , false , sizeof ( sprime ) ) ; for ( int i = 2 ; i < MAX ; i ++ ) { int cnt = 0 ; int num = i ; for ( int j = 2 ; cnt < 2 && j * j <= num ; ++ j ) { while ( num % j == 0 ) { } } if ( num > 1 ) ++ cnt ; if ( cnt == 2 ) { sprime [ i ] = true ; arr . push_back ( i ) ; } } } bool checkSemiPrime ( int n ) { int i = 0 ; while ( arr [ i ] <= n / 2 ) { if ( sprime [ n - arr [ i ] ] ) { return true ; } i ++ ; } return false ; } int main ( ) { computeSemiPrime ( ) ; int n = 30 ; if ( checkSemiPrime ( n ) ) cout << " YES " ; else cout << " NO " ; return 0 ; } |
Check if a number is a Pythagorean Prime or not | CPP program to check if a number is Pythagorean prime or not ; Function to check if a number is prime or not ; Corner cases ; This is checked so that we can skip middle five numbers in below loop ; Driver Program ; Check if number is prime and of the form 4 * n + 1 | #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 main ( ) { int n = 13 ; if ( isPrime ( n ) && ( n % 4 == 1 ) ) { cout << " YES " ; } else { cout << " NO " ; } return 0 ; } |
Divide an isosceles triangle in two parts with ratio of areas as n : m | C ++ program , to find height h which divide isosceles triangle into ratio n : m ; Function to return the height ; type cast the n , m into float ; calculate the height for cut ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; float heightCalculate ( int H , int n , int m ) { float N = n * 1.0 ; float M = m * 1.0 ; float h = H * sqrt ( N / ( N + M ) ) ; return h ; } int main ( ) { int H = 10 , n = 3 , m = 4 ; cout << heightCalculate ( H , n , m ) ; return 0 ; } |
Check n ^ 2 | CPP program to find n ^ 2 - m ^ 2 is prime or not . ; Check a number is prime or not ; run a loop upto square of given number ; Check if n ^ 2 - m ^ 2 is prime ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool isprime ( int x ) { for ( int i = 2 ; i * i <= x ; i ++ ) if ( x % i == 0 ) return false ; return true ; } bool isNSqMinusnMSqPrime ( int m , int n ) { if ( n - m == 1 and isprime ( m + n ) ) return true ; else return false ; } int main ( ) { int m = 13 , n = 16 ; if ( isNSqMinusnMSqPrime ( m , n ) ) cout << " YES " ; else cout << " NO " ; return 0 ; } |
Find ' N ' number of solutions with the given inequality equations | C ++ implementation of above approach ; Function to calculate all the solutions ; there is no solutions ; print first element as y - n + 1 ; print rest n - 1 elements as 1 ; Driver code ; initialize the number of elements and the value of x an y | #include <bits/stdc++.h> NEW_LINE using namespace std ; #define ll long long NEW_LINE void findsolution ( ll n , ll x , ll y ) { if ( ( y - n + 1 ) * ( y - n + 1 ) + n - 1 < x y < n ) { cout << " No ▁ solution " ; return ; } cout << y - n + 1 ; while ( n -- > 1 ) cout << endl << 1 ; } int main ( ) { ll n , x , y ; n = 5 , x = 15 , y = 15 ; findsolution ( n , x , y ) ; return 0 ; } |
Number of different positions where a person can stand | C ++ implementation of above approach ; Function to find the position ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int findPosition ( int n , int f , int b ) { return n - max ( f + 1 , n - b ) + 1 ; } int main ( ) { int n = 5 , f = 2 , b = 3 ; cout << findPosition ( n , f , b ) ; return 0 ; } |
Program for n | CPP program to find the nth odd number ; Function to find the nth odd number ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int nthOdd ( int n ) { return ( 2 * n - 1 ) ; } int main ( ) { int n = 10 ; cout << nthOdd ( n ) ; return 0 ; } |
Program for n | CPP program to find the nth even number ; Function to find the nth even number ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int nthEven ( int n ) { return ( 2 * n ) ; } int main ( ) { int n = 10 ; cout << nthEven ( n ) ; return 0 ; } |
Program to find the Nth Harmonic Number | CPP program to find N - th Harmonic Number ; Function to find N - th Harmonic Number ; H1 = 1 ; loop to apply the forumula Hn = H1 + H2 + H3 ... + Hn - 1 + Hn - 1 + 1 / n ; Driver Code | #include <iostream> NEW_LINE using namespace std ; double nthHarmonic ( int N ) { float harmonic = 1.00 ; for ( int i = 2 ; i <= N ; i ++ ) { harmonic += ( float ) 1 / i ; } return harmonic ; } int main ( ) { int N = 8 ; cout << nthHarmonic ( N ) ; return 0 ; } |
Program to find Nth term of series 0 , 7 , 18 , 33 , 51 , 75 , 102 , 133 , ... . . | C ++ program to find the N - th term of the series : 0 , 7 , 18 , 33 , 51 , 75 , 102 , ... . . ; calculate Nth term of series ; Driver code | #include <iostream> NEW_LINE #include <math.h> NEW_LINE using namespace std ; int nthTerm ( int n ) { return 2 * pow ( n , 2 ) + n - 3 ; } int main ( ) { int N = 4 ; cout << nthTerm ( N ) << endl ; return 0 ; } |
Program to find Nth term of series 0 , 10 , 30 , 60 , 99 , 150 , 210 , 280. ... ... ... . | C ++ program to find the N - th term of the series : 0 , 10 , 30 , 60 , 99 , 150 , 210 , ... . . ; calculate Nth term of series ; Driver code | #include <iostream> NEW_LINE #include <math.h> NEW_LINE using namespace std ; int nthTerm ( int n ) { return 5 * pow ( n , 2 ) - 5 * n ; } int main ( ) { int N = 4 ; cout << nthTerm ( N ) << endl ; return 0 ; } |
Program to find Nth term of series 2 , 12 , 28 , 50 , 77 , 112 , 152 , 198 , ... . . | C ++ program to find the N - th term of the series : 2 , 12 , 28 , 50 , 77 , 112 , 152 , 198 , ... . . ; calculate Nth term of series ; Driver code | #include <iostream> NEW_LINE #include <math.h> NEW_LINE using namespace std ; int nthTerm ( int n ) { return 3 * pow ( n , 2 ) + n - 2 ; } int main ( ) { int N = 4 ; cout << nthTerm ( N ) << endl ; return 0 ; } |
Program to find Nth term of series 4 , 14 , 28 , 46 , 68 , 94 , 124 , 158 , ... . . | CPP program to find the N - th term of the series : 4 , 14 , 28 , 46 , 68 , 94 , 124 , 158 , ... . . ; calculate Nth term of series ; Driver code | #include <iostream> NEW_LINE #include <math.h> NEW_LINE using namespace std ; int nthTerm ( int n ) { return 2 * pow ( n , 2 ) + 4 * n - 2 ; } int main ( ) { int N = 4 ; cout << nthTerm ( N ) << endl ; return 0 ; } |
Program to find Nth term of series 0 , 11 , 28 , 51 , 79 , 115 , 156 , 203 , ... . | CPP program to find the N - th term of the series : 0 , 11 , 28 , 51 , 79 , 115 , 156 , 203. . ... ; calculate Nth term of series ; Driver code | #include <iostream> NEW_LINE #include <math.h> NEW_LINE using namespace std ; int nthTerm ( int n ) { return 3 * pow ( n , 2 ) + 2 * n - 5 ; } int main ( ) { int N = 4 ; cout << nthTerm ( N ) << endl ; return 0 ; } |
Program to find Nth term of series 0 , 9 , 22 , 39 , 60 , 85 , 114 , 147 , ... . . | C ++ program to find the N - th term of the series : 0 , 9 , 22 , 39 , 60 , 85 , 114 , 147. . ... ; calculate Nth term of series ; Driver code | #include <iostream> NEW_LINE #include <math.h> NEW_LINE using namespace std ; int nthTerm ( int n ) { return 2 * pow ( n , 2 ) + 3 * n - 5 ; } int main ( ) { int N = 4 ; cout << nthTerm ( N ) << endl ; return 0 ; } |
Program to find Nth term of series 3 , 12 , 29 , 54 , 86 , 128 , 177 , 234 , ... . . | C ++ program to find the N - th term of the series : 3 , 12 , 29 , 54 , 86 , 128 , 177 , 234 , ... . . ; calculate Nth term of series ; Driver code | #include <iostream> NEW_LINE #include <math.h> NEW_LINE using namespace std ; int nthTerm ( int n ) { return 4 * pow ( n , 2 ) - 3 * n + 2 ; } int main ( ) { int N = 4 ; cout << nthTerm ( N ) << endl ; return 0 ; } |
Find other two sides and angles of a right angle triangle | C ++ program to print all sides and angles of right angle triangle given one side ; Function to find angle A Angle in front of side a ; applied cosine rule ; convert into degrees ; Function to find angle B Angle in front of side b ; applied cosine rule ; convert into degrees and return ; Function to print all angles of the right angled triangle ; for calculate angle A ; for calculate angle B ; Function to find other two sides of the right angled triangle ; if n is odd ; case of n = 1 handled separately ; case of n = 2 handled separately ; Print angles of the triangle ; Driver Program | #include <bits/stdc++.h> NEW_LINE #include <cmath> NEW_LINE using namespace std ; #define PI 3.1415926535 NEW_LINE double findAnglesA ( double a , double b , double c ) { double A = acos ( ( b * b + c * c - a * a ) / ( 2 * b * c ) ) ; return A * 180 / PI ; } double findAnglesB ( double a , double b , double c ) { double B = acos ( ( a * a + c * c - b * b ) / ( 2 * a * c ) ) ; return B * 180 / PI ; } void printAngles ( int a , int b , int c ) { double x = ( double ) a ; double y = ( double ) b ; double z = ( double ) c ; double A = findAnglesA ( x , y , z ) ; double B = findAnglesB ( x , y , z ) ; cout << " Angles ▁ are ▁ A ▁ = ▁ " << A << " , ▁ B ▁ = ▁ " << B << " , ▁ C ▁ = ▁ " << 90 << endl ; } void printOtherSides ( int n ) { int b , c ; if ( n & 1 ) { if ( n == 1 ) cout << -1 << endl ; else { b = ( n * n - 1 ) / 2 ; c = ( n * n + 1 ) / 2 ; cout << " Side ▁ b ▁ = ▁ " << b << " , ▁ Side ▁ c ▁ = ▁ " << c << endl ; } } else { if ( n == 2 ) cout << -1 << endl ; else { b = n * n / 4 - 1 ; c = n * n / 4 + 1 ; cout << " Side ▁ b ▁ = ▁ " << b << " , ▁ Side ▁ c ▁ = ▁ " << c << endl ; } } printAngles ( n , b , c ) ; } int main ( ) { int a = 12 ; printOtherSides ( a ) ; return 0 ; } |
Sum of the first N terms of the series 2 , 6 , 12 , 20 , 30. ... | C ++ program to find sum of first n terms ; Function to calculate the sum ; Driver code ; number of terms to be included in the sum ; find the Sn | #include <bits/stdc++.h> NEW_LINE using namespace std ; int calculateSum ( int n ) { return n * ( n + 1 ) / 2 + n * ( n + 1 ) * ( 2 * n + 1 ) / 6 ; } int main ( ) { int n = 3 ; cout << " Sum ▁ = ▁ " << calculateSum ( n ) ; return 0 ; } |
Program to find the Nth term of the series 0 , 5 , 14 , 27 , 44 , ... ... . . | CPP program to find N - th term of the series : 0 , 5 , 14 , 27 , 44 . . . ; Calculate Nth term of series ; Driver code | #include <iostream> NEW_LINE #include <math.h> NEW_LINE using namespace std ; int nthTerm ( int n ) { return 2 * pow ( n , 2 ) - n - 1 ; } int main ( ) { int N = 4 ; cout << nthTerm ( N ) ; return 0 ; } |
Program to find the Nth term of the series 0 , 5 , 18 , 39 , 67 , 105 , 150 , 203 , ... | CPP program to find the N - th term of the series : 0 , 5 , 18 , 39 , 67 , 105 , 150 , 203 , ... ; calculate Nth term of series ; Driver code | #include <iostream> NEW_LINE #include <math.h> NEW_LINE using namespace std ; int nthTerm ( int n ) { return 4 * pow ( n , 2 ) - 7 * n + 3 ; } int main ( ) { int N = 4 ; cout << nthTerm ( N ) ; return 0 ; } |
Check whether a given Number is Power | C ++ program to find whether a number is power - isolated or not ; for 2 as prime factor ; for odd prime factor ; calculate product of powers and prime factors ; check result for power - isolation ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; void checkIfPowerIsolated ( int num ) { int input = num ; int count = 0 ; int factor [ num + 1 ] = { 0 } ; if ( num % 2 == 0 ) { while ( num % 2 == 0 ) { ++ count ; num /= 2 ; } factor [ 2 ] = count ; } for ( int i = 3 ; i * i <= num ; i += 2 ) { count = 0 ; while ( num % i == 0 ) { ++ count ; num /= i ; } if ( count > 0 ) factor [ i ] = count ; } if ( num > 1 ) factor [ num ] = 1 ; int product = 1 ; for ( int i = 0 ; i < num + 1 ; i ++ ) { if ( factor [ i ] > 0 ) product = product * factor [ i ] * i ; } if ( product == input ) cout << " Power - isolated ▁ Integer STRNEWLINE " ; else cout << " Not ▁ a ▁ Power - isolated ▁ Integer STRNEWLINE " ; } int main ( ) { checkIfPowerIsolated ( 12 ) ; checkIfPowerIsolated ( 18 ) ; checkIfPowerIsolated ( 35 ) ; return 0 ; } |
Program to find the Nth term of the series 3 , 7 , 13 , 21 , 31. ... . | CPP program to find the Nth term of given series . ; Function to calculate sum ; Return Nth term ; driver code ; declaration of number of terms ; Get the Nth term | #include <iostream> NEW_LINE #include <math.h> NEW_LINE using namespace std ; long long int getNthTerm ( long long int N ) { return ( pow ( N , 2 ) + N + 1 ) ; } int main ( ) { long long int N = 11 ; cout << getNthTerm ( N ) ; return 0 ; } |
Sum of the numbers upto N that are divisible by 2 or 5 | C ++ implementation of above approach ; Function to find the sum ; sum2 is sum of numbers divisible by 2 ; sum5 is sum of number divisible by 5 ; sum10 of numbers divisible by 2 and 5 ; Driver code | #include <bits/stdc++.h> NEW_LINE #define ll long long int NEW_LINE using namespace std ; ll findSum ( int n ) { ll sum2 , sum5 , sum10 ; sum2 = ( ( n / 2 ) * ( 4 + ( n / 2 - 1 ) * 2 ) ) / 2 ; sum5 = ( ( n / 5 ) * ( 10 + ( n / 5 - 1 ) * 5 ) ) / 2 ; sum10 = ( ( n / 10 ) * ( 20 + ( n / 10 - 1 ) * 10 ) ) / 2 ; return sum2 + sum5 - sum10 ; } int main ( ) { int n = 5 ; cout << findSum ( n ) << endl ; return 0 ; } |
Find four factors of N with maximum product and sum equal to N | Set 3 | C ++ implementation of above approach ; Function to find primes ; Corner cases ; This is checked so that we can skip middle five numbers in below loop ; Function to find factors ; run a loop upto square root of that number ; if the n is perfect square ; otherwise push it 's two divisors ; sort the divisors ; Function to find max product ; To store factors of ' n ' ; find factors ; if it is divisible by 4. ; if it is prime ; otherwise answer will be possible ; include last third factor ; nested loop to find other two factors ; Driver code ; function call | #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 ; } void factors ( int N , vector < int > & v [ ] ) { for ( int i = 2 ; i < N ; i ++ ) { for ( int j = 1 ; j * j <= i ; j ++ ) { if ( i % j == 0 ) { if ( i / j == j ) v [ i ] . push_back ( j ) ; else { v [ i ] . push_back ( j ) ; v [ i ] . push_back ( i / j ) ; } } } sort ( v [ i ] . begin ( ) , v [ i ] . end ( ) ) ; } } int product ( int n ) { vector < int > v [ n + 100 ] ; factors ( n + 100 , v ) ; if ( n % 4 == 0 ) { int x = n / 4 ; x *= x ; return x * x ; } else { if ( isPrime [ n ] ) return -1 ; else { int ans = -1 ; if ( v [ n ] . size ( ) > 2 ) { int fac = v [ n ] [ v [ n ] . size ( ) - 3 ] ; for ( int i = v [ n ] . size ( ) - 1 ; i >= 0 ; i -- ) { for ( int j = v [ n ] . size ( ) - 1 ; j >= 0 ; j -- ) { if ( ( fac * 2 ) + ( v [ n ] [ j ] + v [ n ] [ i ] ) == n ) ans = max ( ans , fac * fac * v [ n ] [ j ] * v [ n ] [ i ] ) ; } } return ans ; } } } } int main ( ) { int n = 24 ; cout << product ( n ) ; return 0 ; } |
Ratio of mth and nth terms of an A . P . with given ratio of sums | C ++ code to calculate ratio ; function to calculate ratio of mth and nth term ; ratio will be tm / tn = ( 2 * m - 1 ) / ( 2 * n - 1 ) ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; float CalculateRatio ( float m , float n ) { return ( 2 * m - 1 ) / ( 2 * n - 1 ) ; } int main ( ) { float m = 6 , n = 2 ; cout << CalculateRatio ( m , n ) ; return 0 ; } |
Find the sum of n terms of the series 1 , 8 , 27 , 64 ... . | C ++ program to find the sum of n terms ; Function to calculate the sum ; Return total sum ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int calculateSum ( int n ) { return pow ( n * ( n + 1 ) / 2 , 2 ) ; } int main ( ) { int n = 4 ; cout << calculateSum ( n ) ; return 0 ; } |
Sum of Digits in a ^ n till a single digit | CPP program to find single digit sum of a ^ n . ; This function finds single digit sum of n . ; Returns single digit sum of a ^ n . We use modular exponentiation technique . ; Driver code | #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 powerDigitSum ( int a , int n ) { int res = 1 ; while ( n ) { if ( n % 2 == 1 ) { res = res * digSum ( a ) ; res = digSum ( res ) ; } a = digSum ( digSum ( a ) * digSum ( a ) ) ; n /= 2 ; } return res ; } int main ( ) { int a = 9 , n = 4 ; cout << powerDigitSum ( a , n ) ; return 0 ; } |
Program to find total number of edges in a Complete Graph | C ++ implementation to find the number of edges in a complete graph ; Function to find the total number of edges in a complete graph with N vertices ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int totEdge ( int n ) { int result = 0 ; result = ( n * ( n - 1 ) ) / 2 ; return result ; } int main ( ) { int n = 6 ; cout << totEdge ( n ) ; return 0 ; } |
Find minimum number of Log value needed to calculate Log upto N | C ++ program to find number of log values needed to calculate all the log values from 1 to N ; In this vector prime [ i ] will store true if prime [ i ] is prime , else store false ; Using sieve of Eratosthenes to find all prime upto N ; Function to find number of log values needed to calculate all the log values from 1 to N ; calculate primes upto N ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; #define MAX 1000005 NEW_LINE vector < bool > prime ( MAX , true ) ; void sieve ( int N ) { prime [ 0 ] = prime [ 1 ] = false ; for ( int i = 2 ; i <= N ; i ++ ) { if ( prime [ i ] ) { for ( int j = 2 ; i * j <= N ; j ++ ) prime [ i * j ] = false ; } } } int countLogNeeded ( int N ) { int count = 0 ; sieve ( N ) ; for ( int i = 1 ; i <= N ; i ++ ) { if ( prime [ i ] ) count ++ ; } return count ; } int main ( ) { int N = 6 ; cout << countLogNeeded ( N ) << endl ; return 0 ; } |
Program to find the count of coins of each type from the given ratio | C ++ implementation of above approach ; function to calculate coin . ; Converting each of them in rupees . As we are given totalRupees = 1800 ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int coin ( int totalRupees , int X , int Y , int Z ) { float one = 0 , fifty = 0 , twentyfive = 0 , result = 0 , total = 0 ; one = X * 1 ; fifty = ( ( Y * 1 ) / 2.0 ) ; twentyfive = ( ( Z * 1 ) / 4.0 ) ; total = one + fifty + twentyfive ; result = ( ( totalRupees ) / total ) ; return result ; } int main ( ) { int totalRupees = 1800 ; int X = 1 , Y = 2 , Z = 4 ; int Rupees = coin ( totalRupees , X , Y , Z ) ; cout << "1 ▁ rupess ▁ coins ▁ = ▁ " << Rupees * 1 << endl ; cout << "50 ▁ paisa ▁ coins ▁ = ▁ " << Rupees * 2 << endl ; cout << "25 ▁ paisa ▁ coins ▁ = ▁ " << Rupees * 4 << endl ; return 0 ; } |
Find the sum of series 0. X + 0. XX + 0. XXX + ... upto k terms | C ++ program for sum of the series 0. x , 0. xx , 0. xxx , ... upto k terms ; function which return the sum of series ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; float sumOfSeries ( int x , int k ) { return ( float ( x ) / 81 ) * ( 9 * k - 1 + pow ( 10 , ( -1 ) * k ) ) ; } int main ( ) { int x = 9 ; int k = 20 ; cout << sumOfSeries ( x , k ) ; return 0 ; } |
Find four factors of N with maximum product and sum equal to N | Set | C ++ program to find four factors of N with maximum product and sum equal to N ; Function to find factors and to print those four factors ; push all the factors in the container ; number of factors ; Initial maximum ; hash - array to mark the pairs ; form all the pair sums ; if the pair sum is less than n ; push in another container ; mark the sum with the elements formed ; mark in the map that v [ i ] + v [ j ] is present ; new size of all the pair sums ; iterate for all pair sum ; the required part ; if the required part is also present in pair sum ; find the elements with which the first pair is formed ; find the elements with which the second pair is formed ; check for previous maximum ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; void findfactors ( int n ) { unordered_map < int , int > mpp ; vector < int > v , v1 ; for ( int i = 1 ; i <= sqrt ( n ) ; i ++ ) { if ( n % i == 0 ) { v . push_back ( i ) ; if ( i != ( n / i ) && i != 1 ) v . push_back ( n / i ) ; } } int s = v . size ( ) ; int maxi = -1 ; pair < int , int > mp1 [ n + 5 ] ; for ( int i = 0 ; i < s ; i ++ ) { for ( int j = i ; j < s ; j ++ ) { if ( v [ i ] + v [ j ] < n ) { v1 . push_back ( v [ i ] + v [ j ] ) ; mp1 [ v [ i ] + v [ j ] ] = { v [ i ] , v [ j ] } ; mpp [ v [ i ] + v [ j ] ] = 1 ; } } } s = v1 . size ( ) ; for ( int i = 0 ; i < s ; i ++ ) { int el = n - ( v1 [ i ] ) ; if ( mpp [ el ] == 1 ) { int a = mp1 [ v1 [ i ] ] . first ; int b = mp1 [ v1 [ i ] ] . second ; int c = mp1 [ n - v1 [ i ] ] . first ; int d = mp1 [ n - v1 [ i ] ] . second ; maxi = max ( a * b * c * d , maxi ) ; } } if ( maxi == -1 ) cout << " Not ▁ Possible STRNEWLINE " ; else { cout << " The ▁ maximum ▁ product ▁ is ▁ " << maxi << endl ; } } int main ( ) { int n = 50 ; findfactors ( n ) ; return 0 ; } |
Maximize the product of four factors of a Number | C ++ implementation of above approach ; Declare the vector of factors for storing the ; function to find out the factors of a number ; Loop until the i reaches the sqrt ( n ) ; Check if i is a factor of n ; if both the factors are same we only push one factor ; factor1 is pushed ; factor2 is pushed ; Function to find the maximum product ; Initialize the product with - 1 ; Find the sum of factors and store it in s ; Compare whether it is equal to the n ; product of factors ; Check whether we have a better p now if yes update ; Driver code ; initializes the vectors with the divisors of n ; prints out the maximised product . | #include <bits/stdc++.h> NEW_LINE using namespace std ; vector < int > factors ; void findFactors ( int n ) { for ( int i = 1 ; i * i <= n ; i ++ ) { if ( n % i == 0 ) { if ( ( n / i ) == i ) factors . push_back ( i ) ; else { factors . push_back ( n / i ) ; factors . push_back ( i ) ; } } } } int findProduct ( int n ) { int product = -1 ; int si = factors . size ( ) ; for ( int i = 0 ; i < si ; i ++ ) for ( int j = 0 ; j < si ; j ++ ) for ( int k = 0 ; k < si ; k ++ ) for ( int l = 0 ; l < si ; l ++ ) { int s = factors [ i ] + factors [ j ] + factors [ k ] + factors [ l ] ; if ( s == n ) { int p = factors [ i ] * factors [ j ] * factors [ k ] * factors [ l ] ; if ( p > product ) product = p ; } } return product ; } int main ( ) { int n = 10 ; findFactors ( n ) ; cout << findProduct ( n ) ; return 0 ; } |
Maximize the product of four factors of a Number | C ++ implementation of above approach ; For calculation of a ^ b ; Function to check ; every odd and number less than 3. ; every number divisible by 4. ; every number divisible by 6. ; every number divisible by 10. ; for every even number which is not divisible by above values . ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int modExp ( int a , int b ) { int result = 1 ; while ( b > 0 ) { if ( b & 1 ) result = result * a ; a = a * a ; b /= 2 ; } return result ; } int check ( int num ) { if ( num & 1 num < 3 ) return -1 ; else if ( num % 4 == 0 ) return modExp ( num / 4 , 4 ) ; else if ( num % 6 == 0 ) return modExp ( num / 3 , 2 ) * modExp ( num / 6 , 2 ) ; else if ( num % 10 == 0 ) return modExp ( num / 5 , 2 ) * ( num / 10 ) * ( num / 2 ) ; else return -1 ; } int main ( ) { int num = 10 ; cout << check ( num ) ; return 0 ; } |
Check if any large number is divisible by 17 or not | CPP Program to validate the above logic ; Function to check if the number is divisible by 17 or not ; Extracting the last digit ; Truncating the number ; Subtracting the five times the last digit from the remaining number ; Return n is divisible by 17 ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool isDivisible ( long long int n ) { while ( n / 100 ) { int d = n % 10 ; n /= 10 ; n -= d * 5 ; } return ( n % 17 == 0 ) ; } int main ( ) { long long int n = 19877658 ; if ( isDivisible ( n ) ) cout << " Yes " << endl ; else cout << " No " << endl ; return 0 ; } |
Minimum number of elements to be removed to make XOR maximum | C ++ implementation to find minimum number of elements to remove to get maximum XOR value ; First n in the below condition is for the case where n is 0 ; Function to find minimum number of elements to be removed . ; Driver code ; print minimum number of elements to be removed | #include <bits/stdc++.h> NEW_LINE using namespace std ; unsigned int nextPowerOf2 ( unsigned int n ) { unsigned count = 0 ; if ( n && ! ( n & ( n - 1 ) ) ) return n ; while ( n != 0 ) { n >>= 1 ; count += 1 ; } return 1 << count ; } int removeElement ( unsigned int n ) { if ( n == 1 n == 2 ) return 0 ; unsigned int a = nextPowerOf2 ( n ) ; if ( n == a n == a - 1 ) return 1 ; else if ( n == a - 2 ) return 0 ; else if ( n % 2 == 0 ) return 1 ; else return 2 ; } int main ( ) { unsigned int n = 5 ; cout << removeElement ( n ) ; return 0 ; } |
Program to find Length of Bridge using Speed and Length of Train | C ++ Program to implement above code . ; function to calculate the length of bridge . ; Driver Code ; Assuming the input variables | #include <bits/stdc++.h> NEW_LINE using namespace std ; int bridge_length ( int trainLength , int Speed , int Time ) { return ( ( Time * Speed ) - trainLength ) ; } int main ( ) { int trainLength = 120 ; int Speed = 30 ; int Time = 18 ; cout << " Length ▁ of ▁ bridge ▁ = ▁ " << bridge_length ( trainLength , Speed , Time ) << " ▁ meters " ; return 0 ; } |
Program to find sum of the given sequence | CPP program to find the sum of the given sequence ; function to find moudulo inverse under 10 ^ 9 + 7 ; Function to find the sum of the given sequence ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; const long long MOD = 1000000007 ; long long modInv ( long long x ) { long long n = MOD - 2 ; long long result = 1 ; while ( n ) { if ( n & 1 ) result = result * x % MOD ; x = x * x % MOD ; n = n / 2 ; } return result ; } long long getSum ( long long n , long long k ) { long long ans = 1 ; for ( long long i = n + 1 ; i > n - k ; i -- ) ans = ans * i % MOD ; ans = ans * modInv ( k + 1 ) % MOD ; return ans ; } int main ( ) { long long n = 3 , k = 2 ; cout << getSum ( n , k ) ; return 0 ; } |
Triplet with no element divisible by 3 and sum N | C ++ program to print a , b and c such that a + b + c = N ; Function to print a , b and c ; check if n - 2 is divisible by 3 or not ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; void printCombination ( int n ) { cout << 1 << " ▁ " ; if ( ( n - 2 ) % 3 == 0 ) cout << 2 << " ▁ " << n - 3 ; else cout << 1 << " ▁ " << n - 2 ; } int main ( ) { int n = 233 ; printCombination ( n ) ; return 0 ; } |
Pairs with GCD equal to one in the given range | C ++ program to print all pairs ; Function to print all pairs ; check if even ; We can print all adjacent pairs for ( int i = l ; i < r ; i += 2 ) { cout << " { " << i << " , ▁ " << i + 1 << " } , ▁ " ; } ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool checkPairs ( int l , int r ) { if ( ( l - r ) % 2 == 0 ) return false ; return true ; } int main ( ) { int l = 1 , r = 8 ; if ( checkPairs ( l , r ) ) cout << " Yes " ; else cout << " No " ; return 0 ; } |
Check if a number with even number of digits is palindrome or not | C ++ program to find number is palindrome or not without using any extra space ; Function to check if the number is palindrome ; if divisible by 11 then true ; if not divisible by 11 ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool isPalindrome ( int n ) { if ( n % 11 == 0 ) { return true ; } return false ; } int main ( ) { isPalindrome ( 123321 ) ? cout << " Palindrome " : cout << " Not ▁ Palindrome " ; return 0 ; } |
Count number of triplets with product equal to given number with duplicates allowed | C ++ program for above implementation ; The target value for which we have to find the solution ; This variable contains the total count of triplets found ; Loop from the first to the third last integer in the list ; Check if arr [ i ] is a factor of target or not . If not , skip to the next element ; Check if the pair ( arr [ i ] , arr [ j ] ) can be a part of triplet whose product is equal to the target ; Find the remaining element of the triplet ; If element is found . increment the total count of the triplets | #include <iostream> NEW_LINE using namespace std ; int main ( ) { int target = 93 ; int arr [ ] = { 1 , 31 , 3 , 1 , 93 , 3 , 31 , 1 , 93 } ; int length = sizeof ( arr ) / sizeof ( arr [ 0 ] ) ; int totalCount = 0 ; for ( int i = 0 ; i < length - 2 ; i ++ ) { if ( target % arr [ i ] == 0 ) { for ( int j = i + 1 ; j < length - 1 ; j ++ ) { if ( target % ( arr [ i ] * arr [ j ] ) == 0 ) { int toFind = target / ( arr [ i ] * arr [ j ] ) ; for ( int k = j + 1 ; k < length ; k ++ ) { if ( arr [ k ] == toFind ) { totalCount ++ ; } } } } } } cout << " Total ▁ number ▁ of ▁ triplets ▁ found ▁ : ▁ " << totalCount ; return 0 ; } |
Number of Permutations such that no Three Terms forms Increasing Subsequence | C ++ program to find the nth catalan number ; Returns value of Binomial Coefficient C ( n , k ) ; Since C ( n , k ) = C ( n , n - k ) ; Calculate value of [ n * ( n - 1 ) * -- - * ( n - k + 1 ) ] / [ k * ( k - 1 ) * -- - * 1 ] ; A Binomial coefficient based function to find nth catalan number in O ( n ) time ; Calculate value of 2 nCn ; return 2 nCn / ( n + 1 ) ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; unsigned long int binomialCoeff ( unsigned int n , unsigned int k ) { unsigned long int res = 1 ; if ( k > n - k ) k = n - k ; for ( int i = 0 ; i < k ; ++ i ) { res *= ( n - i ) ; res /= ( i + 1 ) ; } return res ; } unsigned long int catalan ( unsigned int n ) { unsigned long int c = binomialCoeff ( 2 * n , n ) ; return c / ( n + 1 ) ; } int main ( ) { int n = 3 ; cout << catalan ( n ) << endl ; return 0 ; } |
Fascinating Number | C ++ program to implement fascinating number ; function to check if number is fascinating or not ; frequency count array using 1 indexing ; obtaining the resultant number using string concatenation ; Traversing the string character by character ; gives integer value of a character digit ; To check if any digit has appeared multiple times ; Traversing through freq array to check if any digit was missing ; Driver code ; Input number ; Not a valid number ; Calling the function to check if input number is fascinating or not | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool isFascinating ( int num ) { int freq [ 10 ] = { 0 } ; string val = " " + to_string ( num ) + to_string ( num * 2 ) + to_string ( num * 3 ) ; for ( int i = 0 ; i < val . length ( ) ; i ++ ) { int digit = val [ i ] - '0' ; if ( freq [ digit ] and digit != 0 > 0 ) return false ; else freq [ digit ] ++ ; } for ( int i = 1 ; i < 10 ; i ++ ) { if ( freq [ i ] == 0 ) return false ; } return true ; } int main ( ) { int num = 192 ; if ( num < 100 ) cout << " No " << endl ; else { bool ans = isFascinating ( num ) ; if ( ans ) cout << " Yes " ; else cout << " No " ; } } |
Count ways to distribute m items among n people | C ++ code for calculating number of ways to distribute m mangoes amongst n people where all mangoes and people are identical ; function used to generate binomial coefficient time complexity O ( m ) ; helper function for generating no of ways to distribute m mangoes amongst n people ; not enough mangoes to be distributed ; ways -> ( n + m - 1 ) C ( n - 1 ) ; Driver function ; m represents number of mangoes n represents number of people | #include <bits/stdc++.h> NEW_LINE using namespace std ; int binomial_coefficient ( int n , int m ) { int res = 1 ; if ( m > n - m ) m = n - m ; for ( int i = 0 ; i < m ; ++ i ) { res *= ( n - i ) ; res /= ( i + 1 ) ; } return res ; } int calculate_ways ( int m , int n ) { if ( m < n ) return 0 ; int ways = binomial_coefficient ( n + m - 1 , n - 1 ) ; return ways ; } int main ( ) { int m = 7 , n = 5 ; int result = calculate_ways ( m , n ) ; printf ( " % d STRNEWLINE " , result ) ; return 0 ; } |
Queries to count the number of unordered co | C ++ program to find number of unordered coprime pairs of integers from 1 to N ; to store euler 's totient function ; to store required answer ; Computes and prints totient of all numbers smaller than or equal to N . ; Initialise the phi [ ] with 1 ; Compute other Phi values ; If phi [ p ] is not computed already , then number p is prime ; Phi of a prime number p is always equal to p - 1. ; Update phi values of all multiples of p ; Add contribution of p to its multiple i by multiplying with ( 1 - 1 / p ) ; function to compute number coprime pairs ; function call to compute euler totient function ; prefix sum of all euler totient function values ; Driver code ; function call | #include <bits/stdc++.h> NEW_LINE using namespace std ; #define N 100005 NEW_LINE int phi [ N ] ; int S [ N ] ; void computeTotient ( ) { for ( int i = 1 ; i < N ; i ++ ) phi [ i ] = i ; for ( int p = 2 ; p < N ; p ++ ) { if ( phi [ p ] == p ) { phi [ p ] = p - 1 ; for ( int i = 2 * p ; i < N ; i += p ) { phi [ i ] = ( phi [ i ] / p ) * ( p - 1 ) ; } } } } void CoPrimes ( ) { computeTotient ( ) ; for ( int i = 1 ; i < N ; i ++ ) S [ i ] = S [ i - 1 ] + phi [ i ] ; } int main ( ) { CoPrimes ( ) ; int q [ ] = { 3 , 4 } ; int n = sizeof ( q ) / sizeof ( q [ 0 ] ) ; for ( int i = 0 ; i < n ; i ++ ) cout << " Number ▁ of ▁ unordered ▁ coprime STRNEWLINE " << " pairs ▁ of ▁ integers ▁ from ▁ 1 ▁ to ▁ " << q [ i ] << " ▁ are ▁ " << S [ q [ i ] ] << endl ; return 0 ; } |
N | C ++ program to find the N - th term in 1 , 11 , 55 , 239 , 991 , ... . ; Function to return the decimal value of a binary number ; Initializing base value to 1 , i . e 2 ^ 0 ; find the binary representation of the N - th number in sequence ; base case ; answer string ; add n - 1 1 's ; add 0 ; add n 1 's at end ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int binaryToDecimal ( string n ) { string num = n ; int dec_value = 0 ; int base = 1 ; int len = num . length ( ) ; for ( int i = len - 1 ; i >= 0 ; i -- ) { if ( num [ i ] == '1' ) dec_value += base ; base = base * 2 ; } return dec_value ; } int numberSequence ( int n ) { if ( n == 1 ) return 1 ; string s = " " ; for ( int i = 1 ; i < n ; i ++ ) s += '1' ; s += '0' ; for ( int i = 1 ; i <= n ; i ++ ) s += '1' ; int num = binaryToDecimal ( s ) ; return num ; } int main ( ) { int n = 4 ; cout << numberSequence ( n ) ; return 0 ; } |
N | C ++ program to find the N - th term in 1 , 11 , 55 , 239 , 991 , ... . ; Function to find the N - th term ; calculates the N - th term ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int numberSequence ( int n ) { int num = pow ( 4 , n ) - pow ( 2 , n ) - 1 ; return num ; } int main ( ) { int n = 4 ; cout << numberSequence ( n ) ; return 0 ; } |
Alternate Primes till N | C ++ program to print all primes smaller than or equal to n using Naive approach . ; Function for checking number is prime or not ; if flag = 0 then number is prime and return 1 otherwise return 0 ; Function for printing alternate prime number ; counter is initialize with 0 ; looping through 2 to n - 1 ; function calling along with if condition ; if counter is multiple of 2 then only print prime number ; Driver code ; Function calling | #include <bits/stdc++.h> NEW_LINE using namespace std ; int prime ( int num ) { int i , flag = 0 ; for ( i = 2 ; i <= num / 2 ; i ++ ) { if ( num % i == 0 ) { flag = 1 ; break ; } } if ( flag == 0 ) return 1 ; else return 0 ; } void print_alternate_prime ( int n ) { int counter = 0 ; for ( int num = 2 ; num < n ; num ++ ) { if ( prime ( num ) == 1 ) { if ( counter % 2 == 0 ) cout << num << " ▁ " ; counter ++ ; } } } int main ( ) { int n = 15 ; cout << " Following ▁ are ▁ the ▁ alternate ▁ prime " << " ▁ number ▁ smaller ▁ than ▁ or ▁ equal ▁ to ▁ " << n << endl ; print_alternate_prime ( n ) ; } |
Alternate Primes till N | C ++ program to print all primes smaller than or equal to n using 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 ; Print all prime numbers ; for next prime to get printed ; Driver Program to test above function | #include <bits/stdc++.h> NEW_LINE using namespace std ; void SieveOfEratosthenes ( 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 ; } } bool flag = true ; for ( int p = 2 ; p <= n ; p ++ ) { if ( prime [ p ] ) { if ( flag ) { cout << p << " ▁ " ; flag = false ; } else { flag = true ; } } } } int main ( ) { int n = 15 ; cout << " Following ▁ are ▁ the ▁ alternate " << " ▁ prime ▁ numbers ▁ smaller ▁ " << " ▁ than ▁ or ▁ equal ▁ to ▁ " << n << endl ; SieveOfEratosthenes ( n ) ; return 0 ; } |
Find maximum among x ^ ( y ^ 2 ) or y ^ ( x ^ 2 ) where x and y are given | C ++ program to find the greater value ; Function to find maximum ; Case 1 ; Case 2 ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool findGreater ( int x , int y ) { if ( x > y ) { return false ; } else { return true ; } } int main ( ) { int x = 4 ; int y = 9 ; findGreater ( x , y ) ? cout << "1 STRNEWLINE " : cout << "2 STRNEWLINE " ; return 0 ; } |
Maximum profit after buying and selling stocks with transaction fees | C ++ implementation of above approach ; b [ 0 ] will contain the maximum profit ; b [ 1 ] will contain the day on which we are getting the maximum profit ; here finding the max profit ; if we get less then or equal to zero it means we are not getting the profit ; check if sum is greater then maximum then store the new maximum ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int max_profit ( int a [ ] , int b [ ] , int n , int fee ) { int i , j , profit ; int l , r , diff_day = 1 , sum = 0 ; b [ 0 ] = 0 ; b [ 1 ] = diff_day ; for ( i = 1 ; i < n ; i ++ ) { l = 0 ; r = diff_day ; sum = 0 ; for ( j = n - 1 ; j >= i ; j -- ) { profit = ( a [ r ] - a [ l ] ) - fee ; if ( profit > 0 ) { sum = sum + profit ; } l ++ ; r ++ ; } if ( b [ 0 ] < sum ) { b [ 0 ] = sum ; b [ 1 ] = diff_day ; } diff_day ++ ; } return 0 ; } int main ( ) { int arr [ ] = { 6 , 1 , 7 , 2 , 8 , 4 } ; int n = sizeof ( arr ) / sizeof ( arr [ 0 ] ) ; int b [ 2 ] ; int tranFee = 2 ; max_profit ( arr , b , n , tranFee ) ; cout << b [ 0 ] << " , ▁ " << b [ 1 ] << endl ; return 0 ; } |
Eggs dropping puzzle ( Binomial Coefficient and Binary Search Solution ) | C ++ program to find minimum number of trials in worst case . ; Find sum of binomial coefficients xCi ( where i varies from 1 to n ) . ; Do binary search to find minimum number of trials in worst case . ; Initialize low and high as 1 st and last floors ; Do binary search , for every mid , find sum of binomial coefficients and check if the sum is greater than k or not . ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int binomialCoeff ( int x , int n , int k ) { int sum = 0 , term = 1 ; for ( int i = 1 ; i <= n ; ++ i ) { term *= x - i + 1 ; term /= i ; sum += term ; if ( sum > k ) return sum ; } return sum ; } int minTrials ( int n , int k ) { int low = 1 , high = k ; while ( low < high ) { int mid = ( low + high ) / 2 ; if ( binomialCoeff ( mid , n , k ) < k ) low = mid + 1 ; else high = mid ; } return low ; } int main ( ) { cout << minTrials ( 2 , 10 ) ; return 0 ; } |
Find next palindrome prime | CPP program to find next palindromic prime for a given number . ; if ( 8 <= N <= 11 ) return 11 ; generate odd length palindrome number which will cover given constraint . ; if y >= N and it is a prime number then return it . ; Driver code | #include <iostream> NEW_LINE #include <string> NEW_LINE using namespace std ; bool isPrime ( int num ) { if ( num < 2 num % 2 == 0 ) return num == 2 ; for ( int i = 3 ; i * i <= num ; i += 2 ) if ( num % i == 0 ) return false ; return true ; } int primePalindrome ( int N ) { if ( 8 <= N && N <= 11 ) return 11 ; for ( int x = 1 ; x < 100000 ; ++ x ) { string s = to_string ( x ) , r ( s . rbegin ( ) , s . rend ( ) ) ; int y = stoi ( s + r . substr ( 1 ) ) ; if ( y >= N && isPrime ( y ) ) return y ; } return -1 ; } int main ( ) { cout << primePalindrome ( 112 ) ; return 0 ; } |
Number of integral solutions for equation x = b * ( sumofdigits ( x ) ^ a ) + c | C ++ program to find the numbers of values that satisfy the equation ; This function returns the sum of the digits of a number ; This function creates the array of valid numbers ; this computes s ( x ) ^ a ; this gives the result of equation ; checking if the sum same as i ; counter to keep track of numbers ; resultant array ; prints the number ; Driver Code ; calculate which value of x are possible | #include <bits/stdc++.h> NEW_LINE using namespace std ; int getsum ( int a ) { int r = 0 , sum = 0 ; while ( a > 0 ) { r = a % 10 ; sum = sum + r ; a = a / 10 ; } return sum ; } void value ( int a , int b , int c ) { int co = 0 , p = 0 ; int no , r = 0 , x = 0 , q = 0 , w = 0 ; vector < int > v ; for ( int i = 1 ; i < 82 ; i ++ ) { no = pow ( ( double ) i , double ( a ) ) ; no = b * no + c ; if ( no > 0 && no < 1000000000 ) { x = getsum ( no ) ; if ( x == i ) { q ++ ; v . push_back ( no ) ; w ++ ; } } } for ( int i = 0 ; i < v . size ( ) ; i ++ ) { cout << v [ i ] << " ▁ " ; } } int main ( ) { int a = 2 , b = 2 , c = -1 ; value ( a , b , c ) ; return 0 ; } |
Cunningham chain | C ++ program for cunningham chain Function to print the series of second kind ; Function to print Cunningham chain of the second kind ; Iterate till all elements are printed ; check prime or not ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; void print ( int p0 ) { int p1 , i = 0 , x , flag , k ; while ( 1 ) { flag = 1 ; x = ( int ) ( pow ( 2 , i ) ) ; p1 = x * p0 - ( x - 1 ) ; for ( k = 2 ; k < p1 ; k ++ ) { if ( p1 % k == 0 ) { flag = 0 ; break ; } } if ( flag == 0 ) break ; printf ( " % d ▁ " , p1 ) ; i ++ ; } } int main ( ) { int p0 = 19 ; print ( p0 ) ; return 0 ; } |
Count number of right triangles possible with a given perimeter | C ++ program to find the number of right triangles with given perimeter ; Function to return the count ; making a list to store ( a , b ) pairs ; no triangle if p is odd ; make ( a , b ) pair in sorted order ; check to avoid duplicates ; store the new pair ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int countTriangles ( int p ) { vector < pair < int , int > > store ; if ( p % 2 != 0 ) return 0 ; else { int count = 1 ; for ( int b = 1 ; b < p / 2 ; b ++ ) { float a = ( float ) p / 2.0f * ( ( float ) ( ( float ) p - 2.0 * ( float ) b ) / ( ( float ) p - ( float ) b ) ) ; int inta = ( int ) ( a ) ; if ( a == inta ) { pair < int , int > ab ; if ( inta < b ) { ab = { inta , b } ; } else { ab = { b , inta } ; } if ( find ( store . begin ( ) , store . end ( ) , ab ) == store . end ( ) ) { count += 1 ; store . push_back ( ab ) ; } } } return count ; } } int main ( ) { int p = 840 ; cout << " number ▁ of ▁ right ▁ triangles ▁ = ▁ " << countTriangles ( p ) ; return 0 ; } |
Count pairs with Bitwise AND as ODD number | C ++ program to count pairs with Odd AND ; Count total odd numbers in ; return count of even pair ; Driver main ; calling function findOddPair and print number of odd pair | #include <iostream> NEW_LINE using namespace std ; int findOddPair ( int A [ ] , int N ) { int count = 0 ; for ( int i = 0 ; i < N ; i ++ ) if ( ( A [ i ] % 2 == 1 ) ) count ++ ; return count * ( count - 1 ) / 2 ; } int main ( ) { int a [ ] = { 5 , 1 , 3 , 2 } ; int n = sizeof ( a ) / sizeof ( a [ 0 ] ) ; cout << findOddPair ( a , n ) << endl ; return 0 ; } |
Surd number | CPP program to find if a number is Surds or not ; Returns true if x is Surd number ; Try all powers of i ; driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool isSurd ( int n ) { for ( int i = 2 ; i * i <= n ; i ++ ) { int j = i ; while ( j < n ) j = j * i ; if ( j == n ) return false ; } return true ; } int main ( ) { int n = 15 ; if ( isSurd ( n ) ) cout << " Yes " ; else cout << " No " ; return 0 ; } |
Program to find last two digits of 2 ^ n | C ++ code to find last 2 digits of 2 ^ n ; Find the first digit ; Get the last digit from the number ; Remove last digit from number ; Get the last digit from the number ( last second of num ) ; Take last digit to ten 's position i.e. last second digit ; Add the value of ones and tens to make it complete 2 digit number ; return the first digit ; Driver program ; pow function used | #include <bits/stdc++.h> NEW_LINE using namespace std ; int LastTwoDigit ( long long int num ) { int one = num % 10 ; num /= 10 ; int tens = num % 10 ; tens *= 10 ; num = tens + one ; return num ; } int main ( ) { int n = 10 ; long long int num = 1 ; num = pow ( 2 , n ) ; cout << " Last ▁ " << 2 ; cout << " ▁ digits ▁ of ▁ " << 2 ; cout << " ^ " << n << " ▁ = ▁ " ; cout << LastTwoDigit ( num ) << endl ; return 0 ; } |
Program to find last two digits of 2 ^ n | C ++ code to find last 2 digits of 2 ^ n ; 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 >> 1 ; y = y / 2 ; C ++ function to calculate number of digits in x ; C ++ function to print last 2 digits of 2 ^ n ; Generating 10 ^ 2 ; Calling modular exponentiation ; Printing leftmost zeros . Since ( 2 ^ n ) % 2 can have digits less then 2. In that case we need to print zeros ; If temp is not zero then print temp If temp is zero then already printed ; Driver program to test above functions | #include <iostream> NEW_LINE using namespace std ; int power ( long long int x , long long int y , long long int p ) { while ( y > 0 ) { if ( y & 1 ) res = ( res * x ) % p ; x = ( x * x ) % p ; } return res ; } int numberOfDigits ( int x ) { int i = 0 ; while ( x ) { x /= 10 ; i ++ ; } return i ; } void LastTwoDigit ( int n ) { cout << " Last ▁ " << 2 ; cout << " ▁ digits ▁ of ▁ " << 2 ; cout << " ^ " << n << " ▁ = ▁ " ; int temp = 1 ; for ( int i = 1 ; i <= 2 ; i ++ ) temp *= 10 ; temp = power ( 2 , n , temp ) ; for ( int i = 0 ; i < 2 - numberOfDigits ( temp ) ; i ++ ) cout << 0 ; if ( temp ) cout << temp ; } int main ( ) { int n = 72 ; LastTwoDigit ( n ) ; return 0 ; } |
Find gcd ( a ^ n , c ) where a , n and c can vary from 1 to 10 ^ 9 | CPP program to find GCD of a ^ n and b . ; 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 >> 1 ; y = y / 2 ; Finds GCD of a and b ; Finds GCD of a ^ n and c ; check if c is a divisor of a ; First compute ( a ^ n ) % c ; Now simply return GCD of modulo power and c . ; Driver code | #include <bits/stdc++.h> NEW_LINE #define ll long long int NEW_LINE using namespace std ; ll modPower ( ll x , ll y , ll p ) { while ( y > 0 ) { if ( y & 1 ) res = ( res * x ) % p ; x = ( x * x ) % p ; } return res ; } ll gcd ( ll a , ll b ) { if ( b == 0 ) return a ; return gcd ( b , a % b ) ; } ll gcdPow ( ll a , ll n , ll c ) { if ( a % c == 0 ) return c ; ll modexpo = modPower ( a , n , c ) ; return gcd ( modexpo , c ) ; } int main ( ) { ll a = 10248585 , n = 1000000 , c = 12564 ; cout << gcdPow ( a , n , c ) ; return 0 ; } |
Number of sub arrays with odd sum | C ++ code to find count of sub - arrays with odd sum ; Find sum of all subarrays and increment result if sum is odd ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int countOddSum ( int ar [ ] , int n ) { int result = 0 ; for ( int i = 0 ; i <= n - 1 ; i ++ ) { int val = 0 ; for ( int j = i ; j <= n - 1 ; j ++ ) { val = val + ar [ j ] ; if ( val % 2 != 0 ) result ++ ; } } return ( result ) ; } int main ( ) { int ar [ ] = { 5 , 4 , 4 , 5 , 1 , 3 } ; int n = sizeof ( ar ) / sizeof ( ar [ 0 ] ) ; cout << " The ▁ Number ▁ of ▁ Subarrays ▁ with ▁ odd " " ▁ sum ▁ is ▁ " << countOddSum ( ar , n ) ; return ( 0 ) ; } |
Number of sub arrays with odd sum | C ++ proggram to find count of sub - arrays with odd sum ; 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 odd 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 ; An odd can be formed by even - odd pair ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int countOddSum ( int ar [ ] , int n ) { int temp [ 2 ] = { 1 , 0 } ; int result = 0 , val = 0 ; for ( int i = 0 ; i <= n - 1 ; i ++ ) { val = ( ( val + ar [ i ] ) % 2 + 2 ) % 2 ; temp [ val ] ++ ; } result = ( temp [ 0 ] * temp [ 1 ] ) ; return ( result ) ; } int main ( ) { int ar [ ] = { 5 , 4 , 4 , 5 , 1 , 3 } ; int n = sizeof ( ar ) / sizeof ( ar [ 0 ] ) ; cout << " The ▁ Number ▁ of ▁ Subarrays ▁ with ▁ odd " " ▁ sum ▁ is ▁ " << countOddSum ( ar , n ) ; return ( 0 ) ; } |
Program to print factors of a number in pairs | CPP program to print prime factors in pairs . ; Driver code | #include <iostream> NEW_LINE using namespace std ; void printPFsInPairs ( int n ) { for ( int i = 1 ; i * i <= n ; i ++ ) if ( n % i == 0 ) cout << i << " * " << n / i << endl ; } int main ( ) { int n = 24 ; printPFsInPairs ( n ) ; return 0 ; } |
Sum of elements in range L | C ++ program to find the sum between L - R range by creating the array Naive Approach ; Function to find the sum between L and R ; array created ; fill the first half of array ; fill the second half of array ; find the sum between range ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int rangesum ( int n , int l , int r ) { int arr [ n ] ; int c = 1 , i = 0 ; while ( c <= n ) { arr [ i ++ ] = c ; c += 2 ; } c = 2 ; while ( c <= n ) { arr [ i ++ ] = c ; c += 2 ; } int sum = 0 ; for ( i = l - 1 ; i < r ; i ++ ) { sum += arr [ i ] ; } return sum ; } int main ( ) { int n = 12 ; int l = 1 , r = 11 ; cout << ( rangesum ( n , l , r ) ) ; } |
Sum of elements in range L | C ++ program to find the sum between L - R range by creating the array Naive Approach ; Function to calculate the sum if n is even ; both l and r are to the left of mid ; first and last element ; Total number of terms in the sequence is r - l + 1 ; use of formula derived ; both l and r are to the right of mid ; first and last element ; Use of formula derived ; left is to the left of mid and right is to the right of mid ; Take two sums i . e left and right differently and add ; first and last element ; total terms ; no of terms ; The first even number is 2 ; The last element is given by 2 * ( r - n / 2 ) ; formula applied ; Function to calculate the sum if n is odd ; take ceil value if n is odd ; both l and r are to the left of mid ; first and last element ; number of terms ; formula ; both l and r are to the right of mid ; first and last term , ; no of terms ; formula used ; If l is on left and r on right ; calculate separate sums ; first half ; calculate terms ; second half ; add both halves ; Function to find the sum between L and R ; If n is even ; If n is odd ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int sumeven ( int n , int l , int r ) { int sum = 0 ; int mid = n / 2 ; if ( r <= mid ) { int first = ( 2 * l - 1 ) ; int last = ( 2 * r - 1 ) ; int no_of_terms = r - l + 1 ; sum = ( ( no_of_terms ) * ( ( first + last ) ) ) / 2 ; } else if ( l >= mid ) { int first = ( 2 * ( l - n / 2 ) ) ; int last = ( 2 * ( r - n / 2 ) ) ; int no_of_terms = r - l + 1 ; sum = ( ( no_of_terms ) * ( ( first + last ) ) ) / 2 ; } else { int sumleft = 0 , sumright = 0 ; int first_term1 = ( 2 * l - 1 ) ; int last_term1 = ( 2 * ( n / 2 ) - 1 ) ; int no_of_terms1 = n / 2 - l + 1 ; sumleft = ( ( no_of_terms1 ) * ( ( first_term1 + last_term1 ) ) ) / 2 ; int first_term2 = 2 ; int last_term2 = ( 2 * ( r - n / 2 ) ) ; int no_of_terms2 = r - mid ; sumright = ( ( no_of_terms2 ) * ( ( first_term2 + last_term2 ) ) ) / 2 ; sum = ( sumleft + sumright ) ; } return sum ; } int sumodd ( int n , int l , int r ) { int mid = n / 2 + 1 ; int sum = 0 ; if ( r <= mid ) { int first = ( 2 * l - 1 ) ; int last = ( 2 * r - 1 ) ; int no_of_terms = r - l + 1 ; sum = ( ( no_of_terms ) * ( ( first + last ) ) ) / 2 ; } else if ( l > mid ) { int first = ( 2 * ( l - mid ) ) ; int last = ( 2 * ( r - mid ) ) ; int no_of_terms = r - l + 1 ; sum = ( ( no_of_terms ) * ( ( first + last ) ) ) / 2 ; } else { int sumleft = 0 , sumright = 0 ; int first_term1 = ( 2 * l - 1 ) ; int last_term1 = ( 2 * mid - 1 ) ; int no_of_terms1 = mid - l + 1 ; sumleft = ( ( no_of_terms1 ) * ( ( first_term1 + last_term1 ) ) ) / 2 ; int first_term2 = 2 ; int last_term2 = ( 2 * ( r - mid ) ) ; int no_of_terms2 = r - mid ; sumright = ( ( no_of_terms2 ) * ( ( first_term2 + last_term2 ) ) ) / 2 ; sum = ( sumleft + sumright ) ; } return sum ; } int rangesum ( int n , int l , int r ) { int sum = 0 ; if ( n % 2 == 0 ) return sumeven ( n , l , r ) ; else return sumodd ( n , l , r ) ; } int main ( ) { int n = 12 ; int l = 1 , r = 11 ; cout << ( rangesum ( n , l , r ) ) ; } |
Program to find the Interior and Exterior Angle of a Regular Polygon | CPP program to find the interior and exterior angle of a given polygon ; function to find the interior and exterior angle ; formula to find the interior angle ; formula to find the exterior angle ; Displaying the output ; Driver code ; Function calling | #include <iostream> NEW_LINE using namespace std ; void findAngle ( int n ) { int interiorAngle , exteriorAngle ; interiorAngle = ( n - 2 ) * 180 / n ; exteriorAngle = 360 / n ; cout << " Interior ▁ angle : ▁ " << interiorAngle << endl ; cout << " Exterior ▁ angle : ▁ " << exteriorAngle ; } int main ( ) { int n = 10 ; findAngle ( n ) ; return 0 ; } |
Program to calculate distance between two points in 3 D | C ++ program to find distance between two points in 3 D . ; function to print distance ; Driver Code ; function call for distance | #include <bits/stdc++.h> NEW_LINE #include <iomanip> NEW_LINE #include <iostream> NEW_LINE #include <math.h> NEW_LINE using namespace std ; void distance ( float x1 , float y1 , float z1 , float x2 , float y2 , float z2 ) { float d = sqrt ( pow ( x2 - x1 , 2 ) + pow ( y2 - y1 , 2 ) + pow ( z2 - z1 , 2 ) * 1.0 ) ; std :: cout << std :: fixed ; std :: cout << std :: setprecision ( 2 ) ; cout << " ▁ Distance ▁ is ▁ " << d ; return ; } int main ( ) { float x1 = 2 ; float y1 = -5 ; float z1 = 7 ; float x2 = 3 ; float y2 = 4 ; float z2 = 5 ; distance ( x1 , y1 , z1 , x2 , y2 , z2 ) ; return 0 ; } |
Check if the large number formed is divisible by 41 or not | C ++ program to check a large number divisible by 41 or not ; Check if a number is divisible by 41 or not ; array to store all the digits ; base values ; calculate remaining digits ; calculate answer ; check for divisibility ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool DivisibleBy41 ( int first , int second , int c , int n ) { int digit [ n ] ; digit [ 0 ] = first ; digit [ 1 ] = second ; for ( int i = 2 ; i < n ; i ++ ) digit [ i ] = ( digit [ i - 1 ] * c + digit [ i - 2 ] ) % 10 ; int ans = digit [ 0 ] ; for ( int i = 1 ; i < n ; i ++ ) ans = ( ans * 10 + digit [ i ] ) % 41 ; if ( ans % 41 == 0 ) return true ; else return false ; } int main ( ) { int first = 1 , second = 2 , c = 1 , n = 3 ; if ( DivisibleBy41 ( first , second , c , n ) ) cout << " YES " ; else cout << " NO " ; return 0 ; } |
Program to print pentatope numbers upto Nth term | C ++ program to generate Pentatope Number series ; Function to generate nth tetrahedral number ; Function to print pentatope number series upto nth term . ; Initialize prev as 0. It store the sum of all previously generated pentatope numbers ; Loop to print pentatope series ; Find ith tetrahedral number ; Add ith tetrahedral number to sum of all previously generated tetrahedral number to get ith pentatope number ; Update sum of all previously generated tetrahedral number ; Driver code ; Function call to print pentatope number series | #include <bits/stdc++.h> NEW_LINE using namespace std ; int findTetrahedralNumber ( int n ) { return ( ( n * ( n + 1 ) * ( n + 2 ) ) / 6 ) ; } void printSeries ( int n ) { int prev = 0 ; int curr ; for ( int i = 1 ; i <= n ; i ++ ) { curr = findTetrahedralNumber ( i ) ; curr = curr + prev ; cout << curr << " ▁ " ; prev = curr ; } } int main ( ) { int n = 10 ; printSeries ( n ) ; return 0 ; } |
Program to print pentatope numbers upto Nth term | C ++ program to print Pentatope number series . ; Function to print pentatope series up to nth term ; Loop to print pentatope number series ; calculate and print ith pentatope number ; Driver code ; Function call to print pentatope number series | #include <bits/stdc++.h> NEW_LINE using namespace std ; void printSeries ( int n ) { for ( int i = 1 ; i <= n ; i ++ ) { int num = ( i * ( i + 1 ) * ( i + 2 ) * ( i + 3 ) / 24 ) ; cout << num << " ▁ " ; } } int main ( ) { int n = 10 ; printSeries ( n ) ; return 0 ; } |
Program to print tetrahedral numbers upto Nth term | C ++ program to generate tetrahedral number series ; function to generate nth triangular number ; function to print tetrahedral number series up to n ; Initialize prev as 0. It stores the sum of all previously generated triangular number ; Loop to print series ; Find ith triangular number ; Add ith triangular number to sum of all previously generated triangular number to get ith tetrahedral number ; Update sum of all previously generated triangular number ; Driver code ; function call to print series | #include <bits/stdc++.h> NEW_LINE using namespace std ; long findTriangularNumber ( int n ) { return ( n * ( n + 1 ) ) / 2 ; } void printSeries ( int n ) { int prev = 0 ; int curr ; for ( int i = 1 ; i <= n ; i ++ ) { curr = findTriangularNumber ( i ) ; curr = curr + prev ; cout << curr << " ▁ " ; prev = curr ; } } int main ( ) { int n = 10 ; printSeries ( n ) ; return 0 ; } |
Program to print tetrahedral numbers upto Nth term | C ++ program to generate series of tetrahedral numbers ; function to print tetrahedral number series up to n ; loop to print series ; Calculate and print ith tetrahedral number ; Driver code ; function call to print series | #include <bits/stdc++.h> NEW_LINE using namespace std ; void printSeries ( int n ) { for ( int i = 1 ; i <= n ; i ++ ) { int num = i * ( i + 1 ) * ( i + 2 ) / 6 ; cout << num << " ▁ " ; } } int main ( ) { int n = 10 ; printSeries ( n ) ; return 0 ; } |
Number of odd and even results for every value of x in range [ min , max ] after performing N steps | C ++ program to print Number of odd / even results for every value of x in range [ min , end ] after performing N steps ; Function that prints the number of odd and even results ; If constant at layer i is even , beven is true , otherwise false . If the coefficient of x at layer i is even , aeven is true , otherwise false . ; If any of the coefficients at any layer is found to be even , then the product of all the coefficients will always be even . ; Checking whether the constant added after all layers is even or odd . ; Assuming input x is even . ; Assuming input x is odd . ; Displaying the counts . ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; void count_even_odd ( int min , int max , int steps [ ] [ 2 ] ) { int a , b , even , odd ; bool beven = true , aeven = false ; int n = 2 ; for ( int i = 0 ; i < n ; i ++ ) { a = steps [ i ] [ 0 ] , b = steps [ i ] [ 1 ] ; if ( ! ( aeven a & 1 ) ) aeven = true ; if ( beven ) { if ( b & 1 ) beven = false ; } else if ( ! ( a & 1 ) ) { if ( ! ( b & 1 ) ) beven = true ; } else { if ( b & 1 ) beven = true ; } } if ( beven ) { even = ( int ) max / 2 - ( int ) ( min - 1 ) / 2 ; odd = 0 ; } else { even = ( int ) max / 2 - ( int ) ( min - 1 ) / 2 ; odd = 0 ; } if ( ! ( beven ^ aeven ) ) even += max - min + 1 - ( int ) max / 2 + ( int ) ( min - 1 ) / 2 ; else odd += max - min + 1 - ( int ) max / 2 + ( int ) ( min - 1 ) / 2 ; cout << " even ▁ = ▁ " << even << " , ▁ odd ▁ = ▁ " << odd << endl ; } int main ( ) { int min = 1 , max = 4 ; int steps [ ] [ 2 ] = { { 1 , 2 } , { 3 , 4 } } ; count_even_odd ( min , max , steps ) ; return 0 ; } |
Maximum number of ones in a N * N matrix with given constraints | C ++ program to get Maximum Number of ones in a matrix with given constraints ; Function that returns the maximum number of ones ; Minimum number of zeroes ; Totol cells = square of the size of the matrices ; Initialising the answer ; Driver code ; Initialising the variables | #include <bits/stdc++.h> NEW_LINE using namespace std ; int getMaxOnes ( int n , int x ) { int zeroes = ( n / x ) ; zeroes = zeroes * zeroes ; int total = n * n ; int ans = total - zeroes ; return ans ; } int main ( ) { int n = 5 ; int x = 2 ; cout << getMaxOnes ( n , x ) ; return 0 ; } |
Minimum operations required to make all the elements distinct in an array | C ++ program to find Minimum number of changes to make array distinct ; Function that returns minimum number of changes ; Hash - table to store frequency ; Increase the frequency of elements ; Traverse in the map to sum up the ( occurrences - 1 ) of duplicate elements ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int minimumOperations ( int a [ ] , int n ) { unordered_map < int , int > mp ; for ( int i = 0 ; i < n ; i ++ ) mp [ a [ i ] ] += 1 ; int count = 0 ; for ( auto it = mp . begin ( ) ; it != mp . end ( ) ; it ++ ) { if ( ( * it ) . second > 1 ) count += ( * it ) . second - 1 ; } return count ; } int main ( ) { int a [ ] = { 2 , 1 , 2 , 3 , 3 , 4 , 3 } ; int n = sizeof ( a ) / sizeof ( a [ 0 ] ) ; cout << minimumOperations ( a , n ) ; return 0 ; } |
Check if a M | C ++ program to check if M - th fibonacci divides N - th fibonacci ; exceptional case for F ( 2 ) ; if none of the above cases , hence not divisible ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; void check ( int n , int m ) { if ( n == 2 m == 2 n % m == 0 ) { cout << " Yes " << endl ; } else { cout << " No " << endl ; } } int main ( ) { int m = 3 , n = 9 ; check ( n , m ) ; return 0 ; } |
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