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Find unit digit of x raised to power y | Efficient C ++ program to find unit digit of x ^ y . ; Returns unit digit of x raised to power y ; Initialize result as 1 to handle case when y is 0. ; One by one multiply with x mod 10 to avoid overflow . ; Driver program | #include <bits/stdc++.h> NEW_LINE using namespace std ; int unitDigitXRaisedY ( int x , int y ) { int res = 1 ; for ( int i = 0 ; i < y ; i ++ ) res = ( res * x ) % 10 ; return res ; } int main ( ) { cout << unitDigitXRaisedY ( 4 , 2 ) ; return 0 ; } |
Evaluation of Risk in Investments | C ++ code for above approach ; First Item in the pair is the value of observation ( xi ) . Second Item in the pair is the frequency of xi ( fi ) ; Vector stores the observation in pairs of format ( xi , fi ) , where xi = value of observation ; Function to calculate the summation of fi * xi ; Function to calculate summation fi ; Function to calculate the mean of the set of observations v ; Function to calculate the std deviation of set of observations v ; Get sum of frequencies ; Get the mean of the set of observations ; Driver Code | #include <iostream> NEW_LINE #include <vector> NEW_LINE #include <algorithm> NEW_LINE #include <cmath> NEW_LINE using namespace std ; typedef pair < float , float > Data ; typedef vector < Data > Vector ; float sigma_fx ( const Vector & v ) { float sum = 0 ; for ( auto i : v ) { sum += i . first * i . second ; } return sum ; } float sigma_f ( const Vector & v ) { float sum = 0.0 ; for ( auto i : v ) { sum += i . second ; } return sum ; } float calculate_mean ( const Vector & v ) { return sigma_fx ( v ) / sigma_f ( v ) ; } float calculate_std ( const Vector & v ) { float f = sigma_f ( v ) ; float mean = sigma_fx ( v ) / f ; float sum = 0 ; for ( auto i : v ) { sum += ( i . first - mean ) * ( i . first - mean ) * i . second ; } return sqrt ( sum / f ) ; } int main ( ) { Vector A = { { 0 , 0.1 } , { 100 , 0.1 } , { 200 , 0.2 } , { 333 , 0.3 } , { 400 , 0.3 } } ; Vector B = { { 100 , 0.1 } , { 200 , 0.5 } , { 700 , 0.4 } } ; float avg_A = calculate_mean ( A ) ; float avg_B = calculate_mean ( B ) ; float std_A = calculate_std ( A ) ; float std_B = calculate_std ( B ) ; cout << " For β Investment β A " << endl ; cout << " Average : β " << avg_A << endl ; cout << " Standard β Deviation : β " << std_A << endl ; cout << " Normalised β Std : β " << std_A / avg_A << endl ; cout << " For β Investment β B " << endl ; cout << " Average : β " << avg_B << endl ; cout << " Standard β Deviation : β " << std_B << endl ; cout << " Normalised β Std : β " << std_B / avg_B << endl ; ( std_B / avg_B ) < ( std_A / avg_A ) ? cout << " Investment β B β is β less β risky STRNEWLINE " : cout << " Investment β A β is β less β risky STRNEWLINE " ; return 0 ; } |
Max occurring divisor in an interval | Efficient C ++ program to find maximum occurring factor in an interval ; function to find max occurring divisor interval [ x , y ] ; if there is only one number in the in the interval , return that number ; otherwise , 2 is the max occurring divisor ; Driver code | #include <iostream> NEW_LINE using namespace std ; int findDivisor ( int x , int y ) { if ( x == y ) return y ; return 2 ; } int main ( ) { int x = 3 , y = 16 ; cout << findDivisor ( x , y ) ; return 0 ; } |
Average of Squares of Natural Numbers | C ++ program to calculate 1 ^ 2 + 2 ^ 2 + 3 ^ 2 + ... average of square number ; Function to calculate average of squares ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; float AvgofSquareN ( int n ) { float sum = 0 ; for ( int i = 1 ; i <= n ; i ++ ) sum += ( i * i ) ; return sum / n ; } int main ( ) { int n = 2 ; cout << AvgofSquareN ( n ) ; return 0 ; } |
Find sum of even factors of a number | Formula based CPP program to find sum of all divisors of n . ; Returns sum of all factors of n . ; If n is odd , then there are no even factors . ; Traversing through all prime factors . ; While i divides n , print i and divide n ; here we remove the 2 ^ 0 that is 1. All other factors ; This condition is to handle the case when n is a prime number . ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int sumofFactors ( int n ) { if ( n % 2 != 0 ) return 0 ; int res = 1 ; for ( int i = 2 ; i <= sqrt ( n ) ; i ++ ) { int count = 0 , curr_sum = 1 , curr_term = 1 ; while ( n % i == 0 ) { count ++ ; n = n / i ; if ( i == 2 && count == 1 ) curr_sum = 0 ; curr_term *= i ; curr_sum += curr_term ; } res *= curr_sum ; } if ( n >= 2 ) res *= ( 1 + n ) ; return res ; } int main ( ) { int n = 18 ; cout << sumofFactors ( n ) ; return 0 ; } |
Find LCM of rational numbers | CPP program to find LCM of given array ; get lcm of two numbers ; Finds LCM of numerators ; calculate the lcm of all numerators ; return all numerator lcm ; Get GCD of all the denominators ; calculate the gcd of all the denominators ; return all denominator gcd ; find lcm of all the rational number ; return the LCM of all numerator / GCD of all denominator ; Driver code ; give rational number 2 / 7 , 3 / 14 , 5 / 3 make pair as a numerator and denominator | #include <bits/stdc++.h> NEW_LINE using namespace std ; int LCM ( int a , int b ) { return ( a * b ) / ( __gcd ( a , b ) ) ; } int lcmOfNumerator ( vector < pair < int , int > > vect ) { int lcm = vect [ 0 ] . first ; for ( int i = 1 ; i < vect . size ( ) ; i ++ ) lcm = LCM ( vect [ i ] . first , lcm ) ; return lcm ; } int gcdOfDemoninators ( vector < pair < int , int > > vect ) { int gcd = vect [ 0 ] . second ; for ( int i = 1 ; i < vect . size ( ) ; i ++ ) gcd = __gcd ( vect [ i ] . second , gcd ) ; return gcd ; } void lcmOfRationals ( vector < pair < int , int > > vect ) { cout << lcmOfNumerator ( vect ) << " / " << gcdOfDemoninators ( vect ) ; } int main ( ) { vector < pair < int , int > > vect ; vect . push_back ( make_pair ( 2 , 7 ) ) ; vect . push_back ( make_pair ( 3 , 14 ) ) ; vect . push_back ( make_pair ( 5 , 3 ) ) ; lcmOfRationals ( vect ) ; return 0 ; } |
Program to determine focal length of a spherical mirror | C ++ program to determine the focal length of a of a spherical mirror ; Determines focal length of a spherical concave mirror ; Determines focal length of a spherical convex mirror ; Driver function | #include <iostream> NEW_LINE using namespace std ; float focal_length_concave ( float R ) { return R / 2 ; } float focal_length_convex ( float R ) { return - ( R / 2 ) ; } int main ( ) { float R = 30 ; cout << " Focal β length β of β spherical " << " concave β mirror β is β : β " << focal_length_concave ( R ) << " β units STRNEWLINE " ; cout << " Focal β length β of β spherical " << " convex β mirror β is β : β " << focal_length_convex ( R ) << " β units " ; return 0 ; } |
Find sum of odd factors of a number | Formula based CPP program to find sum of all divisors of n . ; Returns sum of all factors of n . ; Traversing through all prime factors . ; ignore even factors by removing all powers of 2 ; While i divides n , print i and divide n ; This condition is to handle the case when n is a prime number . ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int sumofoddFactors ( int n ) { int res = 1 ; while ( n % 2 == 0 ) n = n / 2 ; for ( int i = 3 ; i <= sqrt ( n ) ; i ++ ) { int count = 0 , curr_sum = 1 ; int curr_term = 1 ; while ( n % i == 0 ) { count ++ ; n = n / i ; curr_term *= i ; curr_sum += curr_term ; } res *= curr_sum ; } if ( n >= 2 ) res *= ( 1 + n ) ; return res ; } int main ( ) { int n = 30 ; cout << sumofoddFactors ( n ) ; return 0 ; } |
Number of non | CPP program to find the numbers of non negative integral solutions ; return number of non negative integral solutions ; initialize total = 0 ; Base Case if n = 1 and val >= 0 then it should return 1 ; iterate the loop till equal the val ; total solution of equations and again call the recursive function Solutions ( variable , value ) ; return the total no possible solution ; driver code | #include <iostream> NEW_LINE using namespace std ; int countSolutions ( int n , int val ) { int total = 0 ; if ( n == 1 && val >= 0 ) return 1 ; for ( int i = 0 ; i <= val ; i ++ ) { total += countSolutions ( n - 1 , val - i ) ; } return total ; } int main ( ) { int n = 5 ; int val = 20 ; cout << countSolutions ( n , val ) ; } |
Fibonomial coefficient and Fibonomial triangle | CPP Program to print Fibonomial Triangle of height n . ; Function to produce Fibonacci Series . ; 0 th and 1 st number of the series are 0 and 1 ; Add the previous 2 numbers in the series and store it ; Function to produce fibonomial coefficient ; Function to print Fibonomial Triangle . ; Finding the fibonacci series . ; to store triangle value . ; initialising the 0 th element of each row and diagonal element equal to 0. ; for each row . ; for each column . ; finding each element using recurrence relation . ; printing the Fibonomial Triangle . ; Driven Program | #include <bits/stdc++.h> NEW_LINE #define N 6 NEW_LINE using namespace std ; void fib ( int f [ ] , int n ) { int i ; f [ 0 ] = 0 ; f [ 1 ] = 1 ; for ( i = 2 ; i <= n ; i ++ ) f [ i ] = f [ i - 1 ] + f [ i - 2 ] ; } void fibcoef ( int fc [ ] [ N + 1 ] , int f [ ] , int n ) { for ( int i = 0 ; i <= n ; i ++ ) fc [ i ] [ 0 ] = 1 ; for ( int i = 1 ; i <= n ; i ++ ) { for ( int j = 1 ; j <= i ; j ++ ) { int k = j ; while ( k -- ) fc [ i ] [ j ] *= f [ k ] ; k = 1 ; while ( ( j + 1 ) != k ) fc [ i ] [ j ] /= f [ k ++ ] ; } } } void printFibonomialTriangle ( int n ) { int f [ N + 1 ] = { 0 } ; fib ( f , n ) ; int dp [ N + 1 ] [ N + 1 ] = { 0 } ; for ( int i = 0 ; i <= n ; i ++ ) dp [ i ] [ 0 ] = dp [ i ] [ i ] = 1 ; for ( int i = 1 ; i <= n ; i ++ ) { for ( int j = 1 ; j < i ; j ++ ) dp [ i ] [ j ] = f [ i - j + 1 ] * dp [ i - 1 ] [ j - 1 ] + f [ j - 1 ] * dp [ i - 1 ] [ j ] ; } for ( int i = 0 ; i <= n ; i ++ ) { for ( int j = 0 ; j <= i ; j ++ ) cout << dp [ i ] [ j ] << " β " ; cout << endl ; } } int main ( ) { int n = 6 ; printFibonomialTriangle ( n ) ; return 0 ; } |
Sum of Arithmetic Geometric Sequence | CPP Program to find the sum of first n terms . ; Return the sum of first n term of AGP ; finding the each term of AGP and adding it to sum . ; Driven Program | #include <bits/stdc++.h> NEW_LINE using namespace std ; int sumofNterm ( int a , int d , int b , int r , int n ) { int sum = 0 ; for ( int i = 1 ; i <= n ; i ++ ) sum += ( ( a + ( i - 1 ) * d ) * ( b * pow ( r , i - 1 ) ) ) ; return sum ; } int main ( ) { int a = 1 , d = 1 , b = 2 , r = 2 , n = 3 ; cout << sumofNterm ( a , d , b , r , n ) << endl ; return 0 ; } |
Sum of the series 2 + ( 2 + 4 ) + ( 2 + 4 + 6 ) + ( 2 + 4 + 6 + 8 ) + β¦β¦ + ( 2 + 4 + 6 + 8 + β¦ . + 2 n ) | C ++ implementation to find the sum of the given series ; function to find the sum of the given series ; first term of each i - th term ; next term ; required sum ; Driver program to test above | #include <bits/stdc++.h> NEW_LINE using namespace std ; int sumOfTheSeries ( int n ) { int sum = 0 ; for ( int i = 1 ; i <= n ; i ++ ) { int k = 2 ; for ( int j = 1 ; j <= i ; j ++ ) { sum += k ; k += 2 ; } } return sum ; } int main ( ) { int n = 5 ; cout << " Sum β = β " << sumOfTheSeries ( n ) ; return 0 ; } |
Program to find sum of series 1 * 2 * 3 + 2 * 3 * 4 + 3 * 4 * 5 + . . . + n * ( n + 1 ) * ( n + 2 ) | Program to find the sum of series 1 * 2 * 3 + 2 * 3 * 4 + . . . + n * ( n + 1 ) * ( n + 1 ) ; Function to calculate sum of series . ; Driver function | #include <bits/stdc++.h> NEW_LINE using namespace std ; int sumOfSeries ( int n ) { int sum = 0 ; for ( int i = 1 ; i <= n ; i ++ ) sum = sum + i * ( i + 1 ) * ( i + 2 ) ; return sum ; } int main ( ) { int n = 10 ; cout << sumOfSeries ( n ) ; return 0 ; } |
Program to get the Sum of series : 1 | C ++ program to get the sum of the series ; Function to get the series ; Computing sum of remaining n - 1 terms . ; Driver Code | #include <math.h> NEW_LINE #include <stdio.h> NEW_LINE double Series ( double x , int n ) { double sum = 1 , term = 1 , fct = 1 , p = 1 , multi = 1 ; for ( int i = 1 ; i < n ; i ++ ) { fct = fct * multi * ( multi + 1 ) ; p = p * x * x ; term = ( -1 ) * term ; multi += 2 ; sum = sum + ( term * p ) / fct ; } return sum ; } int main ( ) { double x = 9 ; int n = 10 ; printf ( " % .4f " , Series ( x , n ) ) ; return 0 ; } |
Find the number of consecutive zero at the end after multiplying n numbers | CPP program to find the number of consecutive zero at the end after multiplying n numbers ; Function to count two 's factor ; Count number of 2 s present in n ; Function to count five 's factor ; Function to count number of zeros ; Count the two 's factor of n number ; Count the five 's factor of n number ; Return the minimum ; Driver Code | #include <iostream> NEW_LINE using namespace std ; int two_factor ( int n ) { int twocount = 0 ; while ( n % 2 == 0 ) { twocount ++ ; n = n / 2 ; } return twocount ; } int five_factor ( int n ) { int fivecount = 0 ; while ( n % 5 == 0 ) { fivecount ++ ; n = n / 5 ; } return fivecount ; } int find_con_zero ( int arr [ ] , int n ) { int twocount = 0 ; int fivecount = 0 ; for ( int i = 0 ; i < n ; i ++ ) { twocount += two_factor ( arr [ i ] ) ; fivecount += five_factor ( arr [ i ] ) ; } if ( twocount < fivecount ) return twocount ; else return fivecount ; } int main ( ) { int arr [ ] = { 100 , 10 , 5 , 25 , 35 , 14 } ; int n = 6 ; cout << find_con_zero ( arr , n ) ; } |
First occurrence of a digit in a given fraction | CPP program to find first occurrence of c in a / b ; function to print the first digit ; reduce the number to its mod ; traverse for every decimal places ; get every fraction places when ( a * 10 / b ) / c ; check if it is equal to the required integer ; mod the number ; driver program to test the above function | #include <bits/stdc++.h> NEW_LINE using namespace std ; int first ( int a , int b , int c ) { a %= b ; for ( int i = 1 ; i <= b ; i ++ ) { a = a * 10 ; if ( a / b == c ) return i ; a %= b ; } return -1 ; } int main ( ) { int a = 1 , b = 4 , c = 5 ; cout << first ( a , b , c ) ; return 0 ; } |
Minimize the absolute difference of sum of two subsets | CPP program to Minimize the absolute difference of sum of two subsets ; function to print difference ; summation of n elements ; if divisible by 4 ; if remainder 1 or 2. In case of remainder 2 , we divide elements from 3 to n in groups of size 4 and put 1 in one group and 2 in group . This also makes difference 1. ; We put elements from 4 to n in groups of size 4. Remaining elements 1 , 2 and 3 can be divided as ( 1 , 2 ) and ( 3 ) . ; driver program to test the above function | #include <bits/stdc++.h> NEW_LINE using namespace std ; void subsetDifference ( int n ) { int s = n * ( n + 1 ) / 2 ; if ( n % 4 == 0 ) { cout << " First β subset β sum β = β " << s / 2 ; cout << " Second subset sum = " << s / 2 ; cout << " Difference = " } else { if ( n % 4 == 1 n % 4 == 2 ) { cout << " First β subset β sum β = β " << s / 2 ; cout << " Second subset sum = " << s / 2 + 1 ; cout << " Difference = " } else { cout << " First β subset β sum β = β " << s / 2 ; cout << " Second subset sum = " << s / 2 ; cout << " Difference = " } } } int main ( ) { int n = 6 ; subsetDifference ( n ) ; return 0 ; } |
Time required to meet in equilateral triangle | CPP code to find time taken by animals to meet ; function to calculate time to meet ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; void timeToMeet ( double s , double v ) { double V = 3 * v / 2 ; double time = s / V ; cout << time ; } int main ( void ) { double s = 25 , v = 56 ; timeToMeet ( s , v ) ; return 0 ; } |
Sum of the series 1 + ( 1 + 3 ) + ( 1 + 3 + 5 ) + ( 1 + 3 + 5 + 7 ) + Γ’ β¬Β¦ Γ’ β¬Β¦ + ( 1 + 3 + 5 + 7 + Γ’ β¬Β¦ + ( 2 n | C ++ implementation to find the sum of the given series ; functionn to find the sum of the given series ; first term of each i - th term ; next term ; required sum ; Driver program | #include <bits/stdc++.h> NEW_LINE using namespace std ; int sumOfTheSeries ( int n ) { int sum = 0 ; for ( int i = 1 ; i <= n ; i ++ ) { int k = 1 ; for ( int j = 1 ; j <= i ; j ++ ) { sum += k ; k += 2 ; } } return sum ; } int main ( ) { int n = 5 ; cout << " Sum β = β " << sumOfTheSeries ( n ) ; return 0 ; } |
Check if a number can be written as sum of three consecutive integers | CPP Program to check if a number can be written as sum of three consecutive integers . ; function to check if a number can be written as sum of three consecutive integer . ; if n is 0 ; if n is positive , increment loop by 1. ; if n is negative , decrement loop by 1. ; Running loop from 0 to n - 2 ; check if sum of three consecutive integer is equal to n . ; Driver Program | #include <bits/stdc++.h> NEW_LINE using namespace std ; void checksum ( int n ) { if ( n == 0 ) { cout << " - 1 β 0 β 1" << endl ; return ; } int inc ; if ( n > 0 ) inc = 1 ; else inc = -1 ; for ( int i = 0 ; i <= n - 2 ; i += inc ) { if ( i + i + 1 + i + 2 == n ) { cout << i << " β " << i + 1 << " β " << i + 2 ; return ; } } cout << " - 1" ; } int main ( ) { int n = 6 ; checksum ( n ) ; return 0 ; } |
Sum of all divisors from 1 to n | C ++ program to find sum of all divisor of number up to ' n ' ; Utility function to find sum of all divisor of number up to ' n ' ; Find all divisors of i and add them ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int divisorSum ( int n ) { int sum = 0 ; for ( int i = 1 ; i <= n ; ++ i ) { for ( int j = 1 ; j * j <= i ; ++ j ) { if ( i % j == 0 ) { if ( i / j == j ) sum += j ; else sum += j + i / j ; } } } return sum ; } int main ( ) { int n = 4 ; cout << " β " << divisorSum ( n ) << endl ; n = 5 ; cout << " β " << divisorSum ( n ) ; return 0 ; } |
Program for Binomial Coefficients table | C ++ program for binomial coefficients ; Function to print binomial table ; B ( m , x ) is 1 if either m or x is is 0. ; Otherwise using recursive formula B ( m , x ) = B ( m , x - 1 ) * ( m - x + 1 ) / x ; Driver Function | #include <stdio.h> NEW_LINE int printbinomial ( int max ) { for ( int m = 0 ; m <= max ; m ++ ) { printf ( " % 2d " , m ) ; int binom = 1 ; for ( int x = 0 ; x <= m ; x ++ ) { if ( m != 0 && x != 0 ) binom = binom * ( m - x + 1 ) / x ; printf ( " % 4d " , binom ) ; } printf ( " STRNEWLINE " ) ; } } int main ( ) { int max = 10 ; printbinomial ( max ) ; return 0 ; } |
Find largest prime factor of a number | C ++ Program to find largest prime factor of number ; A function to find largest prime factor ; Initialize the maximum prime factor variable with the lowest one ; Print the number of 2 s that divide n ; n >>= 1 ; equivalent to n /= 2 ; n must be odd at this point ; now we have to iterate only for integers who does not have prime factor 2 and 3 ; This condition is to handle the case when n is a prime number greater than 4 ; Driver program to test above function | #include <iostream> NEW_LINE #include <bits/stdc++.h> NEW_LINE using namespace std ; long long maxPrimeFactors ( long long n ) { long long maxPrime = -1 ; while ( n % 2 == 0 ) { maxPrime = 2 ; } while ( n % 3 == 0 ) { maxPrime = 3 ; n = n / 3 ; } for ( int i = 5 ; i <= sqrt ( n ) ; i += 6 ) { while ( n % i == 0 ) { maxPrime = i ; n = n / i ; } while ( n % ( i + 2 ) == 0 ) { maxPrime = i + 2 ; n = n / ( i + 2 ) ; } } if ( n > 4 ) maxPrime = n ; return maxPrime ; } int main ( ) { long long n = 15 ; cout << maxPrimeFactors ( n ) << endl ; n = 25698751364526 ; cout << maxPrimeFactors ( n ) ; } |
Count unset bits in a range | C ++ implementation to count unset bits in the given range ; Function to get no of set bits in the binary representation of ' n ' ; function to count unset bits in the given range ; calculating a number ' num ' having ' r ' number of bits and bits in the range l to r are the only set bits ; returns number of unset bits in the range ' l ' to ' r ' in ' n ' ; Driver program to test above | #include <bits/stdc++.h> NEW_LINE using namespace std ; unsigned int countSetBits ( int n ) { unsigned int count = 0 ; while ( n ) { n &= ( n - 1 ) ; count ++ ; } return count ; } unsigned int countUnsetBitsInGivenRange ( unsigned int n , unsigned int l , unsigned int r ) { int num = ( ( 1 << r ) - 1 ) ^ ( ( 1 << ( l - 1 ) ) - 1 ) ; return ( r - l + 1 ) - countSetBits ( n & num ) ; } int main ( ) { unsigned int n = 80 ; unsigned int l = 1 , r = 4 ; cout << countUnsetBitsInGivenRange ( n , l , r ) ; return 0 ; } |
Midy 's theorem | C ++ implementation as a proof of the Midy 's theorem ; Returns repeating sequence of a fraction . If repeating sequence doesn 't exits, then returns -1 ; Create a map to store already seen remainders remainder is used as key and its position in result is stored as value . Note that we need position for cases like 1 / 6. In this case , the recurring sequence doesn 't start from first remainder. ; Find first remainder ; Keep finding remainder until either remainder becomes 0 or repeats ; Store this remainder ; Multiply remainder with 10 ; Append rem / denr to result ; Update remainder ; Checks whether a number is prime or not ; If all conditions are met , it proves Midy 's theorem ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; string fractionToDecimal ( int numerator , int denominator ) { string res ; map < int , int > mp ; mp . clear ( ) ; int rem = numerator % denominator ; while ( ( rem != 0 ) && ( mp . find ( rem ) == mp . end ( ) ) ) { mp [ rem ] = res . length ( ) ; rem = rem * 10 ; int res_part = rem / denominator ; res += to_string ( res_part ) ; rem = rem % denominator ; } return ( rem == 0 ) ? " - 1" : res . substr ( mp [ rem ] ) ; } bool isPrime ( int n ) { for ( int i = 2 ; i <= n / 2 ; i ++ ) if ( n % i == 0 ) return false ; return true ; } void Midys ( string str , int n ) { int l = str . length ( ) ; int part1 = 0 , part2 = 0 ; if ( ! isPrime ( n ) ) { cout << " Denominator β is β not β prime , β " << " thus β Midy ' s β theorem β is β not β applicable " ; } else if ( l % 2 == 0 ) { for ( int i = 0 ; i < l / 2 ; i ++ ) { part1 = part1 * 10 + ( str [ i ] - '0' ) ; part2 = part2 * 10 + ( str [ l / 2 + i ] - '0' ) ; } cout << part1 << " β + β " << part2 << " β = β " << ( part1 + part2 ) << endl ; cout << " Midy ' s β theorem β holds ! " ; } else { cout << " The β repeating β decimal β is β of β odd β length β " << " thus β Midy ' s β theorem β is β not β applicable " ; } } int main ( ) { int numr = 2 , denr = 11 ; string res = fractionToDecimal ( numr , denr ) ; if ( res == " - 1" ) cout << " The β fraction β does β not β have β repeating β decimal " ; else { cout << " Repeating β decimal β = β " << res << endl ; Midys ( res , denr ) ; } return 0 ; } |
Sum of fourth power of first n even natural numbers | CPP Program to find the sum of fourth powers of first n even natural numbers ; calculate the sum of fourth power of first n even natural numbers ; made even number ; Driven Program | #include <bits/stdc++.h> NEW_LINE using namespace std ; long long int evenPowerSum ( int n ) { long long int sum = 0 ; for ( int i = 1 ; i <= n ; i ++ ) { int j = 2 * i ; sum = sum + ( j * j * j * j ) ; } return sum ; } int main ( ) { int n = 5 ; cout << evenPowerSum ( n ) << endl ; return 0 ; } |
Sum of fourth power of first n even natural numbers | CPP Program to find the sum of fourth powers of first n even natural numbers ; calculate the sum of fourth power of first n even natural numbers ; Driven Program | #include <bits/stdc++.h> NEW_LINE using namespace std ; long long int evenPowerSum ( int n ) { return ( 8 * n * ( n + 1 ) * ( 2 * n + 1 ) * ( 3 * n * n + 3 * n - 1 ) ) / 15 ; } int main ( ) { int n = 4 ; cout << evenPowerSum ( n ) << endl ; return 0 ; } |
Balanced Prime | CPP Program to find Nth Balanced Prime ; Return the Nth balanced prime . ; Sieve of Eratosthenes ; If prime [ p ] is not changed , then it is a prime ; Update all multiples of p ; storing all primes ; Finding the Nth balanced Prime ; Driven Program | #include <bits/stdc++.h> NEW_LINE #define MAX 501 NEW_LINE using namespace std ; int balancedprime ( int n ) { bool prime [ MAX + 1 ] ; memset ( prime , true , sizeof ( prime ) ) ; for ( int p = 2 ; p * p <= MAX ; p ++ ) { if ( prime [ p ] == true ) { for ( int i = p * 2 ; i <= MAX ; i += p ) prime [ i ] = false ; } } vector < int > v ; for ( int p = 3 ; p <= MAX ; p += 2 ) if ( prime [ p ] ) v . push_back ( p ) ; int count = 0 ; for ( int i = 1 ; i < v . size ( ) ; i ++ ) { if ( v [ i ] == ( v [ i + 1 ] + v [ i - 1 ] ) / 2 ) count ++ ; if ( count == n ) return v [ i ] ; } } int main ( ) { int n = 4 ; cout << balancedprime ( n ) << endl ; return 0 ; } |
Smallest integer which has n factors or more | c ++ program to print the smallest integer with n factors or more ; array to store prime factors ; function to generate all prime factors of numbers from 1 to 10 ^ 6 ; Initializes all the positions with their value . ; Initializes all multiples of 2 with 2 ; A modified version of Sieve of Eratosthenes to store the smallest prime factor that divides every number . ; check if it has no prime factor . ; Initializes of j starting from i * i ; if it has no prime factor before , then stores the smallest prime divisor ; function to calculate number of factors ; stores the smallest prime number that divides n ; stores the count of number of times a prime number divides n . ; reduces to the next number after prime factorization of n ; false when prime factorization is done ; if the same prime number is dividing n , then we increase the count ; if its a new prime factor that is factorizing n , then we again set c = 1 and change dup to the new prime factor , and apply the formula explained above . ; prime factorizes a number ; for the last prime factor ; function to find the smallest integer with n factors or more . ; check if no of factors is more than n or not ; Driver program to test above function ; generate prime factors of number upto 10 ^ 6 | #include <bits/stdc++.h> NEW_LINE using namespace std ; const int MAX = 1000001 ; int factor [ MAX ] = { 0 } ; void generatePrimeFactors ( ) { factor [ 1 ] = 1 ; for ( int i = 2 ; i < MAX ; i ++ ) factor [ i ] = i ; for ( int i = 4 ; i < MAX ; i += 2 ) factor [ i ] = 2 ; for ( int i = 3 ; i * i < MAX ; i ++ ) { if ( factor [ i ] == i ) { for ( int j = i * i ; j < MAX ; j += i ) { if ( factor [ j ] == j ) factor [ j ] = i ; } } } } int calculateNoOFactors ( int n ) { if ( n == 1 ) return 1 ; int ans = 1 ; int dup = factor [ n ] ; int c = 1 ; int j = n / factor [ n ] ; while ( j != 1 ) { if ( factor [ j ] == dup ) c += 1 ; else { dup = factor [ j ] ; ans = ans * ( c + 1 ) ; c = 1 ; } j = j / factor [ j ] ; } ans = ans * ( c + 1 ) ; return ans ; } int smallest ( int n ) { for ( int i = 1 ; ; i ++ ) if ( calculateNoOFactors ( i ) >= n ) return i ; } int main ( ) { generatePrimeFactors ( ) ; int n = 4 ; cout << smallest ( n ) ; return 0 ; } |
Sum of squares of first n natural numbers | CPP Program to find sum of square of first n natural numbers ; Return the sum of square of first n natural numbers ; Iterate i from 1 and n finding square of i and add to sum . ; Driven Program | #include <bits/stdc++.h> NEW_LINE using namespace std ; int squaresum ( int n ) { int sum = 0 ; for ( int i = 1 ; i <= n ; i ++ ) sum += ( i * i ) ; return sum ; } int main ( ) { int n = 4 ; cout << squaresum ( n ) << endl ; return 0 ; } |
Break a number such that sum of maximum divisors of all parts is minimum | CPP program to break a number such that sum of maximum divisors of all parts is minimum ; Function to check if a number is prime or not . ; If n is an even number ( we can write it as sum of two primes ) ; If n is odd and n - 2 is prime . ; If n is odd , n - 3 must be even . ; Driver code | #include <iostream> NEW_LINE using namespace std ; bool isPrime ( int n ) { int i = 2 ; while ( i * i <= n ) { if ( n % i == 0 ) return false ; i ++ ; } return true ; } int minimumSum ( int n ) { if ( isPrime ( n ) ) return 1 ; if ( n % 2 == 0 ) return 2 ; if ( isPrime ( n - 2 ) ) return 2 ; return 3 ; } int main ( ) { int n = 27 ; cout << minimumSum ( n ) ; return 0 ; } |
Find first and last digits of a number | Program to find first and last digits of a number ; Find the first digit ; Remove last digit from number till only one digit is left ; return the first digit ; Find the last digit ; return the last digit ; Driver program | #include <bits/stdc++.h> NEW_LINE using namespace std ; int firstDigit ( int n ) { while ( n >= 10 ) n /= 10 ; return n ; } int lastDigit ( int n ) { return ( n % 10 ) ; } int main ( ) { int n = 98562 ; cout << firstDigit ( n ) << " β " << lastDigit ( n ) << endl ; return 0 ; } |
Find first and last digits of a number | Program to find first and last digits of a number ; Find the first digit ; Find total number of digits - 1 ; Find first digit ; Return first digit ; Find the last digit ; return the last digit ; Driver program | #include <bits/stdc++.h> NEW_LINE using namespace std ; int firstDigit ( int n ) { int digits = ( int ) log10 ( n ) ; n = ( int ) ( n / pow ( 10 , digits ) ) ; return n ; } int lastDigit ( int n ) { return ( n % 10 ) ; } int main ( ) { int n = 98562 ; cout << firstDigit ( n ) << " β " << lastDigit ( n ) << endl ; return 0 ; } |
Express an odd number as sum of prime numbers | CPP program to express N as sum of at - most three prime numbers . ; Function to check if a number is prime or not . ; Prints at most three prime numbers whose sum is n . ; if ( isPrime ( n ) ) CASE - I ; else if ( isPrime ( n - 2 ) ) CASE - II ; else CASE - III ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool isPrime ( int x ) { if ( x == 0 x == 1 ) return false ; for ( int i = 2 ; i * i <= x ; ++ i ) if ( x % i == 0 ) return false ; return true ; } void findPrimes ( int n ) { cout << n << endl ; cout << 2 << " β " << n - 2 << endl ; { cout << 3 << " β " ; n = n - 3 ; for ( int i = 0 ; i < n ; i ++ ) { if ( isPrime ( i ) && isPrime ( n - i ) ) { cout << i << " β " << ( n - i ) ; break ; } } } } int main ( ) { int n = 27 ; findPrimes ( n ) ; return 0 ; } |
AKS Primality Test | C ++ code to check if number is prime . This program demonstrates concept behind AKS algorithm and doesn 't implement the actual algorithm (This works only till n = 64) ; array used to store coefficients . ; function to calculate the coefficients of ( x - 1 ) ^ n - ( x ^ n - 1 ) with the help of Pascal 's triangle . ; function to check whether the number is prime or not ; Calculating all the coefficients by the function coef and storing all the coefficients in c array . ; subtracting c [ n ] and adding c [ 0 ] by 1 as ( x - 1 ) ^ n - ( x ^ n - 1 ) , here we are subtracting c [ n ] by 1 and adding 1 in expression . ; checking all the coefficients whether they are divisible by n or not . if n is not prime , then loop breaks and ( i > 0 ) . ; Return true if all coefficients are divisible by n . ; driver program | #include <bits/stdc++.h> NEW_LINE using namespace std ; long long c [ 100 ] ; void coef ( int n ) { c [ 0 ] = 1 ; for ( int i = 0 ; i < n ; c [ 0 ] = - c [ 0 ] , i ++ ) { c [ 1 + i ] = 1 ; for ( int j = i ; j > 0 ; j -- ) c [ j ] = c [ j - 1 ] - c [ j ] ; } } bool isPrime ( int n ) { coef ( n ) ; c [ 0 ] ++ , c [ n ] -- ; int i = n ; while ( i -- && c [ i ] % n == 0 ) ; return i < 0 ; } int main ( ) { int n = 37 ; if ( isPrime ( n ) ) cout << " Prime " << endl ; else cout << " Not β Prime " << endl ; return 0 ; } |
Motzkin number | CPP Program to find Nth Motzkin Number . ; Return the nth Motzkin Number . ; Base Case ; Recursive step ; Driven Program | #include <bits/stdc++.h> NEW_LINE using namespace std ; int motzkin ( int n ) { if ( n == 0 n == 1 ) return 1 ; return ( ( 2 * n + 1 ) * motzkin ( n - 1 ) + ( 3 * n - 3 ) * motzkin ( n - 2 ) ) / ( n + 2 ) ; } int main ( ) { int n = 8 ; cout << motzkin ( n ) << endl ; return 0 ; } |
Sum of the series 0.6 , 0.06 , 0.006 , 0.0006 , ... to n terms | CPP program to find sum of 0.6 , 0.06 , 0.006 , 0.0006 , ... to n terms ; function which return the the sum of series ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; float sumOfSeries ( int n ) { return ( 0.666 ) * ( 1 - 1 / pow ( 10 , n ) ) ; } int main ( ) { int n = 2 ; cout << sumOfSeries ( n ) ; } |
Narcissistic number | CPP program for checking of Narcissistic number ; function to count digits ; Returns true if n is Narcissistic number ; count the number of digits ; calculates the sum of digits raised to power ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int countDigit ( int n ) { if ( n == 0 ) return 0 ; return 1 + countDigit ( n / 10 ) ; } bool check ( int n ) { int l = countDigit ( n ) ; int dup = n ; int sum = 0 ; while ( dup ) { sum += pow ( dup % 10 , l ) ; dup /= 10 ; } return ( n == sum ) ; } int main ( ) { int n = 1634 ; if ( check ( n ) ) cout << " yes " ; else cout << " no " ; return 0 ; } |
Sum of squares of first n natural numbers | CPP program to calculate 1 ^ 2 + 2 ^ 2 + 3 ^ 2 + ... ; Function to calculate sum ; Driver code | #include <iostream> NEW_LINE using namespace std ; int summation ( int n ) { int sum = 0 ; for ( int i = 1 ; i <= n ; i ++ ) sum += ( i * i ) ; return sum ; } int main ( ) { int n = 2 ; cout << summation ( n ) ; return 0 ; } |
Leyland Number | CPP program to print first N Leyland Numbers . ; Print first n Leyland Number . ; Outer loop for x from 2 to n . ; Inner loop for y from 2 to x . ; Calculating x ^ y + y ^ x ; Sorting the all Leyland Number . ; Printing first n Leyland number . ; Driven Program | #include <bits/stdc++.h> NEW_LINE #define MAX 100 NEW_LINE using namespace std ; void leyland ( int n ) { vector < int > ans ; for ( int x = 2 ; x <= n ; x ++ ) { for ( int y = 2 ; y <= x ; y ++ ) { int temp = pow ( x , y ) + pow ( y , x ) ; ans . push_back ( temp ) ; } } sort ( ans . begin ( ) , ans . end ( ) ) ; for ( int i = 0 ; i < n ; i ++ ) cout << ans [ i ] << " β " ; } int main ( ) { int n = 6 ; leyland ( n ) ; return 0 ; } |
NicomachusΓ’ β¬β’ s Theorem ( Sum of k | CPP program to find sum of k - th group of positive odd integers . ; Return the sum of k - th group of positive odd integers . ; Finding first element of kth group . ; Finding the sum . ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int kthgroupsum ( int k ) { int cur = ( k * ( k - 1 ) ) + 1 ; int sum = 0 ; while ( k -- ) { sum += cur ; cur += 2 ; } return sum ; } int main ( ) { int k = 3 ; cout << kthgroupsum ( k ) << endl ; return 0 ; } |
n | CPP program to find n - th term of the series 2 , 12 , 36 , 80 , 150 , . . ; Returns n - th term of the series 2 , 12 , 36 , 80 , 150 ; driver code | #include <iostream> NEW_LINE using namespace std ; int nthTerm ( int n ) { return ( n * n ) + ( n * n * n ) ; } int main ( ) { int n = 4 ; cout << nthTerm ( n ) ; return 0 ; } |
Sum of the series 1 , 3 , 6 , 10. . . ( Triangular Numbers ) | CPP program to find sum series 1 , 3 , 6 , 10 , 15 , 21. . . and then find its sum ; Function to find the sum of series ; Driver code | #include <iostream> NEW_LINE using namespace std ; int seriesSum ( int n ) { return ( n * ( n + 1 ) * ( n + 2 ) ) / 6 ; } int main ( ) { int n = 4 ; cout << seriesSum ( n ) ; return 0 ; } |
Calculate speed , distance and time | C ++ Program to calculate speed distance and time ; Function to calculate speed ; Function to calculate distance traveled ; Function to calculate time taken ; Driver function ; Calling function cal_speed ( ) ; Calling function cal_dis ( ) ; Calling function cal_time ( ) | #include <iostream> NEW_LINE using namespace std ; double cal_speed ( double dist , double time ) { cout << " Distance ( km ) : " cout << " Time ( hr ) : " return dist / time ; } double cal_dis ( double speed , double time ) { cout << " Time ( hr ) : " cout << " Speed ( km / hr ) : " return speed * time ; } double cal_time ( double dist , double speed ) { cout << " Distance ( km ) : " cout << " Speed ( km / hr ) : " return speed * dist ; } int main ( ) { cout << " The calculated Speed ( km / hr ) is : " << cal_speed ( 45.9 , 2.0 ) << endl ; cout << " The calculated Distance ( km ) : " << cal_dis ( 62.9 , 2.5 ) << endl ; cout << " The calculated Time ( hr ) : " << cal_time ( 48.0 , 4.5 ) << endl ; return 0 ; } |
Print factorials of a range in right aligned format | CPP Program to print format of factorial ; vector for store the result ; variable for store the each number factorial ; copy of first number ; found first number factorial ; push the first number in result vector ; incerement the first number ; found the all reaming number factorial loop is working until all required number factorial founded ; store the result of factorial ; incerement the first number ; return the result ; function for print the result ; setw ( ) is used for fill the blank right is used for right justification of data ; Driver function ; number which found the factorial of between range ; store the result of factorial ; function for found factorial ; function for print format | #include <boost/multiprecision/cpp_int.hpp> NEW_LINE #include <iostream> NEW_LINE #include <vector> NEW_LINE using namespace std ; using boost :: multiprecision :: cpp_int ; vector < cpp_int > find_factorial ( int num1 , int num2 ) { vector < cpp_int > vec ; cpp_int fac = 1 ; int temp = num1 ; while ( 1 ) { if ( temp == 1 ) break ; fac *= temp ; temp -- ; } vec . push_back ( fac ) ; num1 ++ ; while ( num1 <= num2 ) { fac *= num1 ; vec . push_back ( fac ) ; num1 ++ ; } return ( vec ) ; } void print_format ( vector < cpp_int > & result ) { int digits = result . back ( ) . str ( ) . size ( ) ; for ( int i = 0 ; i < result . size ( ) ; i ++ ) { cout << setw ( digits + 1 ) << right << result [ i ] << endl ; } } int main ( ) { int m = 10 , n = 20 ; vector < cpp_int > result_fac ; result_fac = find_factorial ( m , n ) ; print_format ( result_fac ) ; return 0 ; } |
Find n | CPP program to find n - th term of series 1 , 3 , 6 , 10 , 15 , 21. . . ; Function to find the nth term of series ; Loop to add numbers ; Driver code | #include <iostream> NEW_LINE using namespace std ; int term ( int n ) { int ans = 0 ; for ( int i = 1 ; i <= n ; i ++ ) ans += i ; return ans ; } int main ( ) { int n = 4 ; cout << term ( n ) ; return 0 ; } |
Find the average of first N natural numbers | CPP Program to find the Average of first n natural numbers ; Return the average of first n natural numbers ; Driven Program | #include <bits/stdc++.h> NEW_LINE using namespace std ; float avgOfFirstN ( int n ) { return ( float ) ( 1 + n ) / 2 ; } int main ( ) { int n = 20 ; cout << avgOfFirstN ( n ) << endl ; return 0 ; } |
Find the sum of the series 1 + 11 + 111 + 1111 + ... . . upto n terms | C ++ program to find the sum of the series 1 + 11 + 111 + 1111 + ... . ; Function for finding summation ; Driver Code | #include <bits/stdc++.h> NEW_LINE int summation ( int n ) { int sum ; sum = ( pow ( 10 , n + 1 ) - 10 - ( 9 * n ) ) / 81 ; return sum ; } int main ( ) { int n = 5 ; printf ( " % d " , summation ( n ) ) ; return 0 ; } |
Sum of the Series 1 + x / 1 + x ^ 2 / 2 + x ^ 3 / 3 + . . + x ^ n / n | C ++ program to find sum of series 1 + x ^ 2 / 2 + x ^ 3 / 3 + ... . + x ^ n / n ; C ++ code to print the sum of the series ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; double sum ( int x , int n ) { double i , total = 1.0 , multi = x ; for ( i = 1 ; i <= n ; i ++ ) { total = total + multi / i ; multi = multi * x ; } return total ; } int main ( ) { int x = 2 ; int n = 5 ; cout << fixed << setprecision ( 2 ) << sum ( x , n ) ; return 0 ; } |
Find n | CPP program to find the nth term of the series 1 2 2 3 3 3 ... ; function to solve the quadratic equation ; calculating the Nth term ; driver code to check the above function | #include <bits/stdc++.h> NEW_LINE using namespace std ; int term ( int n ) { int x = ( ( ( 1 ) + ( double ) sqrt ( 1 + ( 8 * n ) ) ) / 2 ) ; return x ; } int main ( ) { int n = 5 ; cout << term ( n ) ; return 0 ; } |
Deserium Number | C ++ program to check whether a number is Deserium number or not ; Returns count of digits in n . ; Returns true if x is Diserium ; Compute powers of digits from right to left . ; If sum of powers is same as given number . ; Driver code | #include <iostream> NEW_LINE #include <math.h> NEW_LINE using namespace std ; int countDigits ( int n ) { int c = 0 ; do { c ++ ; n = n / 10 ; } while ( n != 0 ) ; return c ; } bool isDeserium ( int x ) { int temp = x ; int p = countDigits ( x ) ; int sum = 0 ; while ( x != 0 ) { int digit = x % 10 ; sum += pow ( digit , p ) ; p -- ; x = x / 10 ; } return ( sum == temp ) ; } int main ( ) { int x = 135 ; if ( isDeserium ( x ) ) cout << " Yes " ; else cout << " No " ; return 0 ; } |
Largest number by which given 3 numbers should be divided such that they leaves same remainder | C ++ program to find the largest numbers that leads to same remainder when divides given three sorted numbers ; __gcd function ; function return number which divides these three number and leaves same remainder . ; We find the differences of all three pairs ; Return GCD of three differences . ; driver program | #include <bits/stdc++.h> NEW_LINE using namespace std ; int gcd ( int a , int b ) { if ( a == 0 ) return b ; return gcd ( b % a , a ) ; } int sameRemainder ( int a , int b , int c ) { int a1 = ( b - a ) , b1 = ( c - b ) , c1 = ( c - a ) ; return gcd ( a1 , gcd ( b1 , c1 ) ) ; } int main ( ) { int a = 62 , b = 132 , c = 237 ; cout << sameRemainder ( a , b , c ) << endl ; return 0 ; } |
Find combined mean and variance of two series | C ++ program to find combined mean and variance of two series . ; Function to find mean of series . ; Function to find the standard deviation of series . ; Function to find combined variance of two different series . ; mean1 and mean2 are the mean of two arrays . ; sd1 and sd2 are the standard deviation of two array . ; combinedMean is variable to store the combined mean of both array . ; d1_square and d2_square are the combined mean deviation . ; combinedVar is variable to store combined variance of both array . ; Driver function . ; Function call to combined mean . | #include <bits/stdc++.h> NEW_LINE using namespace std ; float mean ( int arr [ ] , int n ) { int sum = 0 ; for ( int i = 0 ; i < n ; i ++ ) sum = sum + arr [ i ] ; float mean = ( float ) sum / n ; return mean ; } float sd ( int arr [ ] , int n ) { float sum = 0 ; for ( int i = 0 ; i < n ; i ++ ) sum = sum + ( arr [ i ] - mean ( arr , n ) ) * ( arr [ i ] - mean ( arr , n ) ) ; float sdd = sum / n ; return sdd ; } float combinedVariance ( int arr1 [ ] , int arr2 [ ] , int n , int m ) { float mean1 = mean ( arr1 , n ) ; float mean2 = mean ( arr2 , m ) ; cout << " Mean1 : β " << mean1 << " β mean2 : β " << mean2 << endl ; float sd1 = sd ( arr1 , n ) ; float sd2 = sd ( arr2 , m ) ; cout << " StandardDeviation1 : β " << sd1 << " β StandardDeviation2 : β " << sd2 << endl ; float combinedMean = ( float ) ( n * mean1 + m * mean2 ) / ( n + m ) ; cout << " Combined β Mean : β " << combinedMean << endl ; float d1_square = ( mean1 - combinedMean ) * ( mean1 - combinedMean ) ; float d2_square = ( mean2 - combinedMean ) * ( mean2 - combinedMean ) ; cout << " d1 β square : β " << d1_square << " β d2 _ square : β " << d2_square << endl ; float combinedVar = ( n * ( sd1 + d1_square ) + m * ( sd2 + d2_square ) ) / ( n + m ) ; return combinedVar ; } int main ( ) { int arr1 [ ] = { 23 , 45 , 34 , 78 , 12 , 76 , 34 } ; int arr2 [ ] = { 65 , 67 , 34 , 23 , 45 } ; int n = sizeof ( arr1 ) / sizeof ( arr1 [ 0 ] ) ; int m = sizeof ( arr2 ) / sizeof ( arr2 [ 0 ] ) ; cout << " Combined β Variance : β " << combinedVariance ( arr1 , arr2 , n , m ) ; return 0 ; } |
Check if a large number is divisible by 13 or not | CPP program to check whether a number is divisible by 13 or not . ; Returns true if number is divisible by 13 else returns false ; Append required 0 s . at the beginning . ; Same as strcat ( num , "00" ) ; in c . ; Same as strcat ( num , "0" ) ; in c . ; Alternatively add / subtract digits in group of three to result . ; Store group of three numbers in group variable . ; Generate alternate series of plus and minus ; Driver code | #include <iostream> NEW_LINE using namespace std ; bool checkDivisibility ( string num ) { int length = num . size ( ) ; if ( length == 1 && num [ 0 ] == '0' ) return true ; if ( length % 3 == 1 ) { num += "00" ; length += 2 ; } else if ( length % 3 == 2 ) { num += "0" ; length += 1 ; } int sum = 0 , p = 1 ; for ( int i = length - 1 ; i >= 0 ; i -- ) { int group = 0 ; group += num [ i -- ] - '0' ; group += ( num [ i -- ] - '0' ) * 10 ; group += ( num [ i ] - '0' ) * 100 ; sum = sum + group * p ; p *= ( -1 ) ; } sum = abs ( sum ) ; return ( sum % 13 == 0 ) ; } int main ( ) { string number = "83959092724" ; if ( checkDivisibility ( number ) ) cout << number << " β is β divisible β by β 13 . " ; else cout << number << " β is β not β divisible β by β 13 . " ; return 0 ; } |
Given two numbers a and b find all x such that a % x = b | C ++ program to find x such that a % x is equal to b . ; if a is less than b then no solution ; if a is equal to b then every number greater than a will be the solution so its infinity ; count variable store the number of values possible ; checking for both divisor and quotient whether they divide ( a - b ) completely and greater than b . ; Here y is added twice in the last iteration so 1 y should be decremented to get correct solution ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; void modularEquation ( int a , int b ) { if ( a < b ) { cout << " No β solution β possible β " << endl ; return ; } if ( a == b ) { cout << " Infinite β Solution β possible β " << endl ; return ; } int count = 0 ; int n = a - b ; int y = sqrt ( a - b ) ; for ( int i = 1 ; i <= y ; ++ i ) { if ( n % i == 0 ) { if ( n / i > b ) count ++ ; if ( i > b ) count ++ ; } } if ( y * y == n && y > b ) count -- ; cout << count << endl ; } int main ( ) { int a = 21 , b = 5 ; modularEquation ( a , b ) ; return 0 ; } |
Count different numbers that can be generated such that there digits sum is equal to ' n ' | C ++ program to count ways to write ' n ' as sum of digits ; Function to count ' num ' as sum of digits ( 1 , 2 , 3 , 4 ) ; Initialize dp [ ] array ; Base case ; Initialize the current dp [ ] array as '0' ; if i == j then there is only one way to write with element itself ' i ' ; If j == 1 , then there exist two ways , one from '1' and other from '4' ; if i - j is positive then pick the element from ' i - j ' element of dp [ ] array ; Check for modulas ; return the final answer ; Driver code | #include <iostream> NEW_LINE using namespace std ; int countWays ( int num ) { int dp [ num + 1 ] ; const int MOD = 1e9 + 7 ; dp [ 1 ] = 2 ; for ( int i = 2 ; i <= num ; ++ i ) { dp [ i ] = 0 ; for ( int j = 1 ; j <= 3 ; ++ j ) { if ( i - j == 0 ) dp [ i ] += 1 ; else if ( j == 1 ) dp [ i ] += dp [ i - j ] * 2 ; else if ( i - j > 0 ) dp [ i ] += dp [ i - j ] ; if ( dp [ i ] >= MOD ) dp [ i ] %= MOD ; } } return dp [ num ] ; } int main ( ) { int n = 3 ; cout << countWays ( n ) ; return 0 ; } |
Check whether a number can be represented by sum of two squares | An efficient approach based implementation to find if a number can be written as sum of two squares . ; function to check if there exist two numbers sum of whose squares is n . ; store square value in hashmap ; driver Program | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool sumSquare ( int n ) { unordered_map < int , int > s ; for ( int i = 0 ; i * i <= n ; ++ i ) { s [ i * i ] = 1 ; if ( s . find ( n - i * i ) != s . end ( ) ) { cout << sqrt ( n - i * i ) << " ^ 2 β + β " << i << " ^ 2" << endl ; return true ; } } return false ; } int main ( ) { int n = 169 ; if ( sumSquare ( n ) ) cout << " Yes " ; else cout << " No " ; } |
Check whether a number can be represented by sum of two squares | Check whether a number can be represented by sum of two squares using Fermat Theorem . ; Count all the prime factors . ; Ifany prime factor of the form ( 4 k + 3 ) ( 4 k + 3 ) occurs an odd number of times . ; If n itself is a x prime number and can be expressed in the form of 4 k + 3 we return false . ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool judgeSquareSum ( int n ) { for ( int i = 2 ; i * i <= n ; i ++ ) { int count = 0 ; if ( n % i == 0 ) { while ( n % i == 0 ) { count ++ ; n /= i ; } if ( i % 4 == 3 && count % 2 != 0 ) return false ; } } return n % 4 != 3 ; } int main ( ) { int n = 17 ; if ( judgeSquareSum ( n ) ) cout << " Yes " ; else cout << " No " ; } |
Total no of 1 's in numbers | c ++ code to count the frequency of 1 in numbers less than or equal to the given number . ; function to count the frequency of 1. ; driver function | #include <bits/stdc++.h> NEW_LINE using namespace std ; int countDigitOne ( int n ) { int countr = 0 ; for ( int i = 1 ; i <= n ; i *= 10 ) { int divider = i * 10 ; countr += ( n / divider ) * i + min ( max ( n % divider - i + 1 , 0 ) , i ) ; } return countr ; } int main ( ) { int n = 13 ; cout << countDigitOne ( n ) << endl ; n = 113 ; cout << countDigitOne ( n ) << endl ; n = 205 ; cout << countDigitOne ( n ) << endl ; } |
Largest number with prime digits | CPP program to find largest number smaller than equal to n with all prime digits . ; check if character is prime ; replace with previous prime character ; if 2 erase s [ i ] and replace next with 7 ; find first non prime char ; find first char greater than 2 ; like 20 ; like 7721 ; replace remaining with 7 ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool isPrime ( char c ) { return ( c == '2' c == '3' c == '5' c == '7' ) ; } void decrease ( string & s , int i ) { if ( s [ i ] <= '2' ) { s . erase ( i , 1 ) ; s [ i ] = '7' ; } else if ( s [ i ] == '3' ) s [ i ] = '2' ; else if ( s [ i ] <= '5' ) s [ i ] = '3' ; else if ( s [ i ] <= '7' ) s [ i ] = '5' ; else s [ i ] = '7' ; return ; } string primeDigits ( string s ) { for ( int i = 0 ; i < s . length ( ) ; i ++ ) { if ( ! isPrime ( s [ i ] ) ) { while ( s [ i ] <= '2' && i >= 0 ) i -- ; if ( i < 0 ) { i = 0 ; decrease ( s , i ) ; } else decrease ( s , i ) ; for ( int j = i + 1 ; j < s . length ( ) ; j ++ ) s [ j ] = '7' ; break ; } } return s ; } int main ( ) { string s = "45" ; cout << primeDigits ( s ) << endl ; s = "1000" ; cout << primeDigits ( s ) << endl ; s = "7721" ; cout << primeDigits ( s ) << endl ; s = "7221" ; cout << primeDigits ( s ) << endl ; s = "74545678912345689748593275897894708927680" ; cout << primeDigits ( s ) << endl ; return 0 ; } |
Divisors of n | C ++ program to count number of divisors of n ^ 2 which are not divisible by divisor of n ; Function to count divisors of n ^ 2 having no factors of ' n ' ; Increment count of i - th prime divisor ; Find next prime divisor ; Increment count if divisor still remains ; Initialize variable for counting the factors of n ^ 2 and n as ans1 and ans2 respectively ; Range based for - loop ; Use formula as discussed in above ; return the difference of answers ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int factors ( long long n ) { unordered_map < int , int > prime ; for ( int i = 2 ; i <= sqrt ( n ) ; ++ i ) { while ( n % i == 0 ) { ++ prime [ i ] ; n = n / i ; } } if ( n > 2 ) ++ prime [ n ] ; int ans1 = 1 , ans2 = 1 ; for ( auto it : prime ) { ans1 *= 2 * it . second + 1 ; ans2 *= it . second + 1 ; } return ans1 - ans2 ; } int main ( ) { long long n = 5 ; cout << factors ( n ) << endl ; n = 8 ; cout << factors ( n ) ; return 0 ; } |
Print digit 's position to be removed to make a number divisible by 6 | C ++ program to print digit 's position to be removed to make number divisible by 6 ; function to print the number divisible by 6 after exactly removing a digit ; stores the sum of all elements ; traverses the string and converts string to number array and sums up ; if ( a [ n - 1 ] % 2 ) ODD CHECK ; if second last is odd or sum of n - 1 elements are not divisible by 3. ; second last is even and print n - 1 elements removing last digit ; last digit removed ; counter to check if any element after removing , its sum % 3 == 0 ; traverse till second last element ; to check if any element after removing , its sum % 3 == 0 ; the leftmost element ; break at the leftmost element ; stores the right most element ; if no element has been found as a [ i + 1 ] > a [ i ] ; if second last is even , then remove last if ( sum - last ) % 3 == 0 ; if no element which on removing gives sum % 3 == 0 ; driver program to test the above function | #include <bits/stdc++.h> NEW_LINE using namespace std ; void greatest ( string s ) { int n = s . length ( ) ; int a [ n ] ; int sum = 0 ; for ( int i = 0 ; i < n ; i ++ ) { a [ i ] = s [ i ] - '0' ; sum += a [ i ] ; } { if ( a [ n - 2 ] % 2 != 0 or ( sum - a [ n - 1 ] ) % 3 != 0 ) { cout << " - 1" << endl ; } else { cout << n << endl ; } } else { int re = sum % 3 ; int del = -1 ; int flag = 0 ; for ( int i = 0 ; i < n - 1 ; i ++ ) { if ( ( a [ i ] ) % 3 == re ) { if ( a [ i + 1 ] > a [ i ] ) { del = i ; flag = 1 ; break ; } else { del = i ; } } } if ( flag == 0 ) { if ( a [ n - 2 ] % 2 == 0 and re == a [ n - 1 ] % 3 ) del = n - 1 ; } if ( del == -1 ) cout << -1 << endl ; else { cout << del + 1 << endl ; } } } int main ( ) { string s = "7510222" ; greatest ( s ) ; return 0 ; } |
Representation of a number in powers of other | CPP program to check if m can be represented as powers of w . ; break ; None of 3 worked . ; If m is not zero means , it can 't be represented in terms of powers of w. ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool asPowerSum ( int w , int m ) { while ( m ) { if ( ( m - 1 ) % w == 0 ) m = ( m - 1 ) / w ; else if ( ( m + 1 ) % w == 0 ) m = ( m + 1 ) / w ; else if ( m % w == 0 ) m = m / w ; else } return ( m == 0 ) ; } int main ( ) { int w = 3 , m = 7 ; if ( asPowerSum ( w , m ) ) cout << " Yes " << endl ; else cout << " No " << endl ; return 0 ; } |
Number of digits to be removed to make a number divisible by 3 | CPP program to find the minimum number of digits to be removed to make a large number divisible by 3. ; function to count the no of removal of digits to make a very large number divisible by 3 ; add up all the digits of num ; if num is already is divisible by 3 then no digits are to be removed ; if there is single digit , then it is not possible to remove one digit . ; traverse through the number and find out if any number on removal makes the sum divisible by 3 ; if there are two numbers then it is not possible to remove two digits . ; Otherwise we can always make a number multiple of 2 by removing 2 digits . ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int divisible ( string num ) { int n = num . length ( ) ; int sum = accumulate ( begin ( num ) , end ( num ) , 0 ) - '0' * 1 ; if ( sum % 3 == 0 ) return 0 ; if ( n == 1 ) return -1 ; for ( int i = 0 ; i < n ; i ++ ) if ( sum % 3 == ( num [ i ] - '0' ) % 3 ) return 1 ; if ( n == 2 ) return -1 ; return 2 ; } int main ( ) { string num = "1234" ; cout << divisible ( num ) ; return 0 ; } |
Program for dot product and cross product of two vectors | C ++ implementation for dot product and cross product of two vector . ; Function that return dot product of two vector array . ; Loop for calculate cot product ; Function to find cross product of two vector array . ; Driver function ; dotProduct function call ; crossProduct function call ; Loop that print cross product of two vector array . | #include <bits/stdc++.h> NEW_LINE #define n 3 NEW_LINE using namespace std ; int dotProduct ( int vect_A [ ] , int vect_B [ ] ) { int product = 0 ; for ( int i = 0 ; i < n ; i ++ ) product = product + vect_A [ i ] * vect_B [ i ] ; return product ; } void crossProduct ( int vect_A [ ] , int vect_B [ ] , int cross_P [ ] ) { cross_P [ 0 ] = vect_A [ 1 ] * vect_B [ 2 ] - vect_A [ 2 ] * vect_B [ 1 ] ; cross_P [ 1 ] = vect_A [ 2 ] * vect_B [ 0 ] - vect_A [ 0 ] * vect_B [ 2 ] ; cross_P [ 2 ] = vect_A [ 0 ] * vect_B [ 1 ] - vect_A [ 1 ] * vect_B [ 0 ] ; } int main ( ) { int vect_A [ ] = { 3 , -5 , 4 } ; int vect_B [ ] = { 2 , 6 , 5 } ; int cross_P [ n ] ; cout << " Dot β product : " ; cout << dotProduct ( vect_A , vect_B ) << endl ; cout << " Cross β product : " ; crossProduct ( vect_A , vect_B , cross_P ) ; for ( int i = 0 ; i < n ; i ++ ) cout << cross_P [ i ] << " β " ; return 0 ; } |
Number of n digit numbers that do not contain 9 | CPP program to find number of n digit numbers that do not contain 9 as it 's digit ; function to find number of n digit numbers possible ; driver function | #include <bits/stdc++.h> NEW_LINE using namespace std ; int totalNumber ( int n ) { return 8 * pow ( 9 , n - 1 ) ; } int main ( ) { int n = 3 ; cout << totalNumber ( n ) ; return 0 ; } |
Count ways to express even number Γ’ β¬Λ nΓ’ β¬β’ as sum of even integers | C ++ program to count ways to write number as sum of even integers ; Initialize mod variable as constant ; 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 ; Return number of ways to write ' n ' as sum of even integers ; Driver code | #include <iostream> NEW_LINE using namespace std ; const int MOD = 1e9 + 7 ; int power ( int x , unsigned int y , int p ) { while ( y > 0 ) { if ( y & 1 ) res = ( 1LL * res * x ) % p ; x = ( 1LL * x * x ) % p ; } return res ; } int countEvenWays ( int n ) { return power ( 2 , n / 2 - 1 , MOD ) ; } int main ( ) { int n = 6 ; cout << countEvenWays ( n ) << " STRNEWLINE " ; n = 8 ; cout << countEvenWays ( n ) ; return 0 ; } |
Number of steps to convert to prime factors | CPP program to count number of steps required to convert an integer array to array of factors . ; array to store prime factors ; function to generate all prime factors of numbers from 1 to 10 ^ 6 ; Initializes all the positions with their value . ; Initializes all multiples of 2 with 2 ; A modified version of Sieve of Eratosthenes to store the smallest prime factor that divides every number . ; check if it has no prime factor . ; Initializes of j starting from i * i ; if it has no prime factor before , then stores the smallest prime divisor ; function to calculate the number of representations ; keep an count of prime factors ; traverse for every element ; count the no of factors ; subtract 1 if Ai is not 1 as the last step wont be taken into count ; driver program to test the above function ; call sieve to calculate the factors | #include <iostream> NEW_LINE using namespace std ; const int MAX = 1000001 ; int factor [ MAX ] = { 0 } ; void cal_factor ( ) { factor [ 1 ] = 1 ; for ( int i = 2 ; i < MAX ; i ++ ) factor [ i ] = i ; for ( int i = 4 ; i < MAX ; i += 2 ) factor [ i ] = 2 ; for ( int i = 3 ; i * i < MAX ; i ++ ) { if ( factor [ i ] == i ) { for ( int j = i * i ; j < MAX ; j += i ) { if ( factor [ j ] == j ) factor [ j ] = i ; } } } } int no_of_representations ( int a [ ] , int n ) { int count = 0 ; for ( int i = 0 ; i < n ; i ++ ) { int temp = a [ i ] ; int flag = 0 ; while ( factor [ temp ] != 1 ) { flag = -1 ; count ++ ; temp = temp / factor [ temp ] ; } count += flag ; } return count ; } int main ( ) { cal_factor ( ) ; int a [ ] = { 4 , 4 , 4 } ; int n = sizeof ( a ) / sizeof ( a [ 0 ] ) ; cout << no_of_representations ( a , n ) ; return 0 ; } |
Subsequences of size three in an array whose sum is divisible by m | Brute Force Implementation ( Time Complexity : O ( N ^ 3 ) ) C ++ program to find count of subsequences of size three divisible by M . ; Three nested loop to find all the sub sequences of length three in the given array A [ ] . ; checking if the sum of the chosen three number is divisible by m . ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int coutSubSeq ( int A [ ] , int N , int M ) { int sum = 0 ; int ans = 0 ; for ( int i = 0 ; i < N ; i ++ ) { for ( int j = i + 1 ; j < N ; j ++ ) { for ( int k = j + 1 ; k < N ; k ++ ) { sum = A [ i ] + A [ j ] + A [ k ] ; if ( sum % M == 0 ) ans ++ ; } } } return ans ; } int main ( ) { int M = 3 ; int A [ ] = { 1 , 2 , 4 , 3 } ; int N = sizeof ( A ) / sizeof ( A [ 0 ] ) ; cout << coutSubSeq ( A , N , M ) ; return 0 ; } |
Subsequences of size three in an array whose sum is divisible by m | O ( M ^ 2 ) time complexity C ++ program to find count of subsequences of size three divisible by M . ; Storing frequencies of all remainders when divided by M . ; including i and j in the sum rem calculate the remainder required to make the sum divisible by M ; if the required number is less than j , we skip as we have already calculated for that value before . As j here starts with i and rem is less than j . ; if satisfies the first case . ; if satisfies the second case ; if satisfies the third case ; Driver code | #include <iostream> NEW_LINE using namespace std ; int countSubSeq ( int A [ ] , int N , int M ) { int ans = 0 ; int h [ M ] = { 0 } ; for ( int i = 0 ; i < N ; i ++ ) { A [ i ] = A [ i ] % M ; h [ A [ i ] ] ++ ; } for ( int i = 0 ; i < M ; i ++ ) { for ( int j = i ; j < M ; j ++ ) { int rem = ( M - ( i + j ) % M ) % M ; if ( rem < j ) continue ; if ( i == j && rem == j ) ans += h [ i ] * ( h [ i ] - 1 ) * ( h [ i ] - 2 ) / 6 ; else if ( i == j ) ans += h [ i ] * ( h [ i ] - 1 ) * h [ rem ] / 2 ; else if ( i == rem ) ans += h [ i ] * ( h [ i ] - 1 ) * h [ j ] / 2 ; else if ( rem == j ) ans += h [ j ] * ( h [ j ] - 1 ) * h [ i ] / 2 ; else ans = ans + h [ i ] * h [ j ] * h [ rem ] ; } } return ans ; } int main ( ) { int M = 3 ; int A [ ] = { 1 , 2 , 4 , 3 } ; int N = sizeof ( A ) / sizeof ( A [ 0 ] ) ; cout << countSubSeq ( A , N , M ) ; return 0 ; } |
Find n | CPP program to find nth term ; utility function ; since first element of the series is 7 , we initialise a variable with 7 ; Using iteration to find nth term ; driver function | #include <bits/stdc++.h> NEW_LINE using namespace std ; int findTerm ( int n ) { if ( n == 1 ) return n ; else { int term = 7 ; for ( int i = 2 ; i <= n ; i ++ ) term = term * 2 + ( i - 1 ) ; return term ; } } int main ( ) { int n = 5 ; cout << findTerm ( n ) ; return 0 ; } |
Find n | CPP program to find the value at n - th place in the given sequence ; Returns n - th number in sequence 1 , 1 , 2 , 1 , 2 , 3 , 1 , 2 , 4 , ... ; One by one subtract counts elements in different blocks ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int findNumber ( int n ) { n -- ; int i = 1 ; while ( n >= 0 ) { n -= i ; ++ i ; } return ( n + i ) ; } int main ( ) { int n = 3 ; cout << findNumber ( n ) << endl ; return 0 ; } |
Program to find correlation coefficient | Program to find correlation coefficient ; function that returns correlation coefficient . ; sum of elements of array X . ; sum of elements of array Y . ; sum of X [ i ] * Y [ i ] . ; sum of square of array elements . ; use formula for calculating correlation coefficient . ; Driver function ; Find the size of array . ; Function call to correlationCoefficient . | #include <bits/stdc++.h> NEW_LINE using namespace std ; float correlationCoefficient ( int X [ ] , int Y [ ] , int n ) { int sum_X = 0 , sum_Y = 0 , sum_XY = 0 ; int squareSum_X = 0 , squareSum_Y = 0 ; for ( int i = 0 ; i < n ; i ++ ) { sum_X = sum_X + X [ i ] ; sum_Y = sum_Y + Y [ i ] ; sum_XY = sum_XY + X [ i ] * Y [ i ] ; squareSum_X = squareSum_X + X [ i ] * X [ i ] ; squareSum_Y = squareSum_Y + Y [ i ] * Y [ i ] ; } float corr = ( float ) ( n * sum_XY - sum_X * sum_Y ) / sqrt ( ( n * squareSum_X - sum_X * sum_X ) * ( n * squareSum_Y - sum_Y * sum_Y ) ) ; return corr ; } int main ( ) { int X [ ] = { 15 , 18 , 21 , 24 , 27 } ; int Y [ ] = { 25 , 25 , 27 , 31 , 32 } ; int n = sizeof ( X ) / sizeof ( X [ 0 ] ) ; cout << correlationCoefficient ( X , Y , n ) ; return 0 ; } |
Find the number of spectators standing in the stadium at time t | CPP program to find number of spectators standing at a time ; If the time is less than k then we can print directly t time . ; If the time is n then k spectators are standing . ; Otherwise we calculate the spectators standing . ; Driver code ; Stores the value of n , k and t t is time n & k is the number of specators | #include <bits/stdc++.h> NEW_LINE using namespace std ; void result ( long long n , long long k , long long t ) { if ( t <= k ) cout << t ; else if ( t <= n ) cout << k ; else { long long temp = t - n ; temp = k - temp ; cout << temp ; } } int main ( ) { long long n , k , t ; n = 10 ; k = 5 ; t = 12 ; result ( n , k , t ) ; return 0 ; } |
Program for weighted mean of natural numbers . | Program to find weighted mean of natural numbers . ; Function to calculate weighted mean . ; Driver program to test the function . ; Take num array and corresponding weight array and initialize it . ; Calculate the size of array . ; Check the size of both array is equal or not . | #include <bits/stdc++.h> NEW_LINE using namespace std ; float weightedMean ( int X [ ] , int W [ ] , int n ) { int sum = 0 , numWeight = 0 ; for ( int i = 0 ; i < n ; i ++ ) { numWeight = numWeight + X [ i ] * W [ i ] ; sum = sum + W [ i ] ; } return ( float ) numWeight / sum ; } int main ( ) { int X [ ] = { 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 } ; int W [ ] = { 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 } ; int n = sizeof ( X ) / sizeof ( X [ 0 ] ) ; int m = sizeof ( W ) / sizeof ( W [ 0 ] ) ; if ( n == m ) cout << weightedMean ( X , W , n ) ; else cout << " - 1" ; return 0 ; } |
Program to find GCD of floating point numbers | CPP code for finding the GCD of two floating numbers . ; Recursive function to return gcd of a and b ; base case ; Driver Function . | #include <bits/stdc++.h> NEW_LINE using namespace std ; double gcd ( double a , double b ) { if ( a < b ) return gcd ( b , a ) ; if ( fabs ( b ) < 0.001 ) return a ; else return ( gcd ( b , a - floor ( a / b ) * b ) ) ; } int main ( ) { double a = 1.20 , b = 22.5 ; cout << gcd ( a , b ) ; return 0 ; } |
Program for harmonic mean of numbers | CPP program to find harmonic mean of numbers . ; Function that returns harmonic mean . ; Declare sum variables and initialize with zero . ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; float harmonicMean ( float arr [ ] , int n ) { float sum = 0 ; for ( int i = 0 ; i < n ; i ++ ) sum = sum + ( float ) 1 / arr [ i ] ; return ( float ) n / sum ; } int main ( ) { float arr [ ] = { 13.5 , 14.5 , 14.8 , 15.2 , 16.1 } ; int n = sizeof ( arr ) / sizeof ( arr [ 0 ] ) ; cout << harmonicMean ( arr , n ) ; return 0 ; } |
Program for harmonic mean of numbers | C ++ program to find harmonic mean . ; Function that returns harmonic mean . ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; float harmonicMean ( int arr [ ] , int freq [ ] , int n ) { float sum = 0 , frequency_sum = 0 ; for ( int i = 0 ; i < n ; i ++ ) { sum = sum + ( float ) freq [ i ] / arr [ i ] ; frequency_sum = frequency_sum + freq [ i ] ; } return frequency_sum / sum ; } int main ( ) { int num [ ] = { 13 , 14 , 15 , 16 , 17 } ; int freq [ ] = { 2 , 5 , 13 , 7 , 3 } ; int n = sizeof ( num ) / sizeof ( num [ 0 ] ) ; cout << harmonicMean ( num , freq , n ) ; return 0 ; } |
First collision point of two series | CPP program to calculate the colliding point of two series ; Iterating through n terms of the first series ; x is i - th term of first series ; d is first element of second series and c is common difference for second series . ; If no term of first series is found ; Driver function | #include <bits/stdc++.h> NEW_LINE using namespace std ; void point ( int a , int b , int c , int d , int n ) { int x , flag = 0 ; for ( int i = 0 ; i < n ; i ++ ) { x = b + i * a ; if ( ( x - d ) % c == 0 and x - d >= 0 ) { cout << x << endl ; flag = 1 ; break ; } } if ( flag == 0 ) { cout << " No β collision β point " << endl ; } } int main ( ) { int a = 20 ; int b = 2 ; int c = 9 ; int d = 19 ; int n = 20 ; point ( a , b , c , d , n ) ; return 0 ; } |
Armstrong Numbers between two integers | CPP program to find Armstrong numbers in a range ; Prints Armstrong Numbers in given range ; number of digits calculation ; compute sum of nth power of its digits ; checks if number i is equal to the sum of nth power of its digits ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; void findArmstrong ( int low , int high ) { for ( int i = low + 1 ; i < high ; ++ i ) { int x = i ; int n = 0 ; while ( x != 0 ) { x /= 10 ; ++ n ; } int pow_sum = 0 ; x = i ; while ( x != 0 ) { int digit = x % 10 ; pow_sum += pow ( digit , n ) ; x /= 10 ; } if ( pow_sum == i ) cout << i << " β " ; } } int main ( ) { int num1 = 100 ; int num2 = 400 ; findArmstrong ( num1 , num2 ) ; cout << ' ' ; return 0 ; } |
Lucas Primality Test | C ++ Program for Lucas Primality Test ; function to generate prime factors of n ; if 2 is a factor ; if prime > 2 is factor ; this function produces power modulo some number . It can be optimized to using ; Base cases ; Generating and storing factors of n - 1 ; Array for random generator . This array is to ensure one number is generated only once ; shuffle random array to produce randomness ; Now one by one perform Lucas Primality Test on random numbers generated . ; this is to check if every factor of n - 1 satisfy the condition ; if a ^ ( ( n - 1 ) / q ) equal 1 ; if all condition satisfy ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; void primeFactors ( int n , vector < int > & factors ) { if ( n % 2 == 0 ) factors . push_back ( 2 ) ; while ( n % 2 == 0 ) n = n / 2 ; for ( int i = 3 ; i <= sqrt ( n ) ; i += 2 ) { if ( n % i == 0 ) factors . push_back ( i ) ; while ( n % i == 0 ) n = n / i ; } if ( n > 2 ) factors . push_back ( n ) ; } int power ( int n , int r , int q ) { int total = n ; for ( int i = 1 ; i < r ; i ++ ) total = ( total * n ) % q ; return total ; } string lucasTest ( int n ) { if ( n == 1 ) return " neither β prime β nor β composite " ; if ( n == 2 ) return " prime " ; if ( n % 2 == 0 ) return " composite1" ; vector < int > factors ; primeFactors ( n - 1 , factors ) ; int random [ n - 3 ] ; for ( int i = 0 ; i < n - 2 ; i ++ ) random [ i ] = i + 2 ; shuffle ( random , random + n - 3 , default_random_engine ( time ( 0 ) ) ) ; for ( int i = 0 ; i < n - 2 ; i ++ ) { int a = random [ i ] ; if ( power ( a , n - 1 , n ) != 1 ) return " composite " ; bool flag = true ; for ( int k = 0 ; k < factors . size ( ) ; k ++ ) { if ( power ( a , ( n - 1 ) / factors [ k ] , n ) == 1 ) { flag = false ; break ; } } if ( flag ) return " prime " ; } return " probably β composite " ; } int main ( ) { cout << 7 << " β is β " << lucasTest ( 7 ) << endl ; cout << 9 << " β is β " << lucasTest ( 9 ) << endl ; cout << 37 << " β is β " << lucasTest ( 37 ) << endl ; return 0 ; } |
Pair with maximum GCD from two arrays | CPP program to find maximum GCD pair from two arrays ; Find the maximum GCD pair with maximum sum ; array to keep a count of existing elements ; first [ i ] and second [ i ] are going to store maximum multiples of i in a [ ] and b [ ] respectively . ; traverse through the first array to mark the elements in cnt ; Find maximum multiple of every number in first array ; Find maximum multiple of every number in second array We re - initialise cnt [ ] and traverse through the second array to mark the elements in cnt ; if the multiple is present in the second array then store the max of number or the pre - existing element ; traverse for every elements and checks the maximum N that is present in both the arrays ; driver program to test the above function ; Maximum possible value of elements in both arrays . | #include <bits/stdc++.h> NEW_LINE using namespace std ; void gcdMax ( int a [ ] , int b [ ] , int n , int N ) { int cnt [ N ] = { 0 } ; int first [ N ] = { 0 } , second [ N ] = { 0 } ; for ( int i = 0 ; i < n ; ++ i ) cnt [ a [ i ] ] = 1 ; for ( int i = 1 ; i < N ; ++ i ) for ( int j = i ; j < N ; j += i ) if ( cnt [ j ] ) first [ i ] = max ( first [ i ] , j ) ; memset ( cnt , 0 , sizeof ( cnt ) ) ; for ( int i = 0 ; i < n ; ++ i ) cnt [ b [ i ] ] = true ; for ( int i = 1 ; i < N ; ++ i ) for ( int j = i ; j < N ; j += i ) if ( cnt [ j ] ) second [ i ] = max ( second [ i ] , j ) ; int i ; for ( i = N - 1 ; i >= 0 ; i -- ) if ( first [ i ] && second [ i ] ) break ; cout << " Maximum β GCD β pair β with β maximum β " " sum β is β " << first [ i ] << " β " << second [ i ] << endl ; } int main ( ) { int a [ ] = { 3 , 1 , 4 , 2 , 8 } ; int b [ ] = { 5 , 2 , 12 , 8 , 3 } ; int n = sizeof ( a ) / sizeof ( a [ 0 ] ) ; int N = 20 ; gcdMax ( a , b , n , N ) ; return 0 ; } |
Pierpont Prime | CPP program to print Pierpont prime numbers smaller than n . ; Finding all numbers having factor power of 2 and 3 Using sieve ; Storing number of the form 2 ^ i . 3 ^ k + 1. ; Finding prime number using sieve of Eratosthenes . Reusing same array as result of above computations in v . ; Printing n pierpont primes smaller than n ; Driven Program | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool printPierpont ( int n ) { bool arr [ n + 1 ] ; memset ( arr , false , sizeof arr ) ; int two = 1 , three = 1 ; while ( two + 1 < n ) { arr [ two ] = true ; while ( two * three + 1 < n ) { arr [ three ] = true ; arr [ two * three ] = true ; three *= 3 ; } three = 1 ; two *= 2 ; } vector < int > v ; for ( int i = 0 ; i < n ; i ++ ) if ( arr [ i ] ) v . push_back ( i + 1 ) ; memset ( arr , false , sizeof arr ) ; for ( int p = 2 ; p * p < n ; p ++ ) { if ( arr [ p ] == false ) for ( int i = p * 2 ; i < n ; i += p ) arr [ i ] = true ; } for ( int i = 0 ; i < v . size ( ) ; i ++ ) if ( ! arr [ v [ i ] ] ) cout << v [ i ] << " β " ; } int main ( ) { int n = 200 ; printPierpont ( n ) ; return 0 ; } |
Woodall Number | CPP program to check if a number is Woodball or not . ; If number is even , return false . ; If x is 1 , return true . ; While x is divisible by 2 ; Divide x by 2 ; Count the power ; If at any point power and x became equal , return true . ; Driven Program | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool isWoodall ( int x ) { if ( x % 2 == 0 ) return false ; if ( x == 1 ) return true ; int p = 0 ; while ( x % 2 == 0 ) { x = x / 2 ; p ++ ; if ( p == x ) return true ; } return false ; } int main ( ) { int x = 383 ; ( isWoodall ( x ) ) ? ( cout << " Yes " << endl ) : ( cout << " No " << endl ) ; return 0 ; } |
Print k numbers where all pairs are divisible by m | CPP program to find a list of k elements from an array such that difference between all of them is divisible by m . ; function to generate k numbers whose difference is divisible by m ; Using an adjacency list like representation to store numbers that lead to same remainder . ; stores the modulus when divided by m ; If we found k elements which have same remainder . ; If we could not find k elements ; driver program to test the above function | #include <bits/stdc++.h> NEW_LINE using namespace std ; void print_result ( int a [ ] , int n , int k , int m ) { vector < int > v [ m ] ; for ( int i = 0 ; i < n ; i ++ ) { int rem = a [ i ] % m ; v [ rem ] . push_back ( a [ i ] ) ; if ( v [ rem ] . size ( ) == k ) { for ( int j = 0 ; j < k ; j ++ ) cout << v [ rem ] [ j ] << " β " ; return ; } } cout << " - 1" ; } int main ( ) { int a [ ] = { 1 , 8 , 4 } ; int n = sizeof ( a ) / sizeof ( a [ 0 ] ) ; print_result ( a , n , 2 , 3 ) ; return 0 ; } |
Largest number less than N whose each digit is prime number | C ++ program to find the largest number smaller than N whose all digits are prime . ; Number is given as string . ; We stop traversing digits , once it become smaller than current number . For that purpose we use small variable . ; Array indicating if index i ( represents a digit ) is prime or not . ; Store largest ; If there is only one character , return the largest prime less than the number ; If number starts with 1 , return number consisting of 7 ; Traversing each digit from right to left Continue traversing till the number we are forming will become less . ; If digit is prime , copy it simply . ; If not prime , copy the largest prime less than current number ; If not prime , and there is no largest prime less than current prime ; Make current digit as 7 Go left of the digit and make it largest prime less than number . Continue do that until we found a digit which has some largest prime less than it ; If the given number is itself a prime . ; Make last digit as highest prime less than given digit . ; If there is no highest prime less than current digit . ; Once one digit become less than any digit of input replace 7 ( largest 1 digit prime ) till the end of digits of number ; If number include 0 in the beginning , ignore them . Case like 2200 ; Driver Program | #include <bits/stdc++.h> NEW_LINE using namespace std ; char * PrimeDigitNumber ( char N [ ] , int size ) { char * ans = ( char * ) malloc ( size * sizeof ( char ) ) ; int ns = 0 ; int small = 0 ; int i ; int p [ ] = { 0 , 0 , 1 , 1 , 0 , 1 , 0 , 1 , 0 , 0 } ; int prevprime [ ] = { 0 , 0 , 0 , 2 , 3 , 3 , 5 , 5 , 7 , 7 } ; if ( size == 1 ) { ans [ 0 ] = prevprime [ N [ 0 ] - '0' ] + '0' ; ans [ 1 ] = ' \0' ; return ans ; } if ( N [ 0 ] == '1' ) { for ( int i = 0 ; i < size - 1 ; i ++ ) ans [ i ] = '7' ; ans [ size - 1 ] = ' \0' ; return ans ; } for ( i = 0 ; i < size && small == 0 ; i ++ ) { if ( p [ N [ i ] - '0' ] == 1 ) { ans [ ns ++ ] = N [ i ] ; } else { if ( p [ N [ i ] - '0' ] == 0 && prevprime [ N [ i ] - '0' ] != 0 ) { ans [ ns ++ ] = prevprime [ N [ i ] - '0' ] + '0' ; small = 1 ; } else if ( p [ N [ i ] - '0' ] == 0 && prevprime [ N [ i ] - '0' ] == 0 ) { int j = i ; while ( j > 0 && p [ N [ j ] - '0' ] == 0 && prevprime [ N [ j ] - '0' ] == 0 ) { ans [ j ] = N [ j ] = '7' ; N [ j - 1 ] = prevprime [ N [ j - 1 ] - '0' ] + '0' ; ans [ j - 1 ] = N [ j - 1 ] ; small = 1 ; j -- ; } i = ns ; } } } if ( small == 0 ) { if ( prevprime [ N [ size - 1 ] - '0' ] + '0' != '0' ) ans [ size - 1 ] = prevprime [ N [ size - 1 ] - '0' ] + '0' ; else { int j = size - 1 ; while ( j > 0 && prevprime [ N [ j ] - '0' ] == 0 ) { ans [ j ] = N [ j ] = '7' ; N [ j - 1 ] = prevprime [ N [ j - 1 ] - '0' ] + '0' ; ans [ j - 1 ] = N [ j - 1 ] ; small = 1 ; j -- ; } } } for ( ; ns < size ; ns ++ ) ans [ ns ] = '7' ; ans [ ns ] = ' \0' ; int k = 0 ; while ( ans [ k ] == '0' ) k ++ ; return ans + k ; } int main ( ) { char N [ ] = "1000" ; int size = strlen ( N ) ; cout << PrimeDigitNumber ( N , size ) << endl ; return 0 ; } |
Smallest x such that 1 * n , 2 * n , ... x * n have all digits from 1 to 9 | CPP program to find x such that 1 * n , 2 * n , 3 * n ... x * n have all digits from 1 to 9 at least once ; Returns smallest value x such that 1 * n , 2 * n , 3 * n ... x * n have all digits from 1 to 9 at least once ; taking temporary array and variable . ; iterate till we get all the 10 digits at least once ; checking all the digits ; driver function | #include <bits/stdc++.h> NEW_LINE using namespace std ; int smallestX ( int n ) { int temp [ 10 ] = { 0 } ; if ( n == 0 ) return -1 ; int count = 0 , x = 0 ; for ( x = 1 ; count < 10 ; x ++ ) { int y = x * n ; while ( y ) { if ( temp [ y % 10 ] == false ) { count ++ ; temp [ y % 10 ] = true ; } y /= 10 ; } } return x - 1 ; } int main ( ) { int n = 5 ; cout << smallestX ( n ) ; return 0 ; } |
Find a number x such that sum of x and its digits is equal to given n . | CPP program to find x such that x + digSum ( x ) is equal to n . ; utility function for digit sum ; function for finding x ; iterate from 1 to n . For every no . check if its digit sum with it is equal to n . ; if no such i found return - 1 ; driver function | #include <bits/stdc++.h> NEW_LINE using namespace std ; int digSum ( int n ) { int sum = 0 , rem = 0 ; while ( n ) { rem = n % 10 ; sum += rem ; n /= 10 ; } return sum ; } int findX ( int n ) { for ( int i = 0 ; i <= n ; i ++ ) if ( i + digSum ( i ) == n ) return i ; return -1 ; } int main ( ) { int n = 43 ; cout << " x β = β " << findX ( n ) ; return 0 ; } |
9 's complement of a decimal number | C ++ program to find 9 's complement of a number. ; Driver code | #include <iostream> NEW_LINE using namespace std ; void complement ( string number ) { for ( int i = 0 ; i < number . length ( ) ; i ++ ) if ( number [ i ] != ' . ' ) number [ i ] = '9' - number [ i ] + '0' ; cout << "9 ' s β complement β is β : β " << number ; } int main ( ) { string number = "345.45" ; complement ( number ) ; return 0 ; } |
Ways to express a number as product of two different factors | CPP program to find number of ways in which number expressed as product of two different factors ; To count number of ways in which number expressed as product of two different numbers ; To store count of such pairs ; Counting number of pairs upto sqrt ( n ) - 1 ; To return count of pairs ; Driver program to test countWays ( ) | #include <bits/stdc++.h> NEW_LINE using namespace std ; int countWays ( int n ) { int count = 0 ; for ( int i = 1 ; i * i < n ; i ++ ) if ( n % i == 0 ) count ++ ; return count ; } int main ( ) { int n = 12 ; cout << countWays ( n ) << endl ; return 0 ; } |
Count divisors of n that have at | C ++ program to count divisors of n that have at least one digit common with n ; function to return true if any digit of m is present in hash [ ] . ; check till last digit ; if number is also present in original number then return true ; if no number matches then return 1 ; Count the no of divisors that have at least 1 digits same ; Store digits present in n in a hash [ ] ; marks that the number is present ; last digit removed ; loop to traverse from 1 to sqrt ( n ) to count divisors ; if i is the factor ; call the function to check if any digits match or not ; if n / i != i then a different number , then check it also ; return the answer ; driver program to test the above function | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool isDigitPresent ( int m , bool hash [ ] ) { while ( m ) { if ( hash [ m % 10 ] ) return true ; m = m / 10 ; } return false ; } int countDivisibles ( int n ) { bool hash [ 10 ] = { 0 } ; int m = n ; while ( m ) { hash [ m % 10 ] = true ; m = m / 10 ; } int ans = 0 ; for ( int i = 1 ; i <= sqrt ( n ) ; i ++ ) { if ( n % i == 0 ) { if ( isDigitPresent ( i , hash ) ) ans ++ ; if ( n / i != i ) { if ( isDigitPresent ( n / i , hash ) ) ans ++ ; } } } return ans ; } int main ( ) { int n = 15 ; cout << countDivisibles ( n ) ; return 0 ; } |
Doolittle Algorithm : LU Decomposition | C ++ Program to decompose a matrix into lower and upper triangular matrix ; Decomposing matrix into Upper and Lower triangular matrix ; Upper Triangular ; Summation of L ( i , j ) * U ( j , k ) ; Evaluating U ( i , k ) ; Lower Triangular ; lower [ i ] [ i ] = 1 ; Diagonal as 1 ; Summation of L ( k , j ) * U ( j , i ) ; Evaluating L ( k , i ) ; setw is for displaying nicely ; Displaying the result : ; Lower ; Upper ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; const int MAX = 100 ; void luDecomposition ( int mat [ ] [ MAX ] , int n ) { int lower [ n ] [ n ] , upper [ n ] [ n ] ; memset ( lower , 0 , sizeof ( lower ) ) ; memset ( upper , 0 , sizeof ( upper ) ) ; for ( int i = 0 ; i < n ; i ++ ) { for ( int k = i ; k < n ; k ++ ) { int sum = 0 ; for ( int j = 0 ; j < i ; j ++ ) sum += ( lower [ i ] [ j ] * upper [ j ] [ k ] ) ; upper [ i ] [ k ] = mat [ i ] [ k ] - sum ; } for ( int k = i ; k < n ; k ++ ) { if ( i == k ) else { int sum = 0 ; for ( int j = 0 ; j < i ; j ++ ) sum += ( lower [ k ] [ j ] * upper [ j ] [ i ] ) ; lower [ k ] [ i ] = ( mat [ k ] [ i ] - sum ) / upper [ i ] [ i ] ; } } } cout << setw ( 6 ) << " TABSYMBOL Lower β Triangular " << setw ( 32 ) << " Upper β Triangular " << endl ; for ( int i = 0 ; i < n ; i ++ ) { for ( int j = 0 ; j < n ; j ++ ) cout << setw ( 6 ) << lower [ i ] [ j ] << " TABSYMBOL " ; cout << " TABSYMBOL " ; for ( int j = 0 ; j < n ; j ++ ) cout << setw ( 6 ) << upper [ i ] [ j ] << " TABSYMBOL " ; cout << endl ; } } int main ( ) { int mat [ ] [ MAX ] = { { 2 , -1 , -2 } , { -4 , 6 , 3 } , { -4 , -2 , 8 } } ; luDecomposition ( mat , 3 ) ; return 0 ; } |
Divide number into two parts divisible by given numbers | C ++ code to break the number string into two divisible parts by given numbers ; method prints divisible parts if possible , otherwise prints ' Not β possible ' ; creating arrays to store reminder ; looping over all suffix and storing reminder with f ; getting suffix reminder from previous suffix reminder ; looping over all prefix and storing reminder with s ; getting prefix reminder from next prefix reminder ; updating base value ; now looping over all reminders to check partition condition ; if both reminders are 0 and digit itself is not 0 , then print result and return ; if we reach here , then string can ' be partitioned under constraints ; Driver code to test above methods | #include <bits/stdc++.h> NEW_LINE using namespace std ; void printTwoDivisibleParts ( string num , int f , int s ) { int N = num . length ( ) ; int prefixReminder [ N + 1 ] ; int suffixReminder [ N + 1 ] ; suffixReminder [ 0 ] = 0 ; for ( int i = 1 ; i < N ; i ++ ) suffixReminder [ i ] = ( suffixReminder [ i - 1 ] * 10 + ( num [ i - 1 ] - '0' ) ) % f ; prefixReminder [ N ] = 0 ; int base = 1 ; for ( int i = N - 1 ; i >= 0 ; i -- ) { prefixReminder [ i ] = ( prefixReminder [ i + 1 ] + ( num [ i ] - '0' ) * base ) % s ; base = ( base * 10 ) % s ; } for ( int i = 0 ; i < N ; i ++ ) { if ( prefixReminder [ i ] == 0 && suffixReminder [ i ] == 0 && num [ i ] != '0' ) { cout << num . substr ( 0 , i ) << " β " << num . substr ( i ) << endl ; return ; } } cout << " Not β Possible STRNEWLINE " ; } int main ( ) { string num = "246904096" ; int f = 12345 ; int s = 1024 ; printTwoDivisibleParts ( num , f , s ) ; return 0 ; } |
Number of subarrays whose minimum and maximum are same | CPP program to count number of subarrays having same minimum and maximum . ; calculate the no of contiguous subarrays which has same minimum and maximum ; stores the answer ; loop to traverse from 0 - n ; start checking subarray from next element ; traverse for finding subarrays ; if the elements are same then we check further and keep a count of same numbers in ' r ' ; the no of elements in between r and i with same elements . ; the no of subarrays that can be formed between i and r ; again start checking from the next index ; returns answer ; drive program to test the above function | #include <bits/stdc++.h> NEW_LINE using namespace std ; int calculate ( int a [ ] , int n ) { int ans = 0 ; for ( int i = 0 ; i < n ; i ++ ) { int r = i + 1 ; for ( int j = r ; j < n ; j ++ ) { if ( a [ i ] == a [ j ] ) r += 1 ; else break ; } int d = r - i ; ans += ( d * ( d + 1 ) / 2 ) ; i = r - 1 ; } return ans ; } int main ( ) { int a [ ] = { 2 , 4 , 5 , 3 , 3 , 3 } ; int n = sizeof ( a ) / sizeof ( a [ 0 ] ) ; cout << calculate ( a , n ) ; return 0 ; } |
Count of numbers satisfying m + sum ( m ) + sum ( sum ( m ) ) = N | CPP program to count numbers satisfying equation . ; function that returns sum of digits in a number ; initially sum of digits is 0 ; loop runs till all digits have been extracted ; last digit from backside ; sums up the digits ; the number is reduced to the number removing the last digit ; returns the sum of digits in a number ; function to calculate the count of such occurrences ; counter to calculate the occurrences ; loop to traverse from n - 97 to n ; calls the function to calculate the sum of digits of i ; calls the function to calculate the sum of digits of a ; if the summation is equal to n then increase counter by 1 ; returns the count ; driver program to test the above function ; calls the function to get the answer | #include <bits/stdc++.h> NEW_LINE using namespace std ; int sum ( int n ) { int rem = 0 ; int sum_of_digits = 0 ; while ( n > 0 ) { rem = n % 10 ; sum_of_digits += rem ; n = n / 10 ; } return sum_of_digits ; } int count ( int n ) { int c = 0 ; for ( int i = n - 97 ; i <= n ; i ++ ) { int a = sum ( i ) ; int b = sum ( a ) ; if ( ( i + a + b ) == n ) { c += 1 ; } } return c ; } int main ( ) { int n = 9939 ; cout << count ( n ) << endl ; return 0 ; } |
Check if a number is power of k using base changing method | CPP program to check if a number can be raised to k ; loop to change base n to base = k ; Find current digit in base k ; If digit is neither 0 nor 1 ; Make sure that only one 1 is present . ; Driver code | #include <iostream> NEW_LINE #include <algorithm> NEW_LINE using namespace std ; bool isPowerOfK ( unsigned int n , unsigned int k ) { bool oneSeen = false ; while ( n > 0 ) { int digit = n % k ; if ( digit > 1 ) return false ; if ( digit == 1 ) { if ( oneSeen ) return false ; oneSeen = true ; } n /= k ; } return true ; } int main ( ) { int n = 64 , k = 4 ; if ( isPowerOfK ( n , k ) ) cout << " Yes " ; else cout << " No " ; } |
Check if number is palindrome or not in Octal | C ++ program to check if octal representation of a number is prime ; Function to Check no is in octal or not ; Function To check no is palindrome or not ; If number is already in octal , we traverse digits using repeated division with 10. Else we traverse digits using repeated division with 8 ; To store individual digits ; Traversing all digits ; checking if octal no is palindrome ; Driver code | #include <iostream> NEW_LINE using namespace std ; const int MAX_DIGITS = 20 ; bool isOctal ( long n ) { while ( n ) { if ( ( n % 10 ) >= 8 ) return false ; else n = n / 10 ; } return true ; } int isPalindrome ( long n ) { int divide = ( isOctal ( n ) == false ) ? 8 : 10 ; int octal [ MAX_DIGITS ] ; int i = 0 ; while ( n != 0 ) { octal [ i ++ ] = n % divide ; n = n / divide ; } for ( int j = i - 1 , k = 0 ; k <= j ; j -- , k ++ ) if ( octal [ j ] != octal [ k ] ) return false ; return true ; } int main ( ) { long n = 97 ; if ( isPalindrome ( n ) ) cout << " Yes " ; else cout << " No " ; return 0 ; } |
Find all factorial numbers less than or equal to n | CPP program to find all factorial numbers smaller than or equal to n . ; Compute next factorial using previous ; Driver code | #include <iostream> NEW_LINE using namespace std ; void printFactorialNums ( int n ) { int fact = 1 ; int x = 2 ; while ( fact <= n ) { cout << fact << " β " ; fact = fact * x ; x ++ ; } } int main ( ) { int n = 100 ; printFactorialNums ( n ) ; return 0 ; } |
Happy Numbers | CPP program to check if a number is happy number ; Returns sum of squares of digits of a number n . For example for n = 12 it returns 1 + 4 = 5 ; Returns true if n is Happy number else returns false . ; A set to store numbers during repeated square sum process ; Keep replacing n with sum of squares of digits until we either reach 1 or we endup in a cycle ; Number is Happy if we reach 1 ; Replace n with sum of squares of digits ; If n is already visited , a cycle is formed , means not Happy ; Mark n as visited ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int sumDigitSquare ( int n ) { int sq = 0 ; while ( n ) { int digit = n % 10 ; sq += digit * digit ; n = n / 10 ; } return sq ; } bool isHappy ( int n ) { set < int > s ; s . insert ( n ) ; while ( 1 ) { if ( n == 1 ) return true ; n = sumDigitSquare ( n ) ; if ( s . find ( n ) != s . end ( ) ) return false ; s . insert ( n ) ; } return false ; } int main ( ) { int n = 23 ; if ( isHappy ( n ) ) cout << " Yes " << endl ; else cout << " No " << endl ; return 0 ; } |
Check whether a number has exactly three distinct factors or not | C ++ program to check whether number has exactly three distinct factors or not ; Utility function to check whether a number is prime or not ; Corner cases ; This is checked so that we can skip middle five numbers in below loop ; Function to check whether given number has three distinct factors or not ; Find square root of number ; Check whether number is perfect square or not ; If number is perfect square , check whether square root is prime or not ; Driver program | #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 ; } bool isThreeDisctFactors ( long long n ) { int sq = ( int ) sqrt ( n ) ; if ( 1LL * sq * sq != n ) return false ; return isPrime ( sq ) ? true : false ; } int main ( ) { long long num = 9 ; if ( isThreeDisctFactors ( num ) ) cout << " Yes STRNEWLINE " ; else cout << " No STRNEWLINE " ; num = 15 ; if ( isThreeDisctFactors ( num ) ) cout << " Yes STRNEWLINE " ; else cout << " No STRNEWLINE " ; num = 12397923568441 ; if ( isThreeDisctFactors ( num ) ) cout << " Yes STRNEWLINE " ; else cout << " No STRNEWLINE " ; return 0 ; } |
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