Find the final X and Y when they are Altering under given condition

Given initial values of two positive integers X and Y, Find the final value of X and Y according to below alterations:

1. If X=0 or Y=0, terminate the process. Else, go to step 2;
2. If X >= 2*Y, then change the value of X to X – 2*Y, and repeat step 1. Else, go to step 3;
3. If Y >= 2*X, then assign the value of Y to Y – 2*X, and repeat step 1. Else, end the process.

Constraints: 1<=X, Y<=10^18

Examples:

```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)
```

Approach: Since the initial values of X and Y can be as high as 10^18. Simple brute force approach will not work.
If we observe carefully, the problem statement is nothing but a sort of Euclid Algorithm, where we will replace all subtractions with modulo.

Implementation:

C++

 `// CPP tp implement above approach ` ` `  `#include ` `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

 `// 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 `

Python3

 `# 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 `

PHP

 `= 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 ` `?> `

Output:

```X=0, Y=1
```

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