硬币游戏赢家，每个玩家都有三个选择

A 和 B 正在玩游戏。一开始有n个硬币。给定另外两个数字 x 和 y。在每次移动中，玩家可以选择 x 或 y 或 1 个硬币。 A 总是开始游戏。选择最后一个硬币的玩家赢得游戏，或者无法选择任何硬币的人输掉游戏。对于给定的 n 值，如果双方都以最佳方式进行游戏，请找出 A 是否会赢得比赛。

例子：

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

让我们为 x = 3, y = 4 取一些 n 的示例值。
n = 0 A 不能选择任何硬币，所以他输了
n = 1 A 可以选择 1 个硬币并赢得比赛
n = 2 A 只能捡到 1 个硬币。现在 B 将选择 1 个硬币并赢得游戏
n = 3 4 A 将通过选择 3 或 4 个硬币赢得比赛
n = 5, 6 A 将选择 3 或 4 个硬币。现在 B 必须从 2 个硬币中进行选择，所以 A 将获胜。
我们可以观察到，只有当 B 输掉 n-1 或 nx 或 ny 硬币时，A 才赢了 n 个硬币。

C++

// C++ program to find winner of game
// if player can pick 1, x, y coins
#include 
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

Python3

# Python3 program to find winner of game
# if player can pick 1, x, y coins
 
# To find winner of game
def findWinner(x, y, n):
     
    # To store results
    dp = [0 for i in range(n + 1)]
 
    # Initial values
    dp[0] = False
    dp[1] = True
 
    # Computing other values.
    for i in range(2, n + 1):
 
        # 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 not dp[i - 1]):
            dp[i] = True
        elif (i - x >= 0 and not dp[i - x]):
            dp[i] = True
        elif (i - y >= 0 and not 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 Code
x = 3; y = 4; n = 5
if (findWinner(x, y, n)):
    print('A')
else:
    print('B')
 
# This code is contributed by Azkia Anam

C#

// C# program to find winner of game
// if player can pick 1, x, y coins
using System;
 
public class GFG {
     
    // To find winner of game
    static bool findWinner(int x, int y, int n)
    {
         
        // To store results
        bool[] dp = new bool[n + 1];
     
        for(int i = 0; i < n+1; i++)
            dp[i] =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()
    {
        int x = 3, y = 4, n = 5;
         
        if (findWinner(x, y, n) == true)
            Console.WriteLine('A');
        else
            Console.WriteLine('B');
    }
}
 
// This code is contributed by vt_m.

PHP

= 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();
    $x = 3; $y = 4; $n = 5;
    if (findWinner($x, $y, $n))
        echo 'A';
    else
        echo 'B';
         
// This code is contributed by anuj_67.
?>

Javascript

输出：

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