检查是否可以通过给定的操作同时使x和y为零

给定两个数字X和Y。任务是检查任意多次执行以下操作，是否可以同时将X和Y减小为零：

• 选择任何自然数(例如z )，并在每次操作中将X和Y减少为以下值之一：
  1. X = X – z和Y = Y – 2 * z
  2. X = X – 2 * z和Y = Y – z

例子：

Input: X = 6, Y = 9 
Output: YES 
Explanation: 
We can perform operation in following way: 
if z = 1, then 
X = X – 2*z = 6 – 2*(1) 
Y = Y – z = 9 – 1 
=> X = 4 & Y = 8
Now again if z = 4, then 
X = X – z = 4 – 4 
Y = Y – 2*z = 8 – 2*(4) 
=> X = 0 & Y = 0 
Therefore, X & Y become zero in 2 steps assuming z as 1 and 4 respectively. 
Input: X = 1, Y = 1 
Output: NO 
Explanation: 
We don’t have any possible value for z such that X & Y can become zero simultaneously. 
 

方法：
以下是给定问题陈述的观察结果：

1. 由于X和Y更新为( X – z和Y – 2 * z )或( X – 2 * z和Y – z )，因此在n次运算后( X + Y )被更新为( X + Y – 3 * n * z )。因此，如果(X + Y)％3等于0，则X和Y可以同时减小为零。
2. 在每个步骤中， X或Y之一减小2 * z 。为了同时将X和Y减少为零，必须满足以下条件： max(X，Y)≤2 * min(X，Y) 。
   例如：

Let X = 6 and Y = 15 
Since (X+Y)%3 = (21%3) = 0 
As our first condition is satisfied, 
But by taking z = 6 
X = X – z = 6 – 6 = 0 
Y = Y – 2*z = 15 – 12 = 3 
Since Y is not less than or equals to 2*X, therefore X and Y cannot be reduced to zero at same time. 
 

如果以上两个条件满足X和Y的值，则X和Y可以同时减小为0 。
下面是上述方法的实现：

C++

// C++ program of the above approach
#include 
using namespace std;
 
// Function to check if it is possible to
// make x and y can become 0 at same time
void canBeReduced(int x, int y)
{
    int maxi = max(x, y);
    int mini = min(x, y);
 
    // Check the given conditions
    if (((x + y) % 3) == 0 && maxi <= 2*mini)
        cout << "YES" << endl;
    else
        cout << "NO" << endl;
}
 
// Driver Code
int main()
{
    int x = 6, y = 9;
 
    // Function Call
    canBeReduced(x, y);
    return 0;
}

Java

// Java program of the above approach
import java.util.*;
 
class GFG{
  
// Function to check if it is possible to
// make x and y can become 0 at same time
static void canBeReduced(int x, int y)
{
    int maxi = Math.max(x, y);
    int mini = Math.min(x, y);
  
    // Check the given conditions
    if (((x + y) % 3) == 0 && maxi <= 2*mini)
        System.out.print("YES" +"\n");
    else
        System.out.print("NO" +"\n");
}
  
// Driver Code
public static void main(String[] args)
{
    int x = 6, y = 9;
  
    // Function Call
    canBeReduced(x, y);
}
}
 
// This code is contributed by Rajput-Ji

C#

// C# program of the above approach
using System;
 
class GFG 
{
    // Function to check if it is possible to
    // make x and y can become 0 at same time
    static void canBeReduced(int x, int y)
    {
        int maxi = Math.Max(x, y);
        int mini = Math.Min(x, y);
     
        // Check the given conditions
        if (((x + y) % 3) == 0 && maxi <= 2*mini)
            Console.WriteLine("YES");
        else
            Console.WriteLine("NO");
    }
     
    // Driver Code
    static void Main()
    {
        int x = 6, y = 9;
     
        // Function Call
        canBeReduced(x, y);
    }
}
 
// This code is contributed by shubhamsingh10

Python3

# Python 3 program of the above approach
 
# Function to check if it is possible to
# make x and y can become 0 at same time
def canBeReduced(x,y):
    maxi = max(x, y)
    mini = min(x, y)
 
    # Check the given conditions
    if (((x + y) % 3) == 0 and maxi <= 2*mini):
        print("YES")
    else:
        print("NO")
 
# Driver Code
if __name__ == '__main__':
    x = 6
    y = 9
 
    # Function Call
    canBeReduced(x, y)
     
# This code is contributed by Surendra_Gangwar

Javascript

 
// javascript program of the above approach
 
// Function to check if it is possible to
// make x and y can become 0 at same time
function canBeReduced(x , y)
{
    var maxi = Math.max(x, y);
    var mini = Math.min(x, y);
  
    // Check the given conditions
    if (((x + y) % 3) == 0 && maxi <= 2*mini)
        document.write("YES" +"\n");
    else
        document.write("NO" +"\n");
}
  
// Driver Code
 
var x = 6, y = 9;
  
// Function Call
canBeReduced(x, y);
 
// This code contributed by shikhasingrajput
 

输出：

YES

时间复杂度： O(1)