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".

Output:

1 2

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