Even-odd turn game with two integers

Given three positive integers X, Y and P. Here P denotes the number of turns. Whenever the turn is odd X is multiplied by 2 and in every even turn Y is multiplied by 2. The task is to find the value of max(X, Y) ÷ min(X, Y) after the complete P turns.**Examples :**

Input : X = 1, Y = 2, P = 1 Output : 1 As turn is odd, X is multiplied by 2 and becomes 2. Now, X is 2 and Y is also 2. Therefore, 2 ÷ 2 is 1. Input : X = 3, Y = 7, p = 2 Output : 2 Here we have 2 turns. In the 1st turn which is odd X is multiplied by 2. And the values are 6 and 7. In the next turn which is even Y is multiplied by 2. Now the final values are 6 and 14. Therefore, 14 ÷ 6 is 2.

Lets play the above game for 8 turns :

| i | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | |------|---|----|----|----|----|----|----|-----|-----| | X(i) | X | 2X | 2X | 4X | 4X | 8X | 8X | 16X | 16X | | Y(i) | Y | Y | 2Y | 2Y | 4Y | 4Y | 8Y | 8Y | 16Y |

Here we can easily spot a pattern :

if i is even, then X(i) = z * X and Y(i) = z * Y. if i is odd, then X(i) = 2*z * X and Y(i) = z * Y.

Here z is actually the power of 2. So, we can simply say –

If P is even output will be max(X, Y) ÷ min(X, Y) else output will be max(2*X, Y) ÷ min(2*X, Y).

Below is the implementation :

## C++

`// CPP program to find max(X, Y) / min(X, Y)` `// after P turns` `#include <bits/stdc++.h>` `using` `namespace` `std;` `int` `findValue(` `int` `X, ` `int` `Y, ` `int` `P)` `{` ` ` `if` `(P % 2 == 0)` ` ` `return` `(max(X, Y) / min(X, Y));` ` ` `else` ` ` `return` `(max(2 * X, Y) / min(2 * X, Y));` `}` `// Driver code` `int` `main()` `{` ` ` `// 1st test case` ` ` `int` `X = 1, Y = 2, P = 1;` ` ` `cout << findValue(X, Y, P) << endl;` ` ` `// 2nd test case` ` ` `X = 3, Y = 7, P = 2;` ` ` `cout << findValue(X, Y, P) << endl;` `}` |

## Java

`// Java program to find max(X, Y) / min(X, Y)` `// after P turns` `import` `java.util.*;` `class` `Even_odd{` ` ` `public` `static` `int` `findValue(` `int` `X, ` `int` `Y,` ` ` `int` `P)` ` ` `{` ` ` `if` `(P % ` `2` `== ` `0` `)` ` ` `return` `(Math.max(X, Y) /` ` ` `Math.min(X, Y));` ` ` `else` ` ` `return` `(Math.max(` `2` `* X, Y) /` ` ` `Math.min(` `2` `* X, Y));` ` ` `}` ` ` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `// 1st test case` ` ` `int` `X = ` `1` `, Y = ` `2` `, P = ` `1` `;` ` ` `System.out.println(findValue(X, Y, P));` ` ` ` ` `// 2nd test case` ` ` `X = ` `3` `;` ` ` `Y = ` `7` `;` ` ` `P = ` `2` `;` ` ` `System.out.print(findValue(X, Y, P));` ` ` `}` `}` `//This code is contributed by rishabh_jain` |

## Python3

`# Python3 code to find max(X, Y) / min(X, Y)` `# after P turns` `def` `findValue( X , Y , P ):` ` ` `if` `P ` `%` `2` `=` `=` `0` `:` ` ` `return` `int` `(` `max` `(X, Y) ` `/` `min` `(X, Y))` ` ` `else` `:` ` ` `return` `int` `(` `max` `(` `2` `*` `X, Y) ` `/` `min` `(` `2` `*` `X, Y))` `# Driver code` `# 1st test case` `X ` `=` `1` `Y ` `=` `2` `P ` `=` `1` `print` `(findValue(X, Y, P))` `# 2nd test case` `X ` `=` `3` `Y ` `=` `7` `P ` `=` `2` `print` `((findValue(X, Y, P)))` `# This code is contribted by "Sharad_Bhardwaj".` |

## C#

`// C# program to find max(X, Y) / min(X, Y)` `// after P turns` `using` `System;` `class` `GFG` `{` ` ` `public` `static` `int` `findValue(` `int` `X, ` `int` `Y,` ` ` `int` `P)` ` ` `{` ` ` `if` `(P % 2 == 0)` ` ` `return` `(Math.Max(X, Y) /` ` ` `Math.Min(X, Y));` ` ` `else` ` ` `return` `(Math.Max(2 * X, Y) /` ` ` `Math.Min(2 * X, Y));` ` ` `}` ` ` ` ` `// Driver code` ` ` `public` `static` `void` `Main()` ` ` `{` ` ` `// 1st test case` ` ` `int` `X = 1, Y = 2, P = 1;` ` ` `Console.WriteLine(findValue(X, Y, P));` ` ` ` ` `// 2nd test case` ` ` `X = 3;` ` ` `Y = 7;` ` ` `P = 2;` ` ` `Console.WriteLine(findValue(X, Y, P));` ` ` `}` `}` `//This code is contributed by vt_m` |

## PHP

`<?php` `// PHP program to find` `// max(X, Y) / min(X, Y)` `// after P turns` `function` `findValue(` `$X` `, ` `$Y` `, ` `$P` `)` `{` ` ` `if` `(` `$P` `% 2 == 0)` ` ` `return` `(int)(max(` `$X` `, ` `$Y` `) /` ` ` `min(` `$X` `, ` `$Y` `));` ` ` `else` ` ` `return` `(int)(max(2 * ` `$X` `, ` `$Y` `) /` ` ` `min(2 * ` `$X` `, ` `$Y` `));` `}` `// Driver code` `// 1st test case` `$X` `= 1;` `$Y` `= 2;` `$P` `= 1;` `echo` `findValue(` `$X` `, ` `$Y` `, ` `$P` `), ` `"\n"` `;` `// 2nd test case` `$X` `= 3; ` `$Y` `= 7; ` `$P` `= 2;` `echo` `findValue(` `$X` `, ` `$Y` `, ` `$P` `), ` `"\n"` `;` `// This code is contributed by ajit` `?>` |

## Javascript

`<script>` ` ` `// Javascript program to find max(X, Y) / min(X, Y)` ` ` `// after P turns` ` ` ` ` `function` `findValue(X, Y, P)` ` ` `{` ` ` `if` `(P % 2 == 0)` ` ` `return` `parseInt((Math.max(X, Y) / Math.min(X, Y)), 10);` ` ` `else` ` ` `return` `parseInt((Math.max(2 * X, Y) / Math.min(2 * X, Y)), 10);` ` ` `}` ` ` ` ` `// 1st test case` ` ` `let X = 1, Y = 2, P = 1;` ` ` `document.write(findValue(X, Y, P) + ` `"</br>"` `);` ` ` ` ` `// 2nd test case` ` ` `X = 3, Y = 7, P = 2;` ` ` `document.write(findValue(X, Y, P));` ` ` ` ` `// This code is contributed by divyeshrabadiya07.` `</script>` |

**Output:**

1 2

**Time Complexity:**O(1)