Even-odd turn game with two integers

2

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;
}

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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