Coin game winner where every player has three choices

A and B are playing a game. At the beginning there are n coins. Given two more numbers x and y. In each move a player can pick x or y or l coins. A always starts the game. The player who picks the last coin wins the game. For a given value of n, find whether A will win the game or not if both are playing optimally.

Examples:

Input :  n = 5, x = 3, y = 4
Output : A
There are 5 coins, every player can pick 1 or
3 or 4 coins on his/her turn.
A can win by picking 3 coins in first chance.
Now 2 coins will be left so B will pick one 
coin and now A can win by picking the last coin.

Input : 2 3 4
Output : B

Let us take few example values of n for x = 3, y = 4.
n = 0 A can not pick any coin so he losses
n = 1 A can pick 1 coin and win the game
n = 2 A can pick only 1 coin. Now B will pick 1 coin and win the game
n = 3 4 A will win the game by picking 3 or 4 coins
n = 5, 6 A will choose 3 or 4 coins. Now B will have to choose from 2 coins so A will win.

We can observe that A wins game for n coins only when it loses for coins n-1, n-x and n-y.

C++

// CPP program to find winner of game
// if player can pick 1, x, y coins
#include <bits/stdc++.h>
using namespace std;

// To find winner of game
bool findWinner(int x, int y, int n)
{
    // To store results
    int dp[n + 1];

    // Initial values
    dp[0] = false;
    dp[1] = true;

    // Computing other values.
    for (int i = 2; i <= n; i++) {

        // If A losses any of i-1 or i-x
        // or i-y game then he will
        // definitely win game i
        if (i - 1 >= 0 and !dp[i - 1])
            dp[i] = true;
        else if (i - x >= 0 and !dp[i - x])
            dp[i] = true;
        else if (i - y >= 0 and !dp[i - y])
            dp[i] = true;

        // Else A loses game.
        else
            dp[i] = false;
    }

    // If dp[n] is true then A will
    // game otherwise  he losses
    return dp[n];
}

// Driver program to test findWinner();
int main()
{
    int x = 3, y = 4, n = 5;
    if (findWinner(x, y, n))
        cout << 'A';
    else
        cout << 'B';

    return 0;
}

Java

// Java program to find winner of game
// if player can pick 1, x, y coins
import java.util.Arrays;

public class GFG { 
    
    // To find winner of game
    static boolean findWinner(int x, int y, int n)
    {
        // To store results
        boolean[] dp = new boolean[n + 1];
     
        Arrays.fill(dp, false);
    
        // Initial values
        dp[0] = false;
        dp[1] = true;
     
        // Computing other values.
        for (int i = 2; i <= n; i++) {
     
            // If A losses any of i-1 or i-x
            // or i-y game then he will
            // definitely win game i
            if (i - 1 >= 0 && dp[i - 1] == false)
                dp[i] = true;
            else if (i - x >= 0 && dp[i - x] == false)
                dp[i] = true;
            else if (i - y >= 0 && dp[i - y] == false)
                dp[i] = true;
     
            // Else A loses game.
            else
                dp[i] = false;
        }
     
        // If dp[n] is true then A will
        // game otherwise  he losses
        return dp[n];
    }
     
    // Driver program to test findWinner();
    public static void main(String args[])
    {
        int x = 3, y = 4, n = 5;
        if (findWinner(x, y, n) == true)
            System.out.println('A');
        else
            System.out.println('B');
    }
}
// This code is contributed by Sumit Ghosh


Output:

A

This article is contributed by nuclode. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

GATE CS Corner    Company Wise Coding Practice

Recommended Posts:







Writing code in comment? Please use ide.geeksforgeeks.org, generate link and share the link here.