两个数字A和B的共同除数之和

给定两个数字A和B，任务是找到两个数字A和B的公因子之和。数字A和B小于10 ^ 8。

例子：

Input: A = 10, B = 15
Output: Sum = 6
The common factors are 1, 5, so their sum is 6 

Input: A = 100, B = 150
Output: Sum = 93

天真的方法：从i = 1迭代到A和B的最小值，并检查i是否是A和B的因数。如果i是A和B的因数，则将其加和。在循环末尾显示总和。

下面是上述方法的实现：

C++

// C++ implementation of above approach
#include 
using namespace std;
  
// print the sum of common factors
int sum(int a, int b)
{
    // sum of common factors
    int sum = 0;
  
    // iterate from 1 to minimum of a and b
    for (int i = 1; i <= min(a, b); i++)
  
        // if i is the common factor
        // of both the numbers
        if (a % i == 0 && b % i == 0)
            sum += i;
  
    return sum;
}
  
// Driver code
int main()
{
    int A = 10, B = 15;
  
    // print the sum of common factors
    cout << "Sum = " << sum(A, B) << endl;
  
    return 0;
}

Java

// Java implementation of above approach
import java.io.*;
  
class GFG {
      
  
  
// print the sum of common factors
static int sum(int a, int b)
{
    // sum of common factors
    int sum = 0;
  
    // iterate from 1 to minimum of a and b
    for (int i = 1; i <= Math.min(a, b); i++)
  
        // if i is the common factor
        // of both the numbers
        if (a % i == 0 && b % i == 0)
            sum += i;
  
    return sum;
}
  
// Driver code
  
  
    public static void main (String[] args) {
            int A = 10, B = 15;
  
    // print the sum of common factors
    System.out.print("Sum = " + sum(A, B));
    }
}
// This code is contributed by shs..

Python 3

# Python 3 implementation of
# above approach
  
# print the sum of common factors
def sum(a, b):
  
    # sum of common factors
    sum = 0
  
    # iterate from 1 to minimum of a and b
    for i in range (1, min(a, b)):
  
        # if i is the common factor
        # of both the numbers
        if (a % i == 0 and b % i == 0):
            sum += i
  
    return sum
  
# Driver Code
A = 10
B = 15
  
# print the sum of common factors
print("Sum =", sum(A, B))
  
# This code is contributed
# by Akanksha Rai

C#

// C# implementation of above approach
  
  
using System;
  
class GFG {
      
  
  
// print the sum of common factors
static int sum(int a, int b)
{
    // sum of common factors
    int sum = 0;
  
    // iterate from 1 to minimum of a and b
    for (int i = 1; i <= Math.Min(a, b); i++)
  
        // if i is the common factor
        // of both the numbers
        if (a % i == 0 && b % i == 0)
            sum += i;
  
    return sum;
}
  
// Driver code
  
  
    public static void Main () {
            int A = 10, B = 15;
  
    // print the sum of common factors
    Console.WriteLine("Sum = " + sum(A, B));
    }
}
// This code is contributed by shs..

输出：

Sum = 6

一种有效的方法是使用两个数的公共除数中使用的相同概念。计算给定两个数字的最大公约数(gcd)，然后找到该gcd的公约数之和。

C++

// C++ implementation of above approach
#include 
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 sumcommDiv(int a, int b)
{
    // find gcd of a, b
    int n = gcd(a, b);
  
    // Find the sum of divisors of n.
    int sum = 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)
                sum += i;
            else
                sum += (n / i) + i;
        }
    }
  
    return sum;
}
  
// Driver program to run the case
int main()
{
    int a = 10, b = 15;
    cout << "Sum = " << sumcommDiv(a, b);
  
    return 0;
}

Java

//Java implementation of above approach 
  
import java.io.*;
  
class GFG {
      
// Function to calculate gcd of two numbers 
static 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 
static int sumcommDiv(int a, int b) 
{ 
    // find gcd of a, b 
    int n = gcd(a, b); 
  
    // Find the sum of divisors of n. 
    int sum = 0; 
    for (int i = 1; i <= Math.sqrt(n); i++) { 
  
        // if 'i' is factor of n 
        if (n % i == 0) { 
  
            // check if divisors are equal 
            if (n / i == i) 
                sum += i; 
            else
                sum += (n / i) + i; 
        } 
    } 
  
    return sum; 
} 
  
// Driver program to run the case 
    public static void main (String[] args) {
      
    int a = 10, b = 15; 
    System.out.println("Sum = " + sumcommDiv(a, b)); 
    }
}

Python3

# Python 3 implementation of above approach
from math import gcd,sqrt
  
# Function to calculate all common divisors
# of two given numbers
# a, b --> input integer numbers
def sumcommDiv(a, b):
    # find gcd of a, b
    n = gcd(a, b)
  
    # Find the sum of divisors of n.
    sum = 0
    N = int(sqrt(n))+1
    for i in range(1,N,1):
        # if 'i' is factor of n
        if (n % i == 0):
            # check if divisors are equal
            if (n / i == i):
                sum += i
            else:
                sum += (n / i) + i
          
    return sum
  
# Driver program to run the case
if __name__ == '__main__':
    a = 10
    b = 15
    print("Sum =",int(sumcommDiv(a, b)))
  
# This code is contributed by 
# Surendra_Gangwar

C＃

// C# implementation of above approach 
  
using System;
  
public class GFG{
          
// Function to calculate gcd of two numbers 
static 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 
static int sumcommDiv(int a, int b) 
{ 
    // find gcd of a, b 
    int n = gcd(a, b); 
  
    // Find the sum of divisors of n. 
    int sum = 0; 
    for (int i = 1; i <= Math.Sqrt(n); i++) { 
  
        // if 'i' is factor of n 
        if (n % i == 0) { 
  
            // check if divisors are equal 
            if (n / i == i) 
                sum += i; 
            else
                sum += (n / i) + i; 
        } 
    } 
  
    return sum; 
} 
  
// Driver program to run the case 
    static public void Main (){
        int a = 10, b = 15; 
        Console.WriteLine("Sum = " + sumcommDiv(a, b));
    }
}

的PHP

 input integer numbers
function  sumcommDiv($a, $b)
{
    // find gcd of a, b
$n = gcd($a, $b);
  
    // Find the sum of divisors of n.
    $sum = 0;
    for ($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)
                $sum += $i;
            else
                $sum += ($n / $i) + $i;
        }
    }
  
    return $sum;
}
  
// Driver program to run the case
    $a = 10;
    $b = 15;
    echo "Sum = " , sumcommDiv($a, $b);
  
  
?>

输出：

Sum = 6