用 N 堆盒子找出游戏的赢家

给定N堆仅包含白 (W) 和黑 (B) 框，两个玩家 A 和 B 玩游戏。
玩家 A 可以从最上面有白色框的堆的顶部取出任意数量的框，玩家 B 可以从最上面有黑色框的堆的顶部取出任意数量的框。如果有一堆顶部为 W 的堆，则玩家 A 移除了一个或多个盒子。类似地，如果有一堆顶部为 B 的堆，那么玩家 B 必须移除一个或多个盒子。他们交替进行，玩家A首先移动。两个球员都发挥最佳。
任务是找到游戏的赢家(谁不能走最后一步)。玩家最后移动输掉比赛。
例子：

Input: N = 2 
WBW, BWB 
Output: A 
Player A can remove all boxes from pile 1. Now player B has to make a choice from pile 2. Whatever choice player B makes, he/she has to make the last move.
Input: N = 3 
WWWW, WBWB, WBBW 
Output: B

方法：博弈论问题，即最后一步的玩家输掉游戏，被称为 Misere Nim 的游戏。

1. 如果在任何一堆中连续出现同一个盒子，我们可以简单地用同一个盒子的一个副本替换它们，因为最好删除所有出现在顶部的盒子。所以，我们可以压缩所有的桩，得到形式为 BWBWBW 或 WBWBWB 的桩。
2. 现在，假设有一堆，其最上面的盒子与其最底部的盒子不同，我们可以证明即使没有这一堆，答案仍然相同，因为如果一个玩家在这堆上移动，其他玩家通过移除来进行下一步来自同一堆的盒子，导致第一个玩家的位置完全相同。
3. 如果一堆最上面和最下面的盒子都一样，它需要一个人或玩家多走一步。因此，只需计算每个玩家必须进行的额外移动次数。首先用完额外动作的玩家赢得游戏。

下面是上述方法的实现。

C++

// Program to find winner of the game
#include 
using namespace std;
 
// Function to return find winner of game
string Winner(int n, string pile[])
{
 
    int a = 0, b = 0;
    for (int i = 0; i < n; ++i) {
        int l = pile[i].length();
 
        // Piles begins and ends with White box 'W'
        if (pile[i][0] == pile[i][l - 1] && pile[i][0] == 'W')
            a++;
 
        // Piles begins and ends with Black box 'B'
        if (pile[i][0] == pile[i][l - 1] && pile[i][0] == 'B')
            b++;
    }
    if (a <= b)
        return "A";
    else
        return "B";
}
 
// Driver code
int main()
{
 
    int n = 2;
    string pile[n] = { "WBW", "BWB" };
 
    // function to print required answer
    cout << Winner(n, pile);
 
    return 0;
}

Java

// Program to find winner of the game
class GFG
{
 
// Function to return find winner of game
static String Winner(int n, String []pile)
{
 
    int a = 0, b = 0;
    for (int i = 0; i < n; ++i)
    {
        int l = pile[i].length();
 
        // Piles begins and ends with White box 'W'
        if (pile[i].charAt(0) == pile[i].charAt(l - 1) &&
            pile[i].charAt(0) == 'W')
            a++;
 
        // Piles begins and ends with Black box 'B'
        if (pile[i].charAt(0) == pile[i].charAt(l - 1) &&
            pile[i].charAt(0) == 'B')
            b++;
    }
    if (a <= b)
        return "A";
    else
        return "B";
}
 
// Driver code
public static void main(String[] args)
{
    int n = 2;
    String pile[] = { "WBW", "BWB" };
 
    // function to print required answer
    System.out.println(Winner(n, pile));
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 code to find winner of the game
 
# Function to return find winner of game
def Winner(n, pile):
 
    a, b = 0, 0
    for i in range(0, n):
        l = len(pile[i])
 
        # Piles begins and ends with White box 'W'
        if (pile[i][0] == pile[i][l - 1] and
            pile[i][0] == 'W'):
            a += 1
 
        # Piles begins and ends with Black box 'B'
        if (pile[i][0] == pile[i][l - 1] and
            pile[i][0] == 'B'):
            b += 1
     
    if a <= b:
        return "A"
    else:
        return "B"
 
# Driver code
if __name__ == "__main__":
 
    n = 2
    pile = ["WBW", "BWB"]
 
    # function to print required answer
    print(Winner(n, pile))
 
# This code is contributed by Rituraj Jain

C#

// Program to find winner of the game
using System;
     
class GFG
{
 
// Function to return find winner of game
static String Winner(int n, String []pile)
{
    int a = 0, b = 0;
    for (int i = 0; i < n; ++i)
    {
        int l = pile[i].Length;
 
        // Piles begins and ends with White box 'W'
        if (pile[i][0] == pile[i][l - 1] &&
            pile[i][0] == 'W')
            a++;
 
        // Piles begins and ends with Black box 'B'
        if (pile[i][0] == pile[i][l - 1] &&
            pile[i][0] == 'B')
            b++;
    }
    if (a <= b)
        return "A";
    else
        return "B";
}
 
// Driver code
public static void Main(String[] args)
{
    int n = 2;
    String []pile = { "WBW", "BWB" };
 
    // function to print required answer
    Console.WriteLine(Winner(n, pile));
}
}
 
// This code is contributed by Princi Singh

Javascript

 
// Program to find winner of the game
 
// Function to return find winner of game
function Winner(n, pile)
{
 
    var a = 0, b = 0;
    for (var i = 0; i < n; ++i) {
        var l = pile[i].length;
 
        // Piles begins and ends with White box 'W'
        if (pile[i][0] == pile[i][l - 1] && pile[i][0] == 'W')
            a++;
 
        // Piles begins and ends with Black box 'B'
        if (pile[i][0] == pile[i][l - 1] && pile[i][0] == 'B')
            b++;
    }
    if (a <= b)
        return "A";
    else
        return "B";
}
 
// Driver code
var n = 2;
var pile = ["WBW", "BWB"];
 
// function to print required answer
document.write( Winner(n, pile));
 
// This code is contributed by noob2000.

输出：

A

时间复杂度： O(N)

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