在给定条件下，当 X 和 Y 发生变化时，找到最终的 X 和 Y
原文:https://www . geeksforgeeks . org/find-the-final-x-and-y-当他们在给定条件下改变时/
给定两个正整数 X 和 Y 的初始值，根据以下修改求 X 和 Y 的最终值:
1.如果 X=0 或 Y=0 ，终止该过程。否则，转到步骤 2； 2。如果 X > = 2*Y ，则将 X 的值改为 X–2 * Y，重复步骤 1。否则，转到步骤 3； 3。如果 Y > = 2*X ，则将 Y 的值赋给 Y–2 * X，重复步骤 1。否则，结束这个过程。 约束:1 < =X，Y < =10^18
例:
Input: X=12, Y=5
Output: X=0, Y=1
Explanation:
Initially X = 12, Y = 5
--> X = 2, Y = 5 (as X = X-2*Y)
--> X = 2, Y = 1 (as Y = Y-2*X)
--> X = 0, Y = 1 (as X = X-2*Y)
--> Stop (as X = 0)
Input: X=31, Y=12
Output: X=7, Y=12
Explanation:
Initially X = 31, Y = 12
--> X = 7, Y = 12 (as X = X-2*Y)
--> Stop (as (Y - 2*X) < 0)
方法:由于 x 和 y 的初始值可以高达 10^18.简单的暴力方法是行不通的。 如果我们仔细观察，问题陈述只不过是一种欧几里德算法，在这里我们将用模替换所有的减法。 实施:
C++
// CPP tp implement above approach
#include <iostream>
using namespace std;
// Function to get final value of X and Y
void alter(long long int x, long long int y)
{
// Following the sequence
// but by replacing minus with modulo
while (true) {
// Step 1
if (x == 0 || y == 0)
break;
// Step 2
if (x >= 2 * y)
x = x % (2 * y);
// Step 3
else if (y >= 2 * x)
y = y % (2 * x);
// Otherwise terminate
else
break;
}
cout << "X=" << x << ", "
<< "Y=" << y;
}
// Driver function
int main()
{
// Get the initial X and Y values
long long int x = 12, y = 5;
// Find the result
alter(x, y);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java tp implement above approach
import java.io.*;
class GFG
{
// Function to get final value of X and Y
static void alter(long x, long y)
{
// Following the sequence but by
// replacing minus with modulo
while (true)
{
// Step 1
if (x == 0 || y == 0)
break;
// Step 2
if (x >= 2 * y)
x = x % (2 * y);
// Step 3
else if (y >= 2 * x)
y = y % (2 * x);
// Otherwise terminate
else
break;
}
System.out.println("X = " + x + ", " +
"Y = " + y);
}
// Driver Code
public static void main (String[] args)
{
// Get the initial X and Y values
long x = 12, y = 5;
// Find the result
alter(x, y);
}
}
// This code is contributed by
// shk
Python 3
# Python3 tp implement above approach
import math as mt
# Function to get final value of X and Y
def alter(x, y):
# Following the sequence but by
# replacing minus with modulo
while (True):
# Step 1
if (x == 0 or y == 0):
break
# Step 2
if (x >= 2 * y):
x = x % (2 * y)
# Step 3
elif (y >= 2 * x):
y = y % (2 * x)
# Otherwise terminate
else:
break
print("X =", x, ", ", "Y =", y)
# Driver Code
# Get the initial X and Y values
x, y = 12, 5
# Find the result
alter(x, y)
# This code is contributed by
# Mohit kumar 29
C
// C# implementation of the approach
using System;
class GFG
{
// Function to get final value of X and Y
static void alter(long x, long y)
{
// Following the sequence but by
// replacing minus with modulo
while (true)
{
// Step 1
if (x == 0 || y == 0)
break;
// Step 2
if (x >= 2 * y)
x = x % (2 * y);
// Step 3
else if (y >= 2 * x)
y = y % (2 * x);
// Otherwise terminate
else
break;
}
Console.WriteLine("X = " + x + ", " + "Y = " + y);
}
// Driver Code
public static void Main ()
{
// Get the initial X and Y values
long x = 12, y = 5;
// Find the result
alter(x, y);
}
}
// This code is contributed by aishwarya.27
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of the approach
// Function to get final value of X and Y
function alter($x, $y)
{
// Following the sequence but by
// replacing minus with modulo
while (true)
{
// Step 1
if ($x == 0 || $y == 0)
break;
// Step 2
if ($x >= 2 * $y)
$x = $x % (2 * $y);
// Step 3
else if ($y >= 2 * $x)
$y = $y % (2 * $x);
// Otherwise terminate
else
break;
}
echo "X = ", $x, ", ", "Y = ", $y;
}
// Driver Code
// Get the initial X and Y values
$x = 12 ;
$y = 5 ;
// Find the result
alter($x, $y);
// This code is contributed by Ryuga
?>
java 描述语言
<script>
// Javascript to implement above approach
// Function to get final value of X and Y
function alter(x, y)
{
// Following the sequence but by
// replacing minus with modulo
while (true)
{
// Step 1
if (x == 0 || y == 0)
break;
// Step 2
if (x >= 2 * y)
x = x % (2 * y);
// Step 3
else if (y >= 2 * x)
y = y % (2 * x);
// Otherwise terminate
else
break;
}
document.write("X = " + x + ", " +
"Y = " + y);
}
// Driver Code
// Get the initial X and Y values
var x = 12, y = 5;
// Find the result
alter(x, y);
// This code is contributed by Kirti
</script>
Output:
X=0, Y=1
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