Nim的修改游戏 - 芒果文档

给定一个整数数组arr[] ，两个玩家A和B正在玩一个游戏，其中A可以从数组中删除任意数量的非零元素，这些元素是3 的倍数。类似地， B可以删除5 的倍数。不能移除任何元素的玩家输掉游戏。任务是如果A先开始并且双方都发挥最佳，则找到游戏的获胜者。
例子：

Input: arr[] = {1, 2, 3, 5, 6}
Output: A
3 and 6 are the elements that A can remove.
5 is the only element that B can remove.
A can remove 3 in his first move then B will have to remove 5. In the next turn, A will remove 6 and B will be left with no more moves to make.
Input: arr[] = {3, 5, 15, 20, 6, 9}
Output: A

编程需要懂一点英语

方法：存放在movesA元素只被3整除的计数，在movesB元素只有5整除的数和元素双方在movesBoth整除。现在，

如果movesBoth = 0，那么两个玩家只能删除可以被他们各自的数字整除的元素，只有当movesA > moveB时， A才会赢得游戏。
如果movesBoth > 0，那么为了发挥最佳效果， A将删除所有可以被3和5整除的元素，以便B没有要从公共元素中删除的元素，那么只有当moveA + 1 时A才会获胜> 动作B

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the winner of the game
string getWinner(int arr[], int n)
{
    int movesA = 0, movesB = 0, movesBoth = 0;
 
    for (int i = 0; i < n; i++) {
 
        // Increment common moves
        if (arr[i] % 3 == 0 && arr[i] % 5 == 0)
            movesBoth++;
 
        // Increment A's moves
        else if (arr[i] % 3 == 0)
            movesA++;
 
        // Increment B's moves
        else if (arr[i] % 5 == 0)
            movesB++;
    }
 
    // If there are no common moves
    if (movesBoth == 0) {
        if (movesA > movesB)
            return "A";
        return "B";
    }
 
    // 1 is added because A can remove all the elements
    // that are part of the common moves in a single move
    if (movesA + 1 > movesB)
        return "A";
    return "B";
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 5, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << getWinner(arr, n);
 
    return 0;
}

Java

// Java implementation of the approach
class GfG
{
 
    // Function to return the winner of the game
    static String getWinner(int arr[], int n)
    {
        int movesA = 0, movesB = 0, movesBoth = 0;
     
        for (int i = 0; i < n; i++)
        {
     
            // Increment common moves
            if (arr[i] % 3 == 0 && arr[i] % 5 == 0)
                movesBoth++;
     
            // Increment A's moves
            else if (arr[i] % 3 == 0)
                movesA++;
     
            // Increment B's moves
            else if (arr[i] % 5 == 0)
                movesB++;
        }
     
        // If there are no common moves
        if (movesBoth == 0)
        {
            if (movesA > movesB)
                return "A";
            return "B";
        }
     
        // 1 is added because A can remove
        // all the elements that are part
        // of the common moves in a single move
        if (movesA + 1 > movesB)
            return "A";
        return "B";
    }
 
    // Driver code
    public static void main(String []args)
    {
         
        int arr[] = { 1, 2, 3, 5, 6 };
        int n = arr.length;
        System.out.println(getWinner(arr, n));
    }
}
 
// This code is contributed by Rituraj Jain

Python3

# Python3 implementation of the approach
 
# Function to return the winner of the game
def getWinner(arr, n):
 
    movesA, movesB, movesBoth = 0, 0, 0
    for i in range(0, n):
 
        # Increment common moves
        if arr[i] % 3 == 0 and arr[i] % 5 == 0:
            movesBoth += 1
 
        # Increment A's moves
        elif arr[i] % 3 == 0:
            movesA += 1
 
        # Increment B's moves
        elif arr[i] % 5 == 0:
            movesB += 1
 
    # If there are no common moves
    if movesBoth == 0:
        if movesA > movesB:
            return "A"
        return "B"
 
    # 1 is added because A can
    # remove all the elements
    # that are part of the common
    # moves in a single move
    if movesA + 1 > movesB:
        return "A"
    return "B"
 
# Driver code
if __name__ == "__main__":
 
    arr = [1, 2, 3, 5, 6]
    n = len(arr)
    print(getWinner(arr, n))
 
# This code is contributed by Rituraj Jain

C#

// C# implementation of the approach
using System;
 
class GfG
{
 
    // Function to return the winner of the game
    static String getWinner(int []arr, int n)
    {
        int movesA = 0, movesB = 0, movesBoth = 0;
     
        for (int i = 0; i < n; i++)
        {
     
            // Increment common moves
            if (arr[i] % 3 == 0 && arr[i] % 5 == 0)
                movesBoth++;
     
            // Increment A's moves
            else if (arr[i] % 3 == 0)
                movesA++;
     
            // Increment B's moves
            else if (arr[i] % 5 == 0)
                movesB++;
        }
     
        // If there are no common moves
        if (movesBoth == 0)
        {
            if (movesA > movesB)
                return "A";
            return "B";
        }
     
        // 1 is added because A can remove
        // all the elements that are part
        // of the common moves in a single move
        if (movesA + 1 > movesB)
            return "A";
        return "B";
    }
 
    // Driver code
    public static void Main(String []args)
    {
         
        int []arr = { 1, 2, 3, 5, 6 };
        int n = arr.Length;
        Console.WriteLine(getWinner(arr, n));
    }
}
 
// This code is contributed by
// Rajput-Ji

PHP

 $movesB)
            return "A";
        return "B";
    }
 
    // 1 is added because A can remove all the elements
    // that are part of the common moves in a single move
    if ($movesA + 1 > $movesB)
        return "A";
    return "B";
}
 
    // Driver code
    $arr = array( 1, 2, 3, 5, 6 );
    $n = sizeof($arr) / sizeof($arr[0]);
    echo getWinner($arr, $n);
 
// This code is contributed by ajit.
?>

Javascript

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