找出两个给定和和 GCD 的数字

给定两个数的总和和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它的倍数。

• 选择第一个数字作为gcd然后另一个数字将是sum – gcd 。
• 如果上一步中选择的两个数字的总和等于sum，则打印这两个数字。
• 否则数字不存在并打印-1代替。

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..

Javascript

 
// Javascript program to find two numbers
// whose sum and GCD is given
 
//function to find gcd of two numbers
function __gcd(a, b)
{
    if (b==0) return a;
   return __gcd(b,a%b);
}
     
// Function to find two numbers
// whose sum and gcd is given
function findTwoNumbers(sum, gcd)
{
    // sum != gcd checks that both the
    // numbers are positive or not
    if (__gcd(gcd, sum - gcd) == gcd && sum != gcd)
        document.write(  "a = " + Math.min(gcd, sum - gcd)
            + ", b = " + (sum - Math.min(gcd, sum - gcd)) );
    else
        document.write( -1 );
}
 
// Driver code
 
    let sum = 8;
    let gcd = 2;
 
    findTwoNumbers(sum, gcd);
 

输出：

a = 2, b = 6

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