用N堆箱子找到游戏的赢家

给定N堆仅包含White(W)和Black(B)的盒子，则两个玩家A和B玩游戏。
玩家A可以从具有最高框White的堆的顶部移除任何数量的盒子，而玩家B可以从具有最高框Black的堆的顶部移除任何数量的盒子。如果有一个顶部为W的堆，则玩家A已移除一个或多个盒子。类似地，如果有一个顶部为B的堆，则玩家B必须删除一个或多个盒子。他们轮流玩，玩家A先移动。两名球员都发挥最佳状态。

任务是找到游戏的赢家(谁不能走最后一步)。玩家最后一步输了比赛。

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，即最后采取行动的玩家输掉游戏，被称为米塞雷·尼姆(Misere Nim)的游戏。

1. 如果在同一堆中连续出现同一盒子，我们可以简单地用同一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。因此，我们可以压缩所有桩以得到形式为BWBWBW或WBWBWB的桩。
2. 现在，假设有一个堆，其最上面的箱子与最下面的箱子不同，我们可以证明即使没有这个堆，答案也保持不变，因为如果一个玩家在该堆上进行移动，另一位玩家则通过移除来进行下一移动同一堆箱子，第一位玩家的位置完全相同。
3. 如果一堆的最顶部和最底部的框都相同，则需要一个或玩家额外移动一格。因此，只需计算每个玩家必须进行的额外移动的数量即可。用完多余动作的玩家将首先赢得比赛。

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

输出：

A

时间复杂度： O(N)