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C++ Program To Find All Factors of A Natural Number
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C++ Program for Common Divisors of Two Numbers

Last Updated : 04 Dec, 2018
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Given two integer numbers, the task is to find the count of all common divisors of given numbers. 
  Input : a = 12, b = 24  Output: 6  // all common divisors are 1, 2, 3,   // 4, 6 and 12    Input : a = 3, b = 17  Output: 1  // all common divisors are 1    Input : a = 20, b = 36  Output: 3  // all common divisors are 1, 2, 4
CPP 
// C++ implementation of program #include <bits/stdc++.h> using namespace std;  // Function to calculate gcd of two numbers int gcd(int a, int b) {     if (a == 0)         return b;     return gcd(b % a, a); }  // Function to calculate all common divisors // of two given numbers // a, b --> input integer numbers int commDiv(int a, int b) {     // find gcd of a, b     int n = gcd(a, b);      // Count divisors of n.     int result = 0;     for (int i = 1; i <= sqrt(n); i++) {         // if 'i' is factor of n         if (n % i == 0) {             // check if divisors are equal             if (n / i == i)                 result += 1;             else                 result += 2;         }     }     return result; }  // Driver program to run the case int main() {     int a = 12, b = 24;     cout << commDiv(a, b);     return 0; }
Output: 
  6
Please refer complete article on Common Divisors of Two Numbers for more details!

Next Article
C++ Program To Find All Factors of A Natural Number
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kartik
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Article Tags :
  • Mathematical
  • C++ Programs
  • DSA
  • GCD-LCM
  • divisors
Practice Tags :
  • Mathematical

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