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Surface Area and Volume of Hexagonal Prism | C ++ program to find the Surface Area and Volume of Hexagonal Prism . ; Function to calculate Surface area ; Formula to calculate surface area ; Display surface area ; Function to calculate Volume ; formula to calculate Volume ; Display Volume ; Driver Code ; surface area function call ; volume function call | #include <bits/stdc++.h> NEW_LINE using namespace std ; void findSurfaceArea ( float a , float h ) { float Area ; Area = 6 * a * h + 3 * sqrt ( 3 ) * a * a ; cout << " Surface ▁ Area : ▁ " << Area ; cout << " STRNEWLINE " ; } void findVolume ( float a , float h ) { float Volume ; Volume = 3 * sqrt ( 3 ) * a * a * h / 2 ; cout << " Volume : ▁ " << Volume ; } int main ( ) { float a = 5 , h = 10 ; findSurfaceArea ( a , h ) ; findVolume ( a , h ) ; return 0 ; } |
Minimum number of mails required to distribute all the questions | C ++ code to find the minimum number of mails ; Function returns the min no of mails required ; Using the formula derived above ; Driver Code ; no of questions ; no of students ; maximum no of questions a mail can hold ; Calling function | #include <bits/stdc++.h> NEW_LINE #define ll long long int NEW_LINE using namespace std ; long long int MinimumMail ( int n , int k , int x ) { ll m = ( n - 1 ) + ( ll ) ceil ( ( n - 1 ) * 1.0 / x ) * ( n - 1 ) + ( ll ) ceil ( n * 1.0 / x ) * ( k - n ) ; return m ; } int main ( ) { int N = 4 ; int K = 9 ; int X = 2 ; cout << MinimumMail ( N , K , X ) << endl ; return 0 ; } |
Program to find the Area of an Ellipse | C ++ program to find area of an Ellipse . ; Function to find area of an ellipse . ; formula to find the area of an Ellipse . ; Display the result ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; void findArea ( float a , float b ) { float Area ; Area = 3.142 * a * b ; cout << " Area : ▁ " << Area ; } int main ( ) { float a = 5 , b = 4 ; findArea ( a , b ) ; return 0 ; } |
Compute power of power k times % m | C ++ program for computing x ^ x ^ x ^ x . . % m ; Function to compute the given value ; compute power k times ; Driver Code ; Calling function | #include <bits/stdc++.h> NEW_LINE using namespace std ; int calculate ( int x , int k , int m ) { int result = x ; k -- ; while ( k -- ) { result = pow ( result , x ) ; if ( result > m ) result %= m ; } return result ; } int main ( ) { int x = 5 , k = 2 , m = 3 ; cout << calculate ( x , k , m ) ; return 0 ; } |
Recursive program to check if number is palindrome or not | Recursive C ++ program to check if the number is palindrome or not ; recursive function that returns the reverse of digits ; base case ; stores the reverse of a number ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int rev ( int n , int temp ) { if ( n == 0 ) return temp ; temp = ( temp * 10 ) + ( n % 10 ) ; return rev ( n / 10 , temp ) ; } int main ( ) { int n = 121 ; int temp = rev ( n , 0 ) ; if ( temp == n ) cout << " yes " << endl ; else cout << " no " << endl ; return 0 ; } |
Program to find greater value between a ^ n and b ^ n | C ++ code for finding greater between the a ^ n and b ^ n ; Function to find the greater value ; If n is even ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; void findGreater ( int a , int b , int n ) { if ( ! ( n & 1 ) ) { a = abs ( a ) ; b = abs ( b ) ; } if ( a == b ) cout << " a ^ n ▁ is ▁ equal ▁ to ▁ b ^ n " ; else if ( a > b ) cout << " a ^ n ▁ is ▁ greater ▁ than ▁ b ^ n " ; else cout << " b ^ n ▁ is ▁ greater ▁ than ▁ a ^ n " ; } int main ( ) { int a = 12 , b = 24 , n = 5 ; findGreater ( a , b , n ) ; return 0 ; } |
Print first n Fibonacci Numbers using direct formula | C ++ code to print fibonacci numbers till n using direct formula . ; Function to calculate fibonacci using recurrence relation formula ; Using direct formula ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; void fibonacci ( int n ) { long long int fib ; for ( long long int i = 0 ; i < n ; i ++ ) { fib = ( pow ( ( 1 + sqrt ( 5 ) ) , i ) - pow ( ( 1 - sqrt ( 5 ) ) , i ) ) / ( pow ( 2 , i ) * sqrt ( 5 ) ) ; cout << fib << " ▁ " ; } } int main ( ) { long long int n = 8 ; fibonacci ( n ) ; return 0 ; } |
Centered Hexadecagonal Number | C ++ Program to find nth centered hexadecagonal number ; centered hexadecagonal function ; Formula to calculate nth centered hexadecagonal number ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int center_hexadecagonal_num ( long int n ) { return 8 * n * n - 8 * n + 1 ; } int main ( ) { long int n = 2 ; cout << n << " th ▁ centered ▁ hexadecagonal ▁ number ▁ : ▁ " << center_hexadecagonal_num ( n ) ; cout << endl ; n = 12 ; cout << n << " th ▁ centered ▁ hexadecagonal ▁ numbe ▁ : ▁ " << center_hexadecagonal_num ( n ) ; return 0 ; } |
Check if the n | CPP Program to check if the nth is odd or even in a sequence where each term is sum of previous two term ; Return if the nth term is even or odd . ; Return true if odd ; Driven Program | #include <bits/stdc++.h> NEW_LINE using namespace std ; #define MAX 100 NEW_LINE bool findNature ( int a , int b , int n ) { int seq [ MAX ] = { 0 } ; seq [ 0 ] = a ; seq [ 1 ] = b ; for ( int i = 2 ; i <= n ; i ++ ) seq [ i ] = seq [ i - 1 ] + seq [ i - 2 ] ; return ( seq [ n ] & 1 ) ; } int main ( ) { int a = 2 , b = 4 ; int n = 3 ; ( findNature ( a , b , n ) ? ( cout << " Odd " << " ▁ " ) : ( cout << " Even " << " ▁ " ) ) ; return 0 ; } |
Program to compare m ^ n and n ^ m | CPP program to compare which is greater m ^ n or n ^ m ; function to compare m ^ n and n ^ m ; m ^ n ; n ^ m ; Drivers Code ; function call to compare m ^ n and n ^ m | #include <bits/stdc++.h> NEW_LINE using namespace std ; void check ( unsigned long long m , unsigned long long int n ) { double RHS = m * ( double ) log ( n ) ; double LHS = n * ( double ) log ( m ) ; if ( LHS > RHS ) cout << " m ^ n ▁ > ▁ n ^ m " ; else if ( LHS < RHS ) cout << " m ^ n ▁ < ▁ n ^ m " ; else cout << " m ^ n ▁ = ▁ n ^ m " ; } int main ( ) { unsigned long long m = 987654321 , n = 123456987 ; check ( m , n ) ; return 0 ; } |
Hilbert Matrix | C ++ program for Hilbert Matrix ; Function that generates a Hilbert matrix ; using the formula to generate hilbert matrix ; driver function | #include <bits/stdc++.h> NEW_LINE using namespace std ; void printMatrix ( int n ) { float H [ n ] [ n ] ; for ( int i = 0 ; i < n ; i ++ ) { for ( int j = 0 ; j < n ; j ++ ) { H [ i ] [ j ] = ( float ) 1.0 / ( ( i + 1 ) + ( j + 1 ) - 1.0 ) ; } } for ( int i = 0 ; i < n ; i ++ ) { for ( int j = 0 ; j < n ; j ++ ) cout << H [ i ] [ j ] << " ▁ " ; cout << endl ; } } int main ( ) { int n = 3 ; printMatrix ( n ) ; return 0 ; } |
Find the GCD that lies in given range | CPP Program to find the Greatest Common divisor of two number which is in given range ; Return the greatest common divisor of two numbers ; Return the gretest common divisor of a and b which lie in the given range . ; Loop from 1 to sqrt ( GCD ( a , b ) . ; if i divides the GCD ( a , b ) , then find maximum of three numbers res , i and g / i ; Driven Program | #include <bits/stdc++.h> NEW_LINE using namespace std ; int gcd ( int a , int b ) { if ( b == 0 ) return a ; return gcd ( b , a % b ) ; } int maxDivisorRange ( int a , int b , int l , int h ) { int g = gcd ( a , b ) ; int res = -1 ; for ( int i = l ; i * i <= g && i <= h ; i ++ ) if ( g % i == 0 ) res = max ( { res , i , g / i } ) ; return res ; } int main ( ) { int a = 3 , b = 27 , l = 1 , h = 5 ; cout << maxDivisorRange ( a , b , l , h ) << endl ; return 0 ; } |
Number expressed as sum of five consecutive integers | CPP Program to check if a number can be expressed as sum of five consecutive integers . ; function to check if a number can be expressed as sum of five consecutive integers . ; 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 - 4 ; check if sum of five 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 << " - 2 ▁ - 1 ▁ 0 ▁ 1 ▁ 2" << endl ; return ; } int inc ; if ( n > 0 ) inc = 1 ; else inc = -1 ; for ( int i = 0 ; i <= n - 4 ; i += inc ) { if ( i + i + 1 + i + 2 + i + 3 + i + 4 == n ) { cout << i << " ▁ " << i + 1 << " ▁ " << i + 2 << " ▁ " << i + 3 << " ▁ " << i + 4 ; return ; } } cout << " - 1" ; } int main ( ) { int n = 15 ; checksum ( n ) ; return 0 ; } |
Number expressed as sum of five consecutive integers | CPP Program to check if a number can be expressed as sum of five consecutive integer . ; function to check if a number can be expressed as sum of five consecutive integers . ; if n is multiple of 5 ; else print " - 1" . ; Driver Program | #include <bits/stdc++.h> NEW_LINE using namespace std ; void checksum ( int n ) { if ( n % 5 == 0 ) cout << n / 5 - 2 << " ▁ " << n / 5 - 1 << " ▁ " << n / 5 << " ▁ " << n / 5 + 1 << " ▁ " << n / 5 + 2 ; else cout << " - 1" ; } int main ( ) { int n = 15 ; checksum ( n ) ; return 0 ; } |
Number of Transpositions in a Permutation | CPP Program to find the number of transpositions in a permutation ; This array stores which element goes to which position ; This function returns the size of a component cycle ; If it is already visited ; This functio returns the number of transpositions in the permutation ; Initializing visited [ ] array ; building the goesTo [ ] array ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; #define N 1000001 NEW_LINE int visited [ N ] ; int goesTo [ N ] ; int dfs ( int i ) { if ( visited [ i ] == 1 ) return 0 ; visited [ i ] = 1 ; int x = dfs ( goesTo [ i ] ) ; return ( x + 1 ) ; } int noOfTranspositions ( int P [ ] , int n ) { for ( int i = 1 ; i <= n ; i ++ ) visited [ i ] = 0 ; for ( int i = 0 ; i < n ; i ++ ) goesTo [ P [ i ] ] = i + 1 ; int transpositions = 0 ; for ( int i = 1 ; i <= n ; i ++ ) { if ( visited [ i ] == 0 ) { int ans = dfs ( i ) ; transpositions += ans - 1 ; } } return transpositions ; } int main ( ) { int permutation [ ] = { 5 , 1 , 4 , 3 , 2 } ; int n = sizeof ( permutation ) / sizeof ( permutation [ 0 ] ) ; cout << noOfTranspositions ( permutation , n ) ; return 0 ; } |
n | CPP program to find n - th term of series ; Function to find the nth term of series ; Loop to add 4 th powers ; Driver code | #include <iostream> NEW_LINE using namespace std ; int sumOfSeries ( int n ) { int ans = 0 ; for ( int i = 1 ; i <= n ; i ++ ) ans += i * i * i * i ; return ans ; } int main ( ) { int n = 4 ; cout << sumOfSeries ( n ) ; return 0 ; } |
Number of unmarked integers in a special sieve | C ++ Program to determine the number of unmarked integers in a special sieve ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int countUnmarked ( int N ) { if ( N % 2 == 0 ) return N / 2 ; else return N / 2 + 1 ; } int main ( ) { int N = 4 ; cout << " Number ▁ of ▁ unmarked ▁ elements : ▁ " << countUnmarked ( N ) << endl ; return 0 ; } |
Sum of series 1 * 1 ! + 2 * 2 ! + …… . . + n * n ! | CPP program to find sum of the series . ; Function to calculate required series ; Drivers code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int factorial ( int n ) { int res = 1 ; for ( int i = 2 ; i <= n ; i ++ ) res = res * i ; return res ; } int calculateSeries ( int n ) { return factorial ( n + 1 ) - 1 ; } int main ( ) { int n = 3 ; cout << calculateSeries ( n ) ; return 0 ; } |
Sum of series 1 * 1 * 2 ! + 2 * 2 * 3 ! + …… . . + n * n * ( n + 1 ) ! | CPP program to find sum of the series . ; Function to calculate required series ; Drivers code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int factorial ( int n ) { int res = 1 ; for ( int i = 2 ; i <= n ; i ++ ) res = res * i ; return res ; } int calculateSeries ( int n ) { return 2 + ( n * n + n - 2 ) * factorial ( n + 1 ) ; } int main ( ) { int n = 3 ; cout << calculateSeries ( n ) ; return 0 ; } |
Aspiring Number | C ++ implementation to check whether a number is aspiring or not ; Function to calculate sum of all proper divisors ; Note that this loop runs till square root of n ; If divisors are equal , take only one of them ; else Otherwise take both ; calculate sum of all proper divisors only ; Function to get last number of Aliquot Sequence . ; Calculate next term from previous term ; Returns true if n is perfect ; To store sum of divisors ; Find all divisors and add them ; If sum of divisors is equal to n , then n is a perfect number ; Returns true if n is aspiring else returns false ; checking condition for aspiring ; Driver program | #include <bits/stdc++.h> NEW_LINE using namespace std ; int getSum ( int n ) { for ( int i = 1 ; i <= sqrt ( n ) ; i ++ ) { if ( n % i == 0 ) { if ( n / i == i ) sum = sum + i ; { sum = sum + i ; sum = sum + ( n / i ) ; } } } return sum - n ; } int getAliquot ( int n ) { unordered_set < int > s ; s . insert ( n ) ; int next = 0 ; while ( n > 0 ) { n = getSum ( n ) ; if ( s . find ( n ) != s . end ( ) ) return n ; s . insert ( n ) ; } return 0 ; } bool isPerfect ( int n ) { long long int sum = 1 ; for ( long long int i = 2 ; i * i <= n ; i ++ ) if ( n % i == 0 ) sum = sum + i + n / i ; if ( sum == n && n != 1 ) return true ; return false ; } bool isAspiring ( int n ) { int alq = getAliquot ( n ) ; if ( isPerfect ( alq ) && ! isPerfect ( n ) ) return true ; else return false ; } int main ( ) { int n = 25 ; if ( isAspiring ( n ) ) cout << " Aspiring " << endl ; else cout << " Not ▁ Aspiring " << endl ; return 0 ; } |
Forming smallest array with given constraints | C ++ program to find the length of smallest array begin with x , having y , ends with z and having absolute difference between adjacent elements <= 1. ; Return the size of smallest array with given constraint . ; Drivers code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int minimumLength ( int x , int y , int z ) { return 1 + abs ( x - y ) + abs ( y - z ) ; } int main ( ) { int x = 3 , y = 1 , z = 2 ; cout << minimumLength ( x , y , z ) ; return 0 ; } |
Find the other | CPP program to find the other - end point of diameter ; function to find the other - end point of diameter ; find end point for x coordinates ; find end point for y coordinates ; Driven Program | #include <iostream> NEW_LINE using namespace std ; void endPointOfDiameterofCircle ( int x1 , int y1 , int c1 , int c2 ) { cout << " x2 ▁ = ▁ " << ( float ) ( 2 * c1 - x1 ) << " ▁ " ; cout << " y2 ▁ = ▁ " << ( float ) ( 2 * c2 - y1 ) ; } int main ( ) { int x1 = -4 , y1 = -1 ; int c1 = 3 , c2 = 5 ; endPointOfDiameterofCircle ( x1 , y1 , c1 , c2 ) ; return 0 ; } |
Newton 's Divided Difference Interpolation Formula | CPP program for implementing Newton divided difference formula ; Function to find the product term ; Function for calculating divided difference table ; Function for applying Newton 's divided difference formula ; Function for displaying divided difference table ; Driver Function ; number of inputs given ; y [ ] [ ] is used for divided difference table where y [ ] [ 0 ] is used for input ; calculating divided difference table ; displaying divided difference table ; value to be interpolated ; printing the value | #include <bits/stdc++.h> NEW_LINE using namespace std ; float proterm ( int i , float value , float x [ ] ) { float pro = 1 ; for ( int j = 0 ; j < i ; j ++ ) { pro = pro * ( value - x [ j ] ) ; } return pro ; } void dividedDiffTable ( float x [ ] , float y [ ] [ 10 ] , int n ) { for ( int i = 1 ; i < n ; i ++ ) { for ( int j = 0 ; j < n - i ; j ++ ) { y [ j ] [ i ] = ( y [ j ] [ i - 1 ] - y [ j + 1 ] [ i - 1 ] ) / ( x [ j ] - x [ i + j ] ) ; } } } float applyFormula ( float value , float x [ ] , float y [ ] [ 10 ] , int n ) { float sum = y [ 0 ] [ 0 ] ; for ( int i = 1 ; i < n ; i ++ ) { sum = sum + ( proterm ( i , value , x ) * y [ 0 ] [ i ] ) ; } return sum ; } void printDiffTable ( float y [ ] [ 10 ] , int n ) { for ( int i = 0 ; i < n ; i ++ ) { for ( int j = 0 ; j < n - i ; j ++ ) { cout << setprecision ( 4 ) << y [ i ] [ j ] << " TABSYMBOL ▁ " ; } cout << " STRNEWLINE " ; } } int main ( ) { int n = 4 ; float value , sum , y [ 10 ] [ 10 ] ; float x [ ] = { 5 , 6 , 9 , 11 } ; y [ 0 ] [ 0 ] = 12 ; y [ 1 ] [ 0 ] = 13 ; y [ 2 ] [ 0 ] = 14 ; y [ 3 ] [ 0 ] = 16 ; dividedDiffTable ( x , y , n ) ; printDiffTable ( y , n ) ; value = 7 ; cout << " Value at " ▁ < < ▁ value ▁ < < ▁ " is " << applyFormula ( value , x , y , n ) << endl ; return 0 ; } |
Centered heptagonal number | CPP program to find n - th Centered heptagonal number ; Function to find Centered heptagonal number ; Formula to calculate nth Centered heptagonal number ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int centered_heptagonal_num ( long int n ) { return ( 7 * n * n - 7 * n + 2 ) / 2 ; } int main ( ) { long int n = 5 ; cout << n << " th ▁ Centered ▁ heptagonal ▁ number ▁ : ▁ " ; cout << centered_heptagonal_num ( n ) ; return 0 ; } |
Sum of square | CPP Program to find the sum of sum of squares of first n natural number ; Function to find sum of sum of square of first n natural number ; Driven Program | #include <bits/stdc++.h> NEW_LINE using namespace std ; int findSum ( int n ) { int sum = 0 ; for ( int i = 1 ; i <= n ; i ++ ) sum += ( ( i * ( i + 1 ) * ( 2 * i + 1 ) ) / 6 ) ; return sum ; } int main ( ) { int n = 3 ; cout << findSum ( n ) << endl ; return 0 ; } |
Check if a given matrix is Hankel or not | C ++ Program to check if given matrix is Hankel Matrix or not . ; Function to check if given matrix is Hankel Matrix or not . ; for each row ; for each column ; checking if i + j is less than n ; checking if the element is equal to the corresponding diagonal constant ; checking if the element is equal to the corresponding diagonal constant ; Drivers code | #include <bits/stdc++.h> NEW_LINE using namespace std ; #define N 4 NEW_LINE bool checkHankelMatrix ( int n , int m [ N ] [ N ] ) { for ( int i = 0 ; i < n ; i ++ ) { for ( int j = 0 ; j < n ; j ++ ) { if ( i + j < n ) { if ( m [ i ] [ j ] != m [ i + j ] [ 0 ] ) return false ; } else { if ( m [ i ] [ j ] != m [ i + j - n + 1 ] [ n - 1 ] ) return false ; } } } return true ; } int main ( ) { int n = 4 ; int m [ N ] [ N ] = { { 1 , 2 , 3 , 5 } , { 2 , 3 , 5 , 8 } , { 3 , 5 , 8 , 0 } , { 5 , 8 , 0 , 9 } } ; checkHankelMatrix ( n , m ) ? ( cout << " Yes " ) : ( cout << " No " ) ; return 0 ; } |
Check if a number can be expressed as power | Set 2 ( Using Log ) | CPP program to find if a number can be expressed as x raised to power y . ; Find Log n in different bases and check if the value is an integer ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool isPower ( unsigned int n ) { for ( int x = 2 ; x <= sqrt ( n ) ; x ++ ) { float f = log ( n ) / log ( x ) ; if ( ( f - ( int ) f ) == 0.0 ) return true ; } return false ; } int main ( ) { for ( int i = 2 ; i < 100 ; i ++ ) if ( isPower ( i ) ) cout << i << " ▁ " ; return 0 ; } |
Queries on sum of odd number digit sums of all the factors of a number | CPP Program to answer queries on sum of sum of odd number digits of all the factors of a number ; finding sum of odd digit number in each integer . ; for each number ; using previous number sum , finding the current number num of odd digit also , adding last digit if it is odd . ; finding sum of sum of odd digit of all the factors of a number . ; for each possible factor ; adding the contribution . ; Wrapper function ; Driver Program | #include <bits/stdc++.h> NEW_LINE using namespace std ; #define N 1000005 NEW_LINE void sumOddDigit ( int digitSum [ ] ) { for ( int i = 1 ; i < N ; i ++ ) { digitSum [ i ] = digitSum [ i / 10 ] + ( i & 1 ) * ( i % 10 ) ; } } void sumFactor ( int digitSum [ ] , int factorDigitSum [ ] ) { for ( int i = 1 ; i < N ; i ++ ) { for ( int j = i ; j < N ; j += i ) { factorDigitSum [ j ] += digitSum [ i ] ; } } } void wrapper ( int q , int n [ ] ) { int digitSum [ N ] ; int factorDigitSum [ N ] ; sumOddDigit ( digitSum ) ; sumFactor ( digitSum , factorDigitSum ) ; for ( int i = 0 ; i < q ; i ++ ) cout << factorDigitSum [ n [ i ] ] << " ▁ " ; } int main ( ) { int q = 2 ; int n [ ] = { 10 , 36 } ; wrapper ( q , n ) ; return 0 ; } |
Program for Gauss | C ++ Implementation for Gauss - Jordan Elimination Method ; Function to print the matrix ; function to reduce matrix to reduced row echelon form . ; Performing elementary operations ; Excluding all i == j ; Converting Matrix to reduced row echelon form ( diagonal matrix ) ; Function to print the desired result if unique solutions exists , otherwise prints no solution or infinite solutions depending upon the input given . ; Printing the solution by dividing constants by their respective diagonal elements ; To check whether infinite solutions exists or no solution exists ; flag == 2 for infinite solution flag == 3 for No solution ; Driver code ; Order of Matrix ( n ) ; Performing Matrix transformation ; Printing Final Matrix ; Printing Solutions ( if exist ) | #include <bits/stdc++.h> NEW_LINE using namespace std ; #define M 10 NEW_LINE void PrintMatrix ( float a [ ] [ M ] , int n ) { for ( int i = 0 ; i < n ; i ++ ) { for ( int j = 0 ; j <= n ; j ++ ) cout << a [ i ] [ j ] << " ▁ " ; cout << endl ; } } int PerformOperation ( float a [ ] [ M ] , int n ) { int i , j , k = 0 , c , flag = 0 , m = 0 ; float pro = 0 ; for ( i = 0 ; i < n ; i ++ ) { if ( a [ i ] [ i ] == 0 ) { c = 1 ; while ( ( i + c ) < n && a [ i + c ] [ i ] == 0 ) c ++ ; if ( ( i + c ) == n ) { flag = 1 ; break ; } for ( j = i , k = 0 ; k <= n ; k ++ ) swap ( a [ j ] [ k ] , a [ j + c ] [ k ] ) ; } for ( j = 0 ; j < n ; j ++ ) { if ( i != j ) { float pro = a [ j ] [ i ] / a [ i ] [ i ] ; for ( k = 0 ; k <= n ; k ++ ) a [ j ] [ k ] = a [ j ] [ k ] - ( a [ i ] [ k ] ) * pro ; } } } return flag ; } void PrintResult ( float a [ ] [ M ] , int n , int flag ) { cout << " Result ▁ is ▁ : ▁ " ; if ( flag == 2 ) cout << " Infinite ▁ Solutions ▁ Exists " << endl ; else if ( flag == 3 ) cout << " No ▁ Solution ▁ Exists " << endl ; else { for ( int i = 0 ; i < n ; i ++ ) cout << a [ i ] [ n ] / a [ i ] [ i ] << " ▁ " ; } } int CheckConsistency ( float a [ ] [ M ] , int n , int flag ) { int i , j ; float sum ; flag = 3 ; for ( i = 0 ; i < n ; i ++ ) { sum = 0 ; for ( j = 0 ; j < n ; j ++ ) sum = sum + a [ i ] [ j ] ; if ( sum == a [ i ] [ j ] ) flag = 2 ; } return flag ; } int main ( ) { float a [ M ] [ M ] = { { 0 , 2 , 1 , 4 } , { 1 , 1 , 2 , 6 } , { 2 , 1 , 1 , 7 } } ; int n = 3 , flag = 0 ; flag = PerformOperation ( a , n ) ; if ( flag == 1 ) flag = CheckConsistency ( a , n , flag ) ; cout << " Final ▁ Augumented ▁ Matrix ▁ is ▁ : ▁ " << endl ; PrintMatrix ( a , n ) ; cout << endl ; PrintResult ( a , n , flag ) ; return 0 ; } |
Number of digits in the nth number made of given four digits | C ++ program to count number of digits in n - th number made of given four digits . ; Efficient function to calculate number of digits in the nth number constructed by using 6 , 1 , 4 and 9 as digits in the ascending order . ; Number of digits increase after every i - th number where i increases in powers of 4. ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int number_of_digits ( int n ) { int i , res , sum = 0 ; for ( i = 4 , res = 1 ; ; i *= 4 , res ++ ) { sum += i ; if ( sum >= n ) break ; } return res ; } int main ( ) { int n = 21 ; cout << number_of_digits ( n ) << endl ; return 0 ; } |
Print prime numbers from 1 to N in reverse order | C ++ program to print all primes between 1 to N in reverse order using Sieve of Eratosthenes ; Create a boolean array " prime [ 0 . . n ] " and initialize all entries it as true . A value in prime [ i ] will finally be false if i is Not a prime , else true . ; If prime [ p ] is not changed , then it is a prime ; Update all multiples of p ; Print all prime numbers in reverse order ; Driver Program ; static input ; to display ; Reverseorder ( N ) ; calling the function | #include <bits/stdc++.h> NEW_LINE using namespace std ; void Reverseorder ( int n ) { bool prime [ n + 1 ] ; memset ( prime , true , sizeof ( prime ) ) ; for ( int p = 2 ; p * p <= n ; p ++ ) { if ( prime [ p ] == true ) { for ( int i = p * 2 ; i <= n ; i += p ) prime [ i ] = false ; } } for ( int p = n ; p >= 2 ; p -- ) if ( prime [ p ] ) cout << p << " ▁ " ; } int main ( ) { int N = 25 ; cout << " Prime ▁ number ▁ in ▁ reverse ▁ order " << endl ; if ( N == 1 ) cout << " No ▁ prime ▁ no ▁ exist ▁ in ▁ this ▁ range " ; else return 0 ; } |
Vantieghems Theorem for Primality Test | C ++ code to verify Vantieghem 's Theorem ; Check if above condition is satisfied ; product of previous powers of 2 ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; void checkVantieghemsTheorem ( int limit ) { long long unsigned prod = 1 ; for ( long long unsigned n = 2 ; n < limit ; n ++ ) { if ( ( ( prod - n ) % ( ( 1LL << n ) - 1 ) ) == 0 ) cout << n << " ▁ is ▁ prime STRNEWLINE " ; prod *= ( ( 1LL << n ) - 1 ) ; } } int main ( ) { checkVantieghemsTheorem ( 10 ) ; return 0 ; } |
Count numbers formed by given two digit with sum having given digits | C ++ program to count the number of numbers formed by digits a and b exactly of a length N such that the sum of the digits of the number thus formed is of digits a and b . ; function to check if sum of digits is made of a and b ; sum of digits is 0 ; if any of digits in sum is other than a and b ; calculate the modInverse V / of a number in O ( log n ) ; q is quotient ; m is remainder now , process same as Euclid 's algo ; Update y and x ; Make x positive ; function to pregenerate factorials ; function to pre calculate the modInverse of factorials ; calculates the modInverse of the last factorial ; precalculates the modInverse of all factorials by formulae ; function that returns the value of nCi ; function that returns the count of numbers ; function call to pre - calculate the factorials and modInverse of factorials ; if a and b are same ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; const int mod = 1e9 + 7 ; const int N = 1000005 ; int fact [ N ] , invfact [ N ] ; int check ( int x , int a , int b ) { if ( x == 0 ) return 0 ; while ( x ) { if ( x % 10 != a and x % 10 != b ) return 0 ; x /= 10 ; } return 1 ; } int modInverse ( int a , int m ) { int m0 = m ; int y = 0 , x = 1 ; if ( m == 1 ) return 0 ; while ( a > 1 ) { int q = a / m ; int t = m ; m = a % m , a = t ; t = y ; y = x - q * y ; x = t ; } if ( x < 0 ) x += m0 ; return x ; } void pregenFact ( ) { fact [ 0 ] = fact [ 1 ] = 1 ; for ( int i = 1 ; i <= 1000000 ; ++ i ) fact [ i ] = ( long long ) fact [ i - 1 ] * i % mod ; } void pregenInverse ( ) { invfact [ 0 ] = invfact [ 1 ] = 1 ; invfact [ 1000000 ] = modInverse ( fact [ 1000000 ] , mod ) ; for ( int i = 999999 ; i > 1 ; -- i ) invfact [ i ] = ( ( long long ) invfact [ i + 1 ] * ( long long ) ( i + 1 ) ) % mod ; } int comb ( int big , int small ) { return ( long long ) fact [ big ] * invfact [ small ] % mod * invfact [ big - small ] % mod ; } int count ( int a , int b , int n ) { pregenFact ( ) ; pregenInverse ( ) ; if ( a == b ) return ( check ( a * n , a , b ) ) ; int ans = 0 ; for ( int i = 0 ; i <= n ; ++ i ) if ( check ( i * a + ( n - i ) * b , a , b ) ) ans = ( ans + comb ( n , i ) ) % mod ; return ans ; } int main ( ) { int a = 3 , b = 4 , n = 11028 ; cout << count ( a , b , n ) ; return 0 ; } |
Finding n | C ++ program to find n - th term of series 3 , 13 , 42 , 108 , 235. . . ; Function to generate a fixed number ; Driver Method | #include <bits/stdc++.h> NEW_LINE using namespace std ; int magicOfSequence ( int N ) { int sum = 0 ; for ( int i = 1 ; i <= N ; i ++ ) sum += ( i * i * i + i * 2 ) ; return sum ; } int main ( ) { int N = 4 ; cout << magicOfSequence ( N ) << endl ; return 0 ; } |
Expressing a number as sum of consecutive | Set 2 ( Using odd factors ) | C ++ program to count number of ways to express N as sum of consecutive numbers . ; returns the number of odd factors ; If i is an odd factor and n is a perfect square ; If n is not perfect square ; Driver Code ; N as sum of consecutive numbers | #include <bits/stdc++.h> NEW_LINE using namespace std ; int countOddFactors ( long long n ) { int odd_factors = 0 ; for ( int i = 1 ; 1ll * i * i <= n ; i ++ ) { if ( n % i == 0 ) { if ( 1ll * i * i == n ) { if ( i & 1 ) odd_factors ++ ; } else { if ( i & 1 ) odd_factors ++ ; int factor = n / i ; if ( factor & 1 ) odd_factors ++ ; } } } return odd_factors - 1 ; } int main ( ) { long long int N = 15 ; cout << countOddFactors ( N ) << endl ; N = 10 ; cout << countOddFactors ( N ) << endl ; return 0 ; } |
Making zero array by decrementing pairs of adjacent | CPP program to find if it is possible to make all array elements 0 by decrement operations . ; used for storing the sum of even and odd position element in array . ; if position is odd , store sum value of odd position in odd ; if position is even , store sum value of even position in even ; Driver program | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool isPossibleToZero ( int a [ ] , int n ) { int even = 0 , odd = 0 ; for ( int i = 0 ; i < n ; i ++ ) { if ( i & 1 ) odd += a [ i ] ; else even += a [ i ] ; } return ( odd == even ) ; } int main ( ) { int arr [ ] = { 0 , 1 , 1 , 0 } ; int n = sizeof ( arr ) / sizeof ( arr [ 0 ] ) ; if ( isPossibleToZero ( arr , n ) ) cout << " YES " ; else cout << " NO " ; } |
Program for sum of cos ( x ) series | CPP program to find the sum of cos ( x ) series ; here x is in degree . we have to convert it to radian for using it with series formula , as in series expansion angle is in radian ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; const double PI = 3.142 ; double cosXSertiesSum ( double x , int n ) { x = x * ( PI / 180.0 ) ; double res = 1 ; double sign = 1 , fact = 1 , pow = 1 ; for ( int i = 1 ; i < 5 ; i ++ ) { sign = sign * -1 ; fact = fact * ( 2 * i - 1 ) * ( 2 * i ) ; pow = pow * x * x ; res = res + sign * pow / fact ; } return res ; } int main ( ) { float x = 50 ; int n = 5 ; cout << cosXSertiesSum ( x , 5 ) ; return 0 ; } |
Sum of digits written in different bases from 2 to n | CPP program to find sum of digits of n in different bases from 2 to n - 1. ; function to calculate sum of digit for a given base ; Sum of digits ; Calculating the number ( n ) by taking mod with the base and adding remainder to the result and parallelly reducing the num value . ; returning the result ; function calling for multiple bases ; Driver program | #include <bits/stdc++.h> NEW_LINE using namespace std ; int solve ( int n , int base ) { int result = 0 ; while ( n > 0 ) { int remainder = n % base ; result = result + remainder ; n = n / base ; } return result ; } void printSumsOfDigits ( int n ) { for ( int base = 2 ; base < n ; ++ base ) cout << solve ( n , base ) << " ▁ " ; } int main ( ) { int n = 8 ; printSumsOfDigits ( n ) ; return 0 ; } |
Possible two sets from first N natural numbers difference of sums as D | C ++ program for implementing above approach ; Function returns true if it is possible to split into two sets otherwise returns false ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool check ( int N , int D ) { int temp = ( N * ( N + 1 ) ) / 2 + D ; return ( temp % 2 == 0 ) ; } int main ( ) { int N = 5 ; int M = 7 ; if ( check ( N , M ) ) cout << " yes " ; else cout << " no " ; return 0 ; } |
Minimum digits to remove to make a number Perfect Square | C ++ program to find required minimum digits need to remove to make a number perfect square ; function to check minimum number of digits should be removed to make this number a perfect square ; size of the string ; our final answer ; to store string which is perfect square . ; We make all possible subsequences ; to check jth bit is set or not . ; we do not consider a number with leading zeros ; convert our temporary string into integer ; checking temp is perfect square or not . ; taking maximum sized string ; print PerfectSquare ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int perfectSquare ( string s ) { int n = s . size ( ) ; int ans = -1 ; string num ; for ( int i = 1 ; i < ( 1 << n ) ; i ++ ) { string str = " " ; for ( int j = 0 ; j < n ; j ++ ) { if ( ( i >> j ) & 1 ) { str += s [ j ] ; } } if ( str [ 0 ] != '0' ) { int temp = 0 ; for ( int j = 0 ; j < str . size ( ) ; j ++ ) temp = temp * 10 + ( int ) ( str [ j ] - '0' ) ; int k = sqrt ( temp ) ; if ( k * k == temp ) { if ( ans < ( int ) str . size ( ) ) { ans = ( int ) str . size ( ) ; num = str ; } } } } if ( ans == -1 ) return ans ; else { cout << num << " ▁ " ; return n - ans ; } } int main ( ) { cout << perfectSquare ( "8314" ) << endl ; cout << perfectSquare ( "753" ) << endl ; return 0 ; } |
Lagrange 's four square theorem | CPP program for Lagrange 's four square identity ; Prints all the possible combinations 4 numbers whose sum of squares is equal to the given no . ; loops checking the sum of squares ; if sum of four squares equals the given no . ; printing the numbers ; Driver Code ; 74 = 0 * 0 + 0 * 0 + 5 * 5 + 7 * 7 74 = 0 * 0 + 1 * 1 + 3 * 3 + 8 * 8 74 = 0 * 0 + 3 * 3 + 4 * 4 + 7 * 7 74 = 1 * 1 + 1 * 1 + 6 * 6 + 6 * 6 74 = 2 * 2 + 3 * 3 + 5 * 5 + 6 * 6 | #include <bits/stdc++.h> NEW_LINE using namespace std ; void printFourSquares ( int a ) { for ( int i = 0 ; i * i <= a ; i ++ ) { for ( int j = i ; j * j <= a ; j ++ ) { for ( int k = j ; k * k <= a ; k ++ ) { for ( int l = k ; l * l <= a ; l ++ ) { if ( i * i + j * j + k * k + l * l == a ) { cout << a << " ▁ = ▁ " << i << " * " << i << " ▁ + ▁ " << j << " * " << j << " ▁ + ▁ " ; cout << k << " * " << k << " ▁ + ▁ " << l << " * " << l << " STRNEWLINE " ; } } } } } } int main ( ) { int a = 74 ; printFourSquares ( a ) ; return 0 ; } |
Hardy | CPP program to count all prime factors ; A function to count prime factors of a given number n ; n must be odd at this point . So we can skip one element ( Note i = i + 2 ) ; This condition is to handle the case when n is a prime number greater than 2 ; driver function | #include <bits/stdc++.h> NEW_LINE using namespace std ; int exactPrimeFactorCount ( int n ) { int count = 0 ; if ( n % 2 == 0 ) { count ++ ; while ( n % 2 == 0 ) n = n / 2 ; } for ( int i = 3 ; i <= sqrt ( n ) ; i = i + 2 ) { if ( n % i == 0 ) { count ++ ; while ( n % i == 0 ) n = n / i ; } } if ( n > 2 ) count ++ ; return count ; } int main ( ) { int n = 51242183 ; cout << " The ▁ number ▁ of ▁ distinct ▁ prime ▁ factors ▁ is / are ▁ " << exactPrimeFactorCount ( n ) << endl ; cout << " The ▁ value ▁ of ▁ log ( log ( n ) ) ▁ is ▁ " << log ( log ( n ) ) << endl ; return 0 ; } |
Number of Digits in a ^ b | CPP Program to calculate no . of digits in a ^ b ; function to calculate number of digits in a ^ b ; driver program | #include <iostream> NEW_LINE #include <math.h> NEW_LINE using namespace std ; int no_of_digit ( int a , int b ) { return ( ( int ) ( b * log10 ( a ) ) + 1 ) ; } int main ( ) { int a = 2 , b = 100 ; cout << " no . ▁ of ▁ digits ▁ = ▁ " << no_of_digit ( a , b ) ; } |
Check whether a number is Emirpimes or not | CPP code to check whether a number is Emirpimes or not ; Checking whether a number is semi - prime or not ; Increment count of prime numbers ; If number is still greater than 1 , after exiting the for loop add it to the count variable as it indicates the number is a prime number ; Return '1' if count is equal to '2' else return '0' ; Checking whether a number is emirpimes or not ; Number itself is not semiprime . ; Finding reverse of n . ; The definition of emirpimes excludes palindromes , hence we do not check further , if the number entered is a palindrome ; Checking whether the reverse of the semi prime number entered is also a semi prime number or not ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int checkSemiprime ( int num ) { int cnt = 0 ; for ( int i = 2 ; cnt < 2 && i * i <= num ; ++ i ) { while ( num % i == 0 ) { num /= i ; ++ cnt ; } } if ( num > 1 ) ++ cnt ; return cnt == 2 ; } bool isEmirpimes ( int n ) { if ( checkSemiprime ( n ) == false ) return false ; int r = 0 ; for ( int t = n ; t != 0 ; t = t / n ) r = r * 10 + t % 10 ; if ( r == n ) return false ; return ( checkSemiprime ( r ) ) ; } int main ( ) { int n = 15 ; if ( isEmirpimes ( n ) ) cout << " Yes " ; else cout << " No " ; return 0 ; } |
Booth ’ s Multiplication Algorithm | CPP code to implement booth 's algorithm ; function to perform adding in the accumulator ; updating accumulator with A = A + BR ; function to find the number 's complement ; function to perform right shift ; function to display operations ; accumulator content ; multiplier content ; Function to implement booth 's algo ; SECOND CONDITION ; subtract BR from accumulator ; THIRD CONDITION ; add BR to accumulator ; FIRST CONDITION ; decrement counter ; driver code ; Number of multiplicand bit ; multiplicand ; copy multiplier to temp array mt [ ] ; No . of multiplier bit ; sequence counter ; multiplier | #include <bits/stdc++.h> NEW_LINE using namespace std ; void add ( int ac [ ] , int x [ ] , int qrn ) { int i , c = 0 ; for ( i = 0 ; i < qrn ; i ++ ) { ac [ i ] = ac [ i ] + x [ i ] + c ; if ( ac [ i ] > 1 ) { ac [ i ] = ac [ i ] % 2 ; c = 1 ; } else c = 0 ; } } void complement ( int a [ ] , int n ) { int i ; int x [ 8 ] = { 0 } ; x [ 0 ] = 1 ; for ( i = 0 ; i < n ; i ++ ) { a [ i ] = ( a [ i ] + 1 ) % 2 ; } add ( a , x , n ) ; } void rightShift ( int ac [ ] , int qr [ ] , int & qn , int qrn ) { int temp , i ; temp = ac [ 0 ] ; qn = qr [ 0 ] ; cout << " TABSYMBOL TABSYMBOL rightShift TABSYMBOL " ; for ( i = 0 ; i < qrn - 1 ; i ++ ) { ac [ i ] = ac [ i + 1 ] ; qr [ i ] = qr [ i + 1 ] ; } qr [ qrn - 1 ] = temp ; } void display ( int ac [ ] , int qr [ ] , int qrn ) { int i ; for ( i = qrn - 1 ; i >= 0 ; i -- ) cout << ac [ i ] ; cout << " TABSYMBOL " ; for ( i = qrn - 1 ; i >= 0 ; i -- ) cout << qr [ i ] ; } void boothAlgorithm ( int br [ ] , int qr [ ] , int mt [ ] , int qrn , int sc ) { int qn = 0 , ac [ 10 ] = { 0 } ; int temp = 0 ; cout << " qn TABSYMBOL q [ n + 1 ] TABSYMBOL TABSYMBOL BR TABSYMBOL TABSYMBOL AC TABSYMBOL QR TABSYMBOL TABSYMBOL sc STRNEWLINE " ; cout << " TABSYMBOL TABSYMBOL TABSYMBOL initial TABSYMBOL TABSYMBOL " ; display ( ac , qr , qrn ) ; cout << " TABSYMBOL TABSYMBOL " << sc << " STRNEWLINE " ; while ( sc != 0 ) { cout << qr [ 0 ] << " TABSYMBOL " << qn ; if ( ( qn + qr [ 0 ] ) == 1 ) { if ( temp == 0 ) { add ( ac , mt , qrn ) ; cout << " TABSYMBOL TABSYMBOL A ▁ = ▁ A ▁ - ▁ BR TABSYMBOL " ; for ( int i = qrn - 1 ; i >= 0 ; i -- ) cout << ac [ i ] ; temp = 1 ; } else if ( temp == 1 ) { add ( ac , br , qrn ) ; cout << " TABSYMBOL TABSYMBOL A ▁ = ▁ A ▁ + ▁ BR TABSYMBOL " ; for ( int i = qrn - 1 ; i >= 0 ; i -- ) cout << ac [ i ] ; temp = 0 ; } cout << " STRNEWLINE TABSYMBOL " ; rightShift ( ac , qr , qn , qrn ) ; } else if ( qn - qr [ 0 ] == 0 ) rightShift ( ac , qr , qn , qrn ) ; display ( ac , qr , qrn ) ; cout << " TABSYMBOL " ; sc -- ; cout << " TABSYMBOL " << sc << " STRNEWLINE " ; } } int main ( int argc , char * * arg ) { int mt [ 10 ] , sc ; int brn , qrn ; brn = 4 ; int br [ ] = { 0 , 1 , 1 , 0 } ; for ( int i = brn - 1 ; i >= 0 ; i -- ) mt [ i ] = br [ i ] ; reverse ( br , br + brn ) ; complement ( mt , brn ) ; qrn = 4 ; sc = qrn ; int qr [ ] = { 1 , 0 , 1 , 0 } ; reverse ( qr , qr + qrn ) ; boothAlgorithm ( br , qr , mt , qrn , sc ) ; cout << endl << " Result ▁ = ▁ " ; for ( int i = qrn - 1 ; i >= 0 ; i -- ) cout << qr [ i ] ; } |
Connell Sequence | CPP code to generate first ' n ' terms of Connell Sequence ; Function to generate a fixed number of even or odd terms . The size of r decides whether numbers to be generated even or odd . ; Generating the first ' n ' terms of Connell Sequence ; A dummy 0 is inserted at the beginning for consistency ; Calling function gen ( ) to generate ' k ' number of terms ; Checking if ' n ' terms are already generated ; Removing the previously inserted dummy 0 ; Driver Method | #include <bits/stdc++.h> NEW_LINE using namespace std ; vector < long long int > gen ( long long int n , vector < long long int > r ) { long long int a = r [ r . size ( ) - 1 ] ; a ++ ; for ( int i = 1 ; i <= n ; a += 2 , i ++ ) r . push_back ( a ) ; return r ; } vector < long long int > conell ( long long int n ) { vector < long long int > res ; long long int k = 1 ; res . push_back ( 0 ) ; while ( 1 ) { res = gen ( k , res ) ; k ++ ; int j = res . size ( ) - 1 ; while ( j != n && j + k > n ) k -- ; if ( j >= n ) break ; } res . erase ( res . begin ( ) ) ; return res ; } int main ( ) { long long int n = 10 ; cout << " The ▁ first ▁ " << n << " ▁ terms ▁ are " << endl ; vector < long long int > res = conell ( n ) ; for ( int i = 0 ; i < res . size ( ) ; i ++ ) cout << res [ i ] << " ▁ " ; cout << endl ; return 0 ; } |
Generate a list of n consecutive composite numbers ( An interesting method ) | CPP program to print n consecutive composite numbers . ; function to find factorial of given number ; Prints n consecutive numbers . ; Driver program to test above function | #include <iostream> NEW_LINE using namespace std ; unsigned long long int factorial ( unsigned int n ) { unsigned long long int res = 1 ; for ( int i = 2 ; i <= n ; i ++ ) res *= i ; return res ; } void printNComposite ( int n ) { unsigned long long int fact = factorial ( n + 1 ) ; for ( int i = 2 ; i <= n + 1 ; ++ i ) cout << fact + i << " ▁ " ; } int main ( ) { int n = 4 ; printNComposite ( n ) ; return 0 ; } |
Frugal Number | Program to check for Frugal number ; Finding primes upto entered number ; Finding primes by Sieve of Eratosthenes method ; If prime [ i ] is not changed , then it is prime ; Update all multiples of p ; Forming array of the prime numbers found ; Returns number of digits in n ; Checking whether a number is Frugal or not ; Finding number of digits in prime factorization of the number ; Exponent for current factor ; Counting number of times this prime factor divides ( Finding exponent ) ; Finding number of digits in the exponent Avoiding exponents of value 1 ; Checking condition for frugal number ; Driver Method to check for frugal number | #include <bits/stdc++.h> NEW_LINE using namespace std ; vector < long long int > primes ( long long int n ) { bool prime [ n + 1 ] ; memset ( prime , true , sizeof ( prime ) ) ; for ( int i = 2 ; i * i <= n ; i ++ ) { if ( prime [ i ] == true ) { for ( int j = i * 2 ; j <= n ; j += i ) prime [ j ] = false ; } } vector < long long int > arr ; for ( int i = 2 ; i < n ; i ++ ) if ( prime [ i ] ) arr . push_back ( i ) ; return arr ; } int countDigits ( long long int n ) { long long int temp = n ; int c = 0 ; while ( temp != 0 ) { temp = temp / 10 ; c ++ ; } return c ; } bool frugal ( long long int n ) { vector < long long int > r = primes ( n ) ; long long int t = n ; long long int s = 0 ; for ( int i = 0 ; i < r . size ( ) ; i ++ ) { if ( t % r [ i ] == 0 ) { long long int k = 0 ; while ( t % r [ i ] == 0 ) { t = t / r [ i ] ; k ++ ; } if ( k == 1 ) s = s + countDigits ( r [ i ] ) ; else if ( k != 1 ) s = s + countDigits ( r [ i ] ) + countDigits ( k ) ; } } return ( countDigits ( n ) > s && s != 0 ) ; } int main ( ) { long long int n = 343 ; if ( frugal ( n ) ) cout << " A ▁ Frugal ▁ number STRNEWLINE " ; else cout << " Not ▁ a ▁ frugal ▁ number STRNEWLINE " ; return 0 ; } |
N | C ++ program to find n - th number which is both square and cube . ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int nthSquareCube ( int n ) { return n * n * n * n * n * n ; } int main ( ) { int n = 5 ; cout << nthSquareCube ( n ) ; return 0 ; } |
Squared triangular number ( Sum of cubes ) | C ++ program to check if a given number is sum of cubes of natural numbers . ; Function to find if the given number is sum of the cubes of first n natural numbers ; Start adding cubes of the numbers from 1 ; If sum becomes equal to s return n ; Driver code | #include <iostream> NEW_LINE using namespace std ; int findS ( int s ) { int sum = 0 ; for ( int n = 1 ; sum < s ; n ++ ) { sum += n * n * n ; if ( sum == s ) return n ; } return -1 ; } int main ( ) { int s = 9 ; int n = findS ( s ) ; n == -1 ? cout << " - 1" : cout << n ; return 0 ; } |
Number with even sum of digits | C ++ program to find n - th Good number . ; Function to find kth good number . ; Find the last digit of n . ; If last digit is between 0 to 4 then return 2 * n . ; If last digit is between 5 to 9 then return 2 * n + 1. ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; long long int findKthGoodNo ( long long int n ) { int lastDig = n % 10 ; if ( lastDig >= 0 && lastDig <= 4 ) return n << 1 ; else return ( n << 1 ) + 1 ; } int main ( ) { long long int n = 10 ; cout << findKthGoodNo ( n ) ; return 0 ; } |
Nicomachu 's Theorem | CPP program to verify Nicomachu 's Theorem ; Compute sum of cubes ; Check if sum is equal to given formula . ; driver function | #include <bits/stdc++.h> NEW_LINE using namespace std ; void NicomachuTheorum_sum ( int n ) { int sum = 0 ; for ( int k = 1 ; k <= n ; k ++ ) sum += k * k * k ; int triNo = n * ( n + 1 ) / 2 ; if ( sum == triNo * triNo ) cout << " Yes " ; else cout << " No " ; } int main ( ) { int n = 5 ; NicomachuTheorum_sum ( n ) ; return 0 ; } |
Largest even digit number not greater than N | CPP program to print the largest integer not greater than N with all even digits ; function to check if all digits are even of a given number ; iterate for all digits ; if ( ( n % 10 ) % 2 ) if digit is odd ; all digits are even ; function to return the largest number with all digits even ; iterate till we find a number with all digits even ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int checkDigits ( int n ) { while ( n ) { return 0 ; n /= 10 ; } return 1 ; } int largestNumber ( int n ) { for ( int i = n ; ; i -- ) if ( checkDigits ( i ) ) return i ; } int main ( ) { int N = 23 ; cout << largestNumber ( N ) ; return 0 ; } |
Largest even digit number not greater than N | CPP program to print the largest integer not greater than N with all even digits ; function to return the largest number with all digits even ; convert the number to a string for easy operations ; find first odd digit ; if no digit , then N is the answer ; till first odd digit , add all even numbers ; decrease 1 from the odd digit ; add 0 in the rest of the digits ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int largestNumber ( int n ) { string s = " " ; int duplicate = n ; while ( n ) { s = char ( n % 10 + 48 ) + s ; n /= 10 ; } int index = -1 ; for ( int i = 0 ; i < s . length ( ) ; i ++ ) { if ( ( s [ i ] - '0' ) % 2 & 1 ) { index = i ; break ; } } if ( index == -1 ) return duplicate ; int num = 0 ; for ( int i = 0 ; i < index ; i ++ ) num = num * 10 + ( s [ i ] - '0' ) ; num = num * 10 + ( s [ index ] - '0' - 1 ) ; for ( int i = index + 1 ; i < s . length ( ) ; i ++ ) num = num * 10 + 8 ; return num ; } int main ( ) { int N = 24578 ; cout << largestNumber ( N ) ; return 0 ; } |
Number of digits in 2 raised to power n | CPP program to find number of digits in 2 ^ n ; Function to find number of digits in 2 ^ n ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int countDigits ( int n ) { return ( n * log10 ( 2 ) + 1 ) ; } int main ( ) { int n = 5 ; cout << countDigits ( n ) << endl ; return 0 ; } |
Smallest even digits number not less than N | CPP program to print the smallest integer not less than N with all even digits ; function to check if all digits are even of a given number ; iterate for all digits ; if ( ( n % 10 ) % 2 ) if digit is odd ; all digits are even ; function to return the smallest number with all digits even ; iterate till we find a number with all digits even ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int check_digits ( int n ) { while ( n ) { return 0 ; n /= 10 ; } return 1 ; } int smallest_number ( int n ) { for ( int i = n ; ; i ++ ) if ( check_digits ( i ) ) return i ; } int main ( ) { int N = 2397 ; cout << smallest_number ( N ) ; return 0 ; } |
Smallest triangular number larger than p | CPP code to find the bucket to choose for picking flowers out of it ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int findBucketNo ( int p ) { return ceil ( ( sqrt ( 8 * p + 1 ) - 1 ) / 2 ) ; } int main ( ) { int p = 10 ; cout << findBucketNo ( p ) ; return 0 ; } |
LCM of factorial and its neighbors | CPP program to calculate the LCM of N ! and its neighbor ( N - 1 ) ! and ( N + 1 ) ! ; function to calculate the factorial ; returning the factorial of the largest number in the given three consecutive numbers ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; unsigned int factorial ( unsigned int n ) { if ( n == 0 ) return 1 ; return n * factorial ( n - 1 ) ; } int LCMOfNeighbourFact ( int n ) { return factorial ( n + 1 ) ; } int main ( ) { int N = 5 ; cout << LCMOfNeighbourFact ( N ) << " STRNEWLINE " ; return 0 ; } |
Expressing factorial n as sum of consecutive numbers | CPP program to count number of ways we can express a factorial as sum of consecutive numbers ; sieve of Eratosthenes to compute the prime numbers ; Store all prime numbers ; function to calculate the largest power of a prime in a number ; Modular multiplication to avoid the overflow of multiplication Please see below for details https : www . geeksforgeeks . org / how - to - avoid - overflow - in - modular - multiplication / ; Returns count of ways to express n ! as sum of consecutives . ; We skip 2 ( First prime ) as we need to consider only odd primes ; compute the largest power of prime ; if the power of current prime number is zero in N ! , power of primes greater than current prime number will also be zero , so break out from the loop ; multiply the result at every step ; subtract 1 to exclude the case of 1 being an odd divisor ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; #define MAX 50002 NEW_LINE vector < int > primes ; void sieve ( ) { bool isPrime [ MAX ] ; memset ( isPrime , true , sizeof ( isPrime ) ) ; for ( int p = 2 ; p * p < MAX ; p ++ ) { if ( isPrime [ p ] == true ) { for ( int i = p * 2 ; i < MAX ; i += p ) isPrime [ i ] = false ; } } for ( int p = 2 ; p < MAX ; p ++ ) if ( isPrime [ p ] ) primes . push_back ( p ) ; } long long int power ( long long int x , long long int y ) { long long int count = 0 ; long long int z = y ; while ( x >= z ) { count += ( x / z ) ; z *= y ; } return count ; } long long int modMult ( long long int a , long long int b , long long int mod ) { long long int res = 0 ; a = a % mod ; while ( b > 0 ) { if ( b % 2 == 1 ) res = ( res + a ) % mod ; a = ( a * 2 ) % mod ; b /= 2 ; } return res % mod ; } long long int countWays ( long long int n , long long int m ) { long long int ans = 1 ; for ( int i = 1 ; i < primes . size ( ) ; i ++ ) { long long int powers = power ( n , primes [ i ] ) ; if ( powers == 0 ) break ; ans = modMult ( ans , powers + 1 , m ) % m ; } if ( ( ( ans - 1 ) % m ) < 0 ) return ( ans - 1 + m ) % m ; else return ( ans - 1 ) % m ; } int main ( ) { sieve ( ) ; long long int n = 4 , m = 7 ; cout << countWays ( n , m ) ; return 0 ; } |
Check if the given two numbers are friendly pair or not | Check if the given two number are friendly pair or not . ; Returns sum of all factors of n . ; Traversing through all prime factors . ; THE BELOW STATEMENT MAKES IT BETTER THAN ABOVE METHOD AS WE REDUCE VALUE OF n . ; This condition is to handle the case when n is a prime number greater than 2. ; Function to return gcd of a and b ; Function to check if the given two number are friendly pair or not . ; Finding the sum of factors of n and m ; finding gcd of n and sum of its factors . ; finding gcd of m and sum of its factors . ; checking is numerator and denominator of abundancy index of both number are equal or not . ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int sumofFactors ( int n ) { int res = 1 ; for ( int i = 2 ; 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 gcd ( int a , int b ) { if ( a == 0 ) return b ; return gcd ( b % a , a ) ; } bool checkFriendly ( int n , int m ) { int sumFactors_n = sumofFactors ( n ) ; int sumFactors_m = sumofFactors ( m ) ; int gcd_n = gcd ( n , sumFactors_n ) ; int gcd_m = gcd ( m , sumFactors_m ) ; if ( n / gcd_n == m / gcd_m && sumFactors_n / gcd_n == sumFactors_m / gcd_m ) return true ; else return false ; } int main ( ) { int n = 6 , m = 28 ; checkFriendly ( n , m ) ? ( cout << " Yes STRNEWLINE " ) : ( cout << " No STRNEWLINE " ) ; return 0 ; } |
Find n | C ++ program to find n - th Fortunate number ; Corner cases ; This is checked so that we can skip middle five numbers in below loop ; Function to Find primorial of order n ( product of first n prime numbers ) . ; Function to find next prime number greater than n ; Note that difference ( or m ) should be greater than 1. ; loop continuously until isPrime returns true for a number above n ; Ignoring the prime number that is 1 greater than n ; Returns n - th Fortunate number ; Driver function | #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 ; } long long int primorial ( long long int n ) { long long int p = 2 ; n -- ; for ( int i = 3 ; n != 0 ; i ++ ) { if ( isPrime ( i ) ) { p = p * i ; n -- ; } i ++ ; } return p ; } long long int findNextPrime ( long long int n ) { long long int nextPrime = n + 2 ; while ( true ) { if ( isPrime ( nextPrime ) ) break ; nextPrime ++ ; } return nextPrime ; } long long int fortunateNumber ( int n ) { long long int p = primorial ( n ) ; return findNextPrime ( p ) - p ; } int main ( ) { long long int n = 5 ; cout << fortunateNumber ( n ) << " STRNEWLINE " ; return 0 ; } |
Probability for three randomly chosen numbers to be in AP | CPP program to find probability that 3 randomly chosen numbers form AP . ; function to calculate probability ; Driver code to run above function | #include <bits/stdc++.h> NEW_LINE using namespace std ; double procal ( int n ) { return ( 3.0 * n ) / ( 4.0 * ( n * n ) - 1 ) ; } int main ( ) { int a [ ] = { 1 , 2 , 3 , 4 , 5 } ; int n = sizeof ( a ) / sizeof ( a [ 0 ] ) ; cout << procal ( n ) ; return 0 ; } |
Fermat 's Last Theorem | C ++ program to verify fermat 's last theorem for a given range and n. ; Check if there exists a triplet such that a ^ n + b ^ n = c ^ n ; driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; void testSomeNumbers ( int limit , int n ) { if ( n < 3 ) return ; for ( int a = 1 ; a <= limit ; a ++ ) for ( int b = a ; b <= limit ; b ++ ) { int pow_sum = pow ( a , n ) + pow ( b , n ) ; double c = pow ( pow_sum , 1.0 / n ) ; int c_pow = pow ( ( int ) c , n ) ; if ( c_pow == pow_sum ) { cout << " Count ▁ example ▁ found " ; return ; } } cout << " No ▁ counter ▁ example ▁ within ▁ given " " ▁ range ▁ and ▁ data " ; } int main ( ) { testSomeNumbers ( 10 , 3 ) ; return 0 ; } |
Product of given N fractions in reduced form | CPP program to find product of N fractions in reduced form . ; Function to return gcd of a and b ; Print the Product of N fraction in Reduced Form . ; finding the product of all N numerators and denominators . ; Finding GCD of new numerator and denominator ; Converting into reduced form . ; Driven Program | #include <bits/stdc++.h> NEW_LINE using namespace std ; int gcd ( int a , int b ) { if ( a == 0 ) return b ; return gcd ( b % a , a ) ; } void productReduce ( int n , int num [ ] , int den [ ] ) { int new_num = 1 , new_den = 1 ; for ( int i = 0 ; i < n ; i ++ ) { new_num *= num [ i ] ; new_den *= den [ i ] ; } int GCD = gcd ( new_num , new_den ) ; new_num /= GCD ; new_den /= GCD ; cout << new_num << " / " << new_den << endl ; } int main ( ) { int n = 3 ; int num [ ] = { 1 , 2 , 5 } ; int den [ ] = { 2 , 1 , 6 } ; productReduce ( n , num , den ) ; return 0 ; } |
Divide a big number into two parts that differ by k | C ++ program to Divide a Big Number into two parts ; Function to adds two Numbers represented as array of character . ; length of string ; initializing extra character position to 0 ; Adding each element of character and storing the carry . ; If remainder remains . ; Function to subtracts two numbers represented by string . ; Finding the length of the string . ; initializing extra character position to 0. ; Substrating each element of character . ; Function divides a number represented by character array a constant . ; Dividing each character element by constant . ; Function to reverses the character array . ; Reversing the array . ; Wrapper Function ; Reversing the character array . ; Adding the each element of both array and storing the sum in array a [ ] . ; Dividing the array a [ ] by 2. ; Reversing the character array to get output . ; Substracting each element of array i . e calculating a = a - b ; Reversing the character array to get output . ; Driven Program | #include <bits/stdc++.h> NEW_LINE using namespace std ; #define MAX 100 NEW_LINE void add ( char v1 [ ] , char v2 [ ] ) { int i , d , c = 0 ; int l1 = strlen ( v1 ) ; int l2 = strlen ( v2 ) ; for ( i = l1 ; i < l2 ; i ++ ) v1 [ i ] = '0' ; for ( i = l2 ; i < l1 ; i ++ ) v2 [ i ] = '0' ; for ( i = 0 ; i < l1 i < l2 ; i ++ ) { d = ( v1 [ i ] - '0' ) + ( v2 [ i ] - '0' ) + c ; c = d / 10 ; d %= 10 ; v1 [ i ] = '0' + d ; } while ( c ) { v1 [ i ] = '0' + ( c % 10 ) ; c /= 10 ; i ++ ; } v1 [ i ] = ' \0' ; v2 [ l2 ] = ' \0' ; } void subs ( char v1 [ ] , char v2 [ ] ) { int i , d , c = 0 ; int l1 = strlen ( v1 ) ; int l2 = strlen ( v2 ) ; for ( i = l2 ; i < l1 ; i ++ ) v2 [ i ] = '0' ; for ( i = 0 ; i < l1 ; i ++ ) { d = ( v1 [ i ] - '0' - c ) - ( v2 [ i ] - '0' ) ; if ( d < 0 ) { d += 10 ; c = 1 ; } else c = 0 ; v1 [ i ] = '0' + d ; } v2 [ l2 ] = ' \0' ; i = l1 - 1 ; while ( i > 0 && v1 [ i ] == '0' ) i -- ; v1 [ i + 1 ] = ' \0' ; } int divi ( char v [ ] , int q ) { int i , l = strlen ( v ) ; int c = 0 , d ; for ( i = l - 1 ; i >= 0 ; i -- ) { d = c * 10 + ( v [ i ] - '0' ) ; c = d % q ; d /= q ; v [ i ] = '0' + d ; } i = l - 1 ; while ( i > 0 && v [ i ] == '0' ) i -- ; v [ i + 1 ] = ' \0' ; return c ; } void rev ( char v [ ] ) { int l = strlen ( v ) ; int i ; char cc ; for ( i = 0 ; i < l - 1 - i ; i ++ ) { cc = v [ i ] ; v [ i ] = v [ l - 1 - i ] ; v [ l - i - 1 ] = cc ; } } void divideWithDiffK ( char a [ ] , char k [ ] ) { rev ( a ) ; rev ( k ) ; add ( a , k ) ; divi ( a , 2 ) ; rev ( a ) ; cout << " ▁ " << a ; rev ( a ) ; subs ( a , k ) ; rev ( a ) ; cout << " ▁ " << a ; } int main ( ) { char a [ MAX ] = "100" , k [ MAX ] = "20" ; divideWithDiffK ( a , k ) ; return 0 ; } |
Find value of ( n ^ 1 + n ^ 2 + n ^ 3 + n ^ 4 ) mod 5 for given n | finding the value of f ( n ) mod 5 for given n . ; function for f ( n ) mod 5 ; if n % 5 == 1 return 4 ; else return 0 ; driver program | #include <bits/stdc++.h> NEW_LINE using namespace std ; int fnMod ( int n ) { if ( n % 5 == 1 ) return 4 ; else return 0 ; } int main ( ) { int n = 10 ; cout << fnMod ( n ) << endl ; n = 11 ; cout << fnMod ( n ) << endl ; return 0 ; } |
Recursive sum of digits of a number formed by repeated appends | C ++ program to find Sum of digits of a number formed by repeating a number X number of times until sum become single digit . ; return single digit sum of a number . ; Returns recursive sum of digits of a number formed by repeating a number X number of times until sum become single digit . ; Driver program | #include <bits/stdc++.h> NEW_LINE using namespace std ; int digSum ( int n ) { if ( n == 0 ) return 0 ; return ( n % 9 == 0 ) ? 9 : ( n % 9 ) ; } int repeatedNumberSum ( int n , int x ) { int sum = x * digSum ( n ) ; return digSum ( sum ) ; } int main ( ) { int n = 24 , x = 3 ; cout << repeatedNumberSum ( n , x ) << endl ; return 0 ; } |
Sum of n digit numbers divisible by a given number | Simple CPP program to sum of n digit divisible numbers . ; Returns sum of n digit numbers divisible by ' number ' ; compute the first and last term ; sum of number which having n digit and divisible by number ; Driver code | #include <cmath> NEW_LINE #include <iostream> NEW_LINE using namespace std ; int totalSumDivisibleByNum ( int n , int number ) { int firstnum = pow ( 10 , n - 1 ) ; int lastnum = pow ( 10 , n ) ; int sum = 0 ; for ( int i = firstnum ; i < lastnum ; i ++ ) if ( i % number == 0 ) sum += i ; return sum ; } int main ( ) { int n = 3 , num = 7 ; cout << totalSumDivisibleByNum ( n , num ) << " STRNEWLINE " ; return 0 ; } |
Extended Midy 's theorem | C ++ program to demonstrate extended 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 Extended Midy 's theorem ; Dividing repeated part into m parts ; Computing sum of parts . ; Checking for Extended Midy ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; string fractionToDecimal ( int numerator , int denominator ) { string res ; unordered_map < int , int > mp ; 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 ExtendedMidys ( string str , int n , int m ) { if ( ! isPrime ( n ) ) { cout << " Denominator ▁ is ▁ not ▁ prime , ▁ " << " thus ▁ Extended ▁ Midy ' s ▁ " << " theorem ▁ is ▁ not ▁ applicable " ; return ; } int l = str . length ( ) ; int part1 = 0 , part2 = 0 ; if ( l % 2 == 0 && l % m == 0 ) { int part [ m ] = { 0 } , sum = 0 , res = 0 ; for ( int i = 0 ; i < l ; i ++ ) { int var = i / m ; part [ var ] = part [ var ] * 10 + ( str [ i ] - '0' ) ; } for ( int i = 0 ; i < m ; i ++ ) { sum = sum + part [ i ] ; cout << part [ i ] << " ▁ " ; } cout << endl ; res = pow ( 10 , m ) - 1 ; if ( sum % res == 0 ) cout << " Extended ▁ Midy ' s ▁ theorem ▁ holds ! " ; else cout << " Extended ▁ Midy ' s ▁ theorem " << " ▁ doesn ' t ▁ hold ! " ; } else if ( l % 2 != 0 ) { cout << " The ▁ repeating ▁ decimal ▁ is " << " ▁ of ▁ odd ▁ length ▁ thus ▁ Extended ▁ " << " Midy ' s ▁ theorem ▁ is ▁ not ▁ applicable " ; } else if ( l % m != 0 ) { cout << " The ▁ repeating ▁ decimal ▁ can ▁ " << " not ▁ be ▁ divided ▁ into ▁ m ▁ digits " ; } } int main ( ) { int numr = 1 , denr = 17 , m = 4 ; string res = fractionToDecimal ( numr , denr ) ; if ( res == " - 1" ) cout << " The ▁ fraction ▁ does ▁ not " << " ▁ have ▁ repeating ▁ decimal " ; else { cout << " Repeating ▁ decimal ▁ = ▁ " << res << endl ; ExtendedMidys ( res , denr , m ) ; } return 0 ; } |
Count n digit numbers divisible by given number | Simple CPP program to count n digit divisible numbers . ; Returns count of n digit numbers divisible by ' number ' ; compute the first and last term ; count total number of which having n digit and divisible by number ; Driver code | #include <cmath> NEW_LINE #include <iostream> NEW_LINE using namespace std ; int numberofterm ( int n , int number ) { int firstnum = pow ( 10 , n - 1 ) ; int lastnum = pow ( 10 , n ) ; int count = 0 ; for ( int i = firstnum ; i < lastnum ; i ++ ) if ( i % number == 0 ) count ++ ; return count ; } int main ( ) { int n = 3 , num = 7 ; cout << numberofterm ( n , num ) << " STRNEWLINE " ; return 0 ; } |
N | C ++ program to find N - th term in George Cantor set of rational numbers ; let i = numerator ; let j = denominator ; to keep the check of no . of terms ; loop till k is not equal to n ; check if k is already equal to N then the first term is the required rational number ; loop for traversing from right to left downwards diagonally ; loop for traversing from left to right upwards diagonally ; driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; void georgeCantor ( int n ) { int i = 1 ; int j = 1 ; int k = 1 ; while ( k < n ) { j ++ , k ++ ; if ( k == n ) break ; while ( j > 1 && k < n ) { i ++ , j -- , k ++ ; } if ( k == n ) break ; i ++ , k ++ ; if ( k == n ) break ; while ( i > 1 && k < n ) { i -- , j ++ , k ++ ; } } cout << " N - th ▁ term ▁ : ▁ " << i << " ▁ / ▁ " << j ; } int main ( ) { int n = 15 ; georgeCantor ( n ) ; return 0 ; } |
Number is divisible by 29 or not | CPP program to demonstrate above method to check divisibility by 29. ; Returns true if n is divisible by 29 else returns false . ; add the lastdigit * 3 to renaming number until number comes only 2 digit ; return true if number is divisible by 29 another ; Driver Code | #include <iostream> NEW_LINE using namespace std ; bool isDivisible ( long long int n ) { while ( n / 100 ) { int last_digit = n % 10 ; n /= 10 ; n += last_digit * 3 ; } return ( n % 29 == 0 ) ; } int main ( ) { long long int n = 348 ; if ( isDivisible ( n ) ) cout << " Yes " << endl ; else cout << " No " << endl ; return 0 ; } |
Solve the Linear Equation of Single Variable | CPP program to solve the given equation ; Function to solve the given equation ; Traverse the equation ; For cases such as : x , - x , + x ; Flip sign once ' = ' is seen ; There may be a number left in the end ; For infinite solutions ; For no solution ; x = total sum / coeff of x ' - ' sign indicates moving numeric value to right hand side ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; string solveEquation ( string equation ) { int n = equation . size ( ) , sign = 1 , coeff = 0 ; int total = 0 , i = 0 ; for ( int j = 0 ; j < n ; j ++ ) { if ( equation [ j ] == ' + ' equation [ j ] == ' - ' ) { if ( j > i ) total += sign * stoi ( equation . substr ( i , j - i ) ) ; i = j ; } else if ( equation [ j ] == ' x ' ) { if ( ( i == j ) equation [ j - 1 ] == ' + ' ) coeff += sign ; else if ( equation [ j - 1 ] == ' - ' ) coeff -= sign ; else coeff += sign * stoi ( equation . substr ( i , j - i ) ) ; i = j + 1 ; } else if ( equation [ j ] == ' = ' ) { if ( j > i ) total += sign * stoi ( equation . substr ( i , j - i ) ) ; sign = -1 ; i = j + 1 ; } } if ( i < n ) total += sign * stoi ( equation . substr ( i ) ) ; if ( coeff == 0 && total == 0 ) return " Infinite ▁ solutions " ; if ( coeff == 0 && total ) return " No ▁ solution " ; int ans = - total / coeff ; return " x = " + to_string ( ans ) ; } int main ( ) { string equation = " x + 5-3 + x = 6 + x - 2" ; cout << solveEquation ( equation ) ; return 0 ; } |
Check if a given number is Pronic | Efficient Approach | C / C ++ program to check if a number is pronic or not ; function to check Pronic Number ; Checking Pronic Number by multiplying consecutive numbers ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool pronic_check ( int n ) { int x = ( int ) ( sqrt ( n ) ) ; if ( x * ( x + 1 ) == n ) return true ; else return false ; } int main ( void ) { int n = 56 ; pronic_check ( n ) == true ? cout << " YES " : cout << " NO " ; return 0 ; } |
Check if given number is perfect square | C ++ program for the above approach ; If ceil and floor are equal the number is a perfect square ; Driver Code | #include <iostream> NEW_LINE #include <math.h> NEW_LINE using namespace std ; void checkperfectsquare ( int n ) { if ( ceil ( ( double ) sqrt ( n ) ) == floor ( ( double ) sqrt ( n ) ) ) { cout << " perfect ▁ square " ; } else { cout << " not ▁ a ▁ perfect ▁ square " ; } } int main ( ) { int n = 49 ; checkperfectsquare ( n ) ; return 0 ; } |
Writing power function for large numbers | C ++ program to compute factorial of big numbers ; Maximum number of digits in output ; This function multiplies x with the number represented by res [ ] . res_size is size of res [ ] or number of digits in the number represented by res [ ] . This function uses simple school mathematics for multiplication . This function may value of res_size and returns the new value of res_size ; Initialize carry ; One by one multiply n with individual digits of res [ ] ; Store last digit of ' prod ' in res [ ] ; Put rest in carry ; Put carry in res and increase result size ; This function finds power of a number x ; printing value "1" for power = 0 ; Initialize result ; Multiply x n times ( x ^ n = x * x * x ... . n times ) ; Driver program | #include <iostream> NEW_LINE using namespace std ; #define MAX 100000 NEW_LINE int multiply ( int x , int res [ ] , int res_size ) { int carry = 0 ; for ( int i = 0 ; i < res_size ; i ++ ) { int prod = res [ i ] * x + carry ; res [ i ] = prod % 10 ; carry = prod / 10 ; } while ( carry ) { res [ res_size ] = carry % 10 ; carry = carry / 10 ; res_size ++ ; } return res_size ; } void power ( int x , int n ) { if ( n == 0 ) { cout << "1" ; return ; } int res [ MAX ] ; int res_size = 0 ; int temp = x ; while ( temp != 0 ) { res [ res_size ++ ] = temp % 10 ; temp = temp / 10 ; } for ( int i = 2 ; i <= n ; i ++ ) res_size = multiply ( x , res , res_size ) ; cout << x << " ^ " << n << " ▁ = ▁ " ; for ( int i = res_size - 1 ; i >= 0 ; i -- ) cout << res [ i ] ; } int main ( ) { int exponent = 100 ; int base = 20 ; power ( base , exponent ) ; return 0 ; } |
P | CPP program to check if a number is a p - smooth number or not ; function to check if number n is a P - smooth number or not ; prime factorise it by 2 ; if the number is divisible by 2 ; check for all the possible numbers that can divide it ; prime factorize it by i ; stores the maximum if maximum and i , if i divides the number ; if n at the end is a prime number , then it a divisor itself ; Driver program to test above function | #include <iostream> NEW_LINE #include <math.h> NEW_LINE using namespace std ; bool check ( int n , int p ) { int maximum = -1 ; while ( ! ( n % 2 ) ) { maximum = max ( maximum , 2 ) ; n = n / 2 ; } for ( int i = 3 ; i <= sqrt ( n ) ; i += 2 ) { while ( n % i == 0 ) { maximum = max ( maximum , i ) ; n = n / i ; } } if ( n > 2 ) maximum = max ( maximum , n ) ; return ( maximum <= p ) ; } int main ( ) { int n = 24 , p = 7 ; if ( check ( n , p ) ) cout << " yes " ; else cout << " no " ; return 0 ; } |
Time when minute hand and hour hand coincide | CPP code to find the minute at which the minute hand and hour hand coincide ; function to find the minute ; finding the angle between minute hand and the first hour hand ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; void find_time ( int h1 ) { int theta = 30 * h1 ; cout << " ( " << ( theta * 2 ) << " / " << "11" << " ) " << " ▁ minutes " ; } int main ( ) { int h1 = 3 ; find_time ( h1 ) ; return 0 ; } |
Sum of Series ( n ^ 2 | CPP Program to finding the sum of the nth series ; function that calculate the sum of the nth series ; using formula of the nth term ; driver function | #include <bits/stdc++.h> NEW_LINE using namespace std ; int sum_series ( int n ) { int nSquare = n * n ; return nSquare * ( nSquare - 1 ) / 4 ; } int main ( ) { int n = 2 ; cout << sum_series ( n ) << endl ; return 0 ; } |
Check if a number is sandwiched between primes | CPP Program to check whether a number is sandwiched between two primes or not ; returns true if number n is prime ; 0 and 1 both are non - primes ; finding square root of n ; checking if n has any factor upto square root of n if yes its not prime ; Driver 's Code | #include <iostream> NEW_LINE #include <cmath> NEW_LINE using namespace std ; bool isPrime ( int n ) { if ( n == 0 n == 1 ) return false ; int root = sqrt ( n ) ; for ( int i = 2 ; i <= root ; i ++ ) if ( n % i == 0 ) return false ; return true ; } bool isSandwitched ( int n ) { return ( isPrime ( n - 1 ) && isPrime ( n + 1 ) ) ; } int main ( ) { int n = 642 ; cout << n << " ▁ : ▁ " ; if ( isSandwitched ( n ) ) cout << " Yes STRNEWLINE " ; else cout << " No STRNEWLINE " ; n = 9 ; cout << n << " ▁ : ▁ " ; if ( isSandwitched ( n ) ) cout << " Yes STRNEWLINE " ; else cout << " No STRNEWLINE " ; return 0 ; } |
Tomohiko Sakamoto 's Algorithm | A CPP program to implement the Tomohiko Sakamoto Algorithm ; function to implement tomohiko sakamoto algorithm ; array with leading number of days values ; if month is less than 3 reduce year by 1 ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int day_of_the_week ( int y , int m , int d ) { int t [ ] = { 0 , 3 , 2 , 5 , 0 , 3 , 5 , 1 , 4 , 6 , 2 , 4 } ; if ( m < 3 ) y -= 1 ; return ( ( y + y / 4 - y / 100 + y / 400 + t [ m - 1 ] + d ) % 7 ) ; } int main ( void ) { int day = 13 , month = 7 , year = 2017 ; cout << ( day_of_the_week ( year , month , day ) ) ; return 0 ; } |
Recursive program for prime number | CPP Program to find whether a Number is Prime or Not using Recursion ; Returns true if n is prime , else return false . i is current divisor to check . ; Base cases ; Check for next divisor ; Driver Program | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool isPrime ( int n , int i = 2 ) { if ( n <= 2 ) return ( n == 2 ) ? true : false ; if ( n % i == 0 ) return false ; if ( i * i > n ) return true ; return isPrime ( n , i + 1 ) ; } int main ( ) { int n = 15 ; if ( isPrime ( n ) ) cout << " Yes " ; else cout << " No " ; return 0 ; } |
Square Free Number | C ++ Program to print all prime factors ; Returns true if n is a square free number , else returns false . ; If 2 again divides n , then n is not a square free number . ; n must be odd at this point . So we can skip one element ( Note i = i + 2 ) ; Check if i is a prime factor ; If i again divides , then n is not square free ; Driver program to test above function | # include <bits/stdc++.h> NEW_LINE using namespace std ; bool isSquareFree ( int n ) { if ( n % 2 == 0 ) n = n / 2 ; if ( n % 2 == 0 ) return false ; for ( int i = 3 ; i <= sqrt ( n ) ; i = i + 2 ) { if ( n % i == 0 ) { n = n / i ; if ( n % i == 0 ) return false ; } } return true ; } int main ( ) { int n = 10 ; if ( isSquareFree ( n ) ) cout << " Yes " ; else cout << " No " ; return 0 ; } |
Area of a square from diagonal length | C ++ Program to find the area of square when its diagonal is given . ; Returns area of square from given diagonal ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; double findArea ( double d ) { return ( d * d ) / 2.0 ; } int main ( ) { double d = 10 ; cout << ( findArea ( d ) ) ; return 0 ; } |
Sum of series 1 ^ 2 + 3 ^ 2 + 5 ^ 2 + . . . + ( 2 * n | Program to find sum of series 1 ^ 2 + 3 ^ 2 + 5 ^ 2 + . . . + ( 2 * n - 1 ) ^ 2. ; Function to find sum of series . ; Driver code | #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 + ( 2 * i - 1 ) * ( 2 * i - 1 ) ; return sum ; } int main ( ) { int n = 10 ; cout << sumOfSeries ( n ) ; return 0 ; } |
Sum of series 1 ^ 2 + 3 ^ 2 + 5 ^ 2 + . . . + ( 2 * n | Program to find sum of series 1 ^ 2 + 3 ^ 2 + 5 ^ 2 + . . . + ( 2 * n - 1 ) ^ 2. ; Function that find sum of series . ; Formula to find sum of series . ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int sumOfSeries ( int n ) { return ( n * ( 2 * n - 1 ) * ( 2 * n + 1 ) ) / 3 ; } int main ( ) { int n = 10 ; cout << sumOfSeries ( n ) ; return 0 ; } |
Program to implement standard error of mean | C ++ Program to implement standard error of mean . ; Function to find sample mean . ; loop to calculate sum of array elements . ; Function to calculate sample standard deviation . ; Function to calculate sample error . ; Formula to find sample error . ; Driver function | #include <bits/stdc++.h> NEW_LINE using namespace std ; float mean ( float arr [ ] , int n ) { float sum = 0 ; for ( int i = 0 ; i < n ; i ++ ) sum = sum + arr [ i ] ; return sum / n ; } float SSD ( float 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 ) ) ; return sqrt ( sum / ( n - 1 ) ) ; } float sampleError ( float arr [ ] , int n ) { return SSD ( arr , n ) / sqrt ( n ) ; } int main ( ) { float arr [ ] = { 78.53 , 79.62 , 80.25 , 81.05 , 83.21 , 83.46 } ; int n = sizeof ( arr ) / sizeof ( arr [ 0 ] ) ; cout << sampleError ( arr , n ) ; return 0 ; } |
Minimum moves to reach target on a infinite line | Set 2 | CPP code to find minimum moves to reach target ; Function to find minimum steps to reach target ; Handling negatives by symmetry ; Keep moving while sum is smaller i . e calculating n ; case 1 : d is even ; d is odd ; Driver code ; Function call | #include <bits/stdc++.h> NEW_LINE using namespace std ; int StepstoReachTarget ( int target ) { target = abs ( target ) ; int n = ceil ( ( -1.0 + sqrt ( 1 + 8.0 * target ) ) / 2 ) ; int sum = n * ( n + 1 ) / 2 ; if ( sum == target ) return n ; int d = sum - target ; if ( ( d & 1 ) == 0 ) return n ; else return n + ( ( n & 1 ) ? 2 : 1 ) ; } int main ( ) { int target = 5 ; cout << StepstoReachTarget ( target ) ; return 0 ; } |
Sum of series 2 / 3 | C ++ program to find sum of given series ; Function to find sum of series up - to n terms ; initializing counter by 1 ; variable to calculate result ; while loop until nth term is not reached ; boolean type variable for checking validation ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; double seriesSum ( int n ) { int i = 1 ; double res = 0.0 ; bool sign = true ; while ( n > 0 ) { n -- ; if ( sign ) { sign = ! sign ; res = res + ( double ) ++ i / ++ i ; } else { sign = ! sign ; res = res - ( double ) ++ i / ++ i ; } } return res ; } int main ( ) { int n = 5 ; cout << seriesSum ( n ) ; return 0 ; } |
Number of Symmetric Relations on a Set | C ++ program to count total symmetric relations on a set of natural numbers . ; function find the square of n ; Base case ; Return 2 ^ ( n ( n + 1 ) / 2 ) ; Driver code | #include <bits/stdc++.h> NEW_LINE unsigned int countSymmetric ( unsigned int n ) { if ( n == 0 ) return 1 ; return 1 << ( ( n * ( n + 1 ) ) / 2 ) ; } int main ( ) { unsigned int n = 3 ; printf ( " % u " , countSymmetric ( n ) ) ; return 0 ; } |
Program for centered nonagonal number | CPP Program to find nth centered nonagonal number . ; Function to find nth centered nonagonal number . ; Formula to find nth centered nonagonal number . ; Driver function . | #include <bits/stdc++.h> NEW_LINE using namespace std ; int centeredNonagonal ( int n ) { return ( 3 * n - 2 ) * ( 3 * n - 1 ) / 2 ; } int main ( ) { int n = 10 ; cout << centeredNonagonal ( n ) ; return 0 ; } |
Program for Mean Absolute Deviation | C ++ Program to find mean absolute deviation of given array . ; Function to find mean of the array elements . ; Calculate sum of all elements . ; Function to find mean absolute deviation of given elements . ; Calculate the sum of absolute deviation about mean . ; Return mean absolute deviation about mean . ; Driver function . | #include <bits/stdc++.h> NEW_LINE using namespace std ; float Mean ( float arr [ ] , int n ) { float sum = 0 ; for ( int i = 0 ; i < n ; i ++ ) sum = sum + arr [ i ] ; return sum / n ; } float meanAbsoluteDeviation ( float arr [ ] , int n ) { float absSum = 0 ; for ( int i = 0 ; i < n ; i ++ ) absSum = absSum + abs ( arr [ i ] - Mean ( arr , n ) ) ; return absSum / n ; } int main ( ) { float arr [ ] = { 10 , 15 , 15 , 17 , 18 , 21 } ; int n = sizeof ( arr ) / sizeof ( arr [ 0 ] ) ; cout << meanAbsoluteDeviation ( arr , n ) ; return 0 ; } |
Find if it is possible to get a ratio from given ranges of costs and quantities | C ++ program to find if it is possible to get the ratio r ; Returns true if it is possible to get ratio r from given cost and quantity ranges . ; Calculating cost corresponding to value of i ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool isRatioPossible ( int lowCost , int upCost , int lowQuant , int upQuant , int r ) { for ( int i = lowQuant ; i <= upQuant ; i ++ ) { int ans = i * r ; if ( lowCost <= ans && ans <= upCost ) return true ; } return false ; } int main ( ) { int lowCost = 14 , upCost = 30 , lowQuant = 5 , upQuant = 12 , r = 9 ; if ( isRatioPossible ( lowCost , upCost , lowQuant , upQuant , r ) ) cout << " Yes " ; else cout << " No " ; return 0 ; } |
Find N integers with given difference between product and sum | CPP code to generate numbers with difference between product and sum is D ; Function to implement calculation ; Driver code | #include <iostream> NEW_LINE using namespace std ; void findNumbers ( int n , int d ) { for ( int i = 0 ; i < n - 2 ; i ++ ) cout << "1" << " ▁ " ; cout << "2" << " ▁ " ; cout << n + d << endl ; } int main ( ) { int N = 3 , D = 5 ; findNumbers ( N , D ) ; return 0 ; } |
Sum of fourth powers of first n odd natural numbers | CPP Program to find the sum of fourth powers of first n odd natural numbers ; calculate the sum of fourth power of first n odd natural numbers ; Driven Program | #include <bits/stdc++.h> NEW_LINE using namespace std ; long long int oddNumSum ( int n ) { return ( n * ( 2 * n + 1 ) * ( 24 * n * n * n - 12 * n * n - 14 * n + 7 ) ) / 15 ; } int main ( ) { int n = 4 ; cout << oddNumSum ( n ) << endl ; return 0 ; } |
Trailing number of 0 s in product of two factorials | CPP program for count number of trailing zeros in N ! * M ! ; Returns number of zeros in factorial n ; dividing x by powers of 5 and update count ; Returns count of trailing zeros in M ! x N ! ; Driver program | #include <iostream> NEW_LINE using namespace std ; int trailingZero ( int x ) { int i = 5 , count = 0 ; while ( x > i ) { count = count + x / i ; i = i * 5 ; } return count ; } int countProductTrailing ( int M , int N ) { return trailingZero ( N ) + trailingZero ( M ) ; } int main ( ) { int N = 67 , M = 98 ; cout << countProductTrailing ( N , M ) ; return 0 ; } |
Trimorphic Number | C ++ program to check if a number is Trimorphic ; Function to check Trimorphic number ; Store the cube ; Start Comparing digits ; Return false , if any digit of N doesn ' t ▁ match ▁ with ▁ ▁ its ▁ cube ' s digits from last ; Reduce N and cube ; Driver code | #include <iostream> NEW_LINE using namespace std ; bool isTrimorphic ( int N ) { int cube = N * N * N ; while ( N > 0 ) { if ( N % 10 != cube % 10 ) return false ; N /= 10 ; cube /= 10 ; } return true ; } int main ( ) { int N = 24 ; isTrimorphic ( N ) ? cout << " trimorphic " : cout << " not ▁ trimporphic " ; return 0 ; } |
Trimorphic Number | CPP program to find nth Trimorphic number ; Functions to find nth Trimorphic number ; Comparing the digits ; Return false , if any digit of num doesn ' t ▁ match ▁ with ▁ ▁ its ▁ cube ' s digits from last ; Reduce num and cube ; Check in max int size ; check number is Trimorphic or not ; if counter is equal to the n then return nth number ; Driver code | #include <iostream> NEW_LINE using namespace std ; # define INT_MAX 2147483647 NEW_LINE bool checkTrimorphic ( int num ) { int cube = num * num * num ; while ( num > 0 ) { if ( num % 10 != cube % 10 ) return false ; num /= 10 ; cube /= 10 ; } return true ; } int nthTrimorphic ( int n ) { int count = 0 ; for ( int i = 0 ; i < INT_MAX ; i ++ ) { if ( checkTrimorphic ( i ) ) count ++ ; if ( count == n ) return i ; } } int main ( ) { int n = 9 ; cout << nthTrimorphic ( n ) ; return 0 ; } |
Find minimum moves to reach target on an infinite line | CPP program to find minimum moves to reach target if we can move i steps in i - th move . ; Handling negatives by symmetry ; Keep moving while sum is smaller or difference is odd . ; Driver code | #include <iostream> NEW_LINE using namespace std ; int reachTarget ( int target ) { target = abs ( target ) ; int sum = 0 , step = 0 ; while ( sum < target || ( sum - target ) % 2 != 0 ) { step ++ ; sum += step ; } return step ; } int main ( ) { int target = 5 ; cout << reachTarget ( target ) ; return 0 ; } |
Sum of fifth powers of the first n natural numbers | CPP Program to find the sum of fifth power of first n natural numbers ; calculate the sum of fifth power of first n natural numbers ; Driven Program | #include <bits/stdc++.h> NEW_LINE using namespace std ; long long int fifthPowerSum ( int n ) { return ( ( 2 * n * n * n * n * n * n ) + ( 6 * n * n * n * n * n ) + ( 5 * n * n * n * n ) - ( n * n ) ) / 12 ; } int main ( ) { int n = 5 ; cout << fifthPowerSum ( n ) << endl ; return 0 ; } |
Subsets and Splits