找到两个总和和GCD均给定的数字

给定两个数字的和和gcd $a$和$b$ 。任务是找到数字a和b 。如果数字不存在，则打印$-1$ 。

例子：

Input: sum = 6, gcd = 2
Output: a = 4, b = 2
4 + 2 = 6 and GCD(4, 2) = 2

Input: sum = 7, gcd = 2
Output: -1
There are no such numbers whose sum is 7 and GCD is 2

方法：由于给出了GCD，因此已知这两个数字都是其倍数。

• 选择第一个数字作为gcd，然后选择另一个数字求和– gcd 。
• 如果在上一步中选择的两个数字的总和等于总和，则打印两个数字。
• 其他数字不存在，而是打印-1 。

下面是上述方法的实现：

C++

// C++ program to find two numbers
// whose sum and GCD is given
#include 
using namespace std;
  
// Function to find two numbers
// whose sum and gcd is given
void findTwoNumbers(int sum, int gcd)
{
    // sum != gcd checks that both the
    // numbers are positive or not
    if (__gcd(gcd, sum - gcd) == gcd && sum != gcd)
        cout << "a = " << min(gcd, sum - gcd)
             << ", b = " << sum - min(gcd, sum - gcd)
             << endl;
    else
        cout << -1 << endl;
}
  
// Driver code
int main()
{
    int sum = 8;
    int gcd = 2;
  
    findTwoNumbers(sum, gcd);
  
    return 0;
}

Java

// Java program to find two numbers 
// whose sum and GCD is given 
import java.util.*;
class Solution{
  
//function to find gcd of two numbers
static int __gcd(int a,int b)
{
    if (b==0) return a;
   return __gcd(b,a%b);
}
      
// Function to find two numbers 
// whose sum and gcd is given 
static void findTwoNumbers(int sum, int gcd) 
{ 
    // sum != gcd checks that both the 
    // numbers are positive or not 
    if (__gcd(gcd, sum - gcd) == gcd && sum != gcd) 
        System.out.println(  "a = " + Math.min(gcd, sum - gcd) 
            + ", b = " + (int)(sum - Math.min(gcd, sum - gcd)) ); 
    else
        System.out.println( -1 ); 
} 
  
// Driver code 
public static void main(String args[]) 
{ 
    int sum = 8; 
    int gcd = 2; 
  
    findTwoNumbers(sum, gcd); 
  
} 
  
  
}
//contributed by Arnab Kundu

Python3

# Python 3 program to find two numbers
# whose sum and GCD is given
from math import gcd as __gcd
  
# Function to find two numbers
# whose sum and gcd is given
def findTwoNumbers(sum, gcd):
      
    # sum != gcd checks that both the
    # numbers are positive or not
    if (__gcd(gcd, sum - gcd) == gcd and
                          sum != gcd):
        print("a =", min(gcd, sum - gcd), 
              ", b =", sum - min(gcd, sum - gcd))
    else:
        print(-1)
          
# Driver code
if __name__ == '__main__':
    sum = 8
    gcd = 2
  
    findTwoNumbers(sum, gcd)
  
# This code is contributed by
# Surendra_Gangwar

C#

// C# program to find two numbers 
// whose sum and GCD is given 
using System;
class GFG
{
  
// function to find gcd of two numbers
static int __gcd(int a, int b)
{
    if (b == 0) 
        return a;
    return __gcd(b, a % b);
}
      
// Function to find two numbers 
// whose sum and gcd is given 
static void findTwoNumbers(int sum, int gcd) 
{ 
    // sum != gcd checks that both the 
    // numbers are positive or not 
    if (__gcd(gcd, sum - gcd) == gcd && sum != gcd) 
        Console.WriteLine("a = " + Math.Min(gcd, sum - gcd) + 
            ", b = " + (int)(sum - Math.Min(gcd, sum - gcd))); 
    else
        Console.WriteLine( -1 ); 
} 
  
// Driver code 
public static void Main() 
{ 
    int sum = 8; 
    int gcd = 2; 
  
    findTwoNumbers(sum, gcd); 
} 
}
  
// This code is contributed by anuj_67..

输出：

a = 2, b = 6