hdu 5011 Game Nim博弈 2014 ACM/ICPC Asia Regional Xi'an Online

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本文介绍了一款名为Nim的两人对战游戏，并详细解析了其背后的博弈策略。玩家轮流从多堆珠子中移除至少一颗珠子，可选择不分堆或分成两堆。文章通过样例输入输出展示了如何判断先手玩家是否能赢得游戏。

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5011

Problem Description

Here is a game for two players. The rule of the game is described below:

● In the beginning of the game, there are a lot of piles of beads.

● Players take turns to play. Each turn, player choose a pile i and remove some (at least one) beads from it. Then he could do nothing or split pile i into two piles with a beads and b beads.(a,b > 0 and a + b equals to the number of beads of pile i after removing)

● If after a player's turn, there is no beads left, the player is the winner.

Suppose that the two players are all very clever and they will use optimal game strategies. Your job is to tell whether the player who plays first can win the game.

Input

There are multiple test cases. Please process till EOF.

For each test case, the first line contains a postive integer n(n < 10 ⁵) means there are n piles of beads. The next line contains n postive integer, the i-th postive integer a _i(a _i < 2 ³¹) means there are a _i beads in the i-th pile.

Output

For each test case, if the first player can win the game, ouput "Win" and if he can't, ouput "Lose"

Sample Input

Sample Output

  
   Win
Lose
Lose

Source

2014 ACM/ICPC Asia Regional Xi'an Online

Nim博弈..

代码：

#include <algorithm>
#include <cstdlib>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cctype>
#include <cmath>
#include <stack>
#include <queue>
#include <list>
#include <map>
#include <set>

using namespace std;

#define min2(x, y)     min(x, y)
#define max2(x, y)     max(x, y)
#define min3(x, y, z)  min(x, min(y, z))
#define max3(x, y, z)  max3(x, max(y, z))
#define clr(x, y)      memset(x, y, sizeof(x))
#define fr(i,n)        for(int i = 0; i < n; i++)
#define fr1(i,n)       for(int i = 1; i < n; i++)
#define upfr(i,j,n)    for(int i = j; i <= n; i++)
#define dowfr(i,j,n)   for(int i = n; i >= j; i--)
#define scf(n)         scanf("%d", &n)
#define ptf(n)         printf("%d",n)
#define ptfs(s)        printf("%s",s)
#define ptln()         printf("\n")
#define srt(a,n)       sort(a,n)
#define LL long long
#define pi acos(-1.0)
#define inf 1 << 31-1
#define eps 0.00001
#define maxn 100005

int main()
{

        int n;
        while(scf(n) != EOF)
        {
            int k;
            __int64 ans = 0;
            fr(i,n)
            {
                scf(k);
                ans = ans ^ k;
            }
            if(!ans)
                printf("Lose\n");
            else
                printf("Win\n");
        }
    return 0;
}