HDU 5011 Game（尼姆博弈）

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探讨了两人游戏中的策略选择与概率理论相结合的应用，通过分析初始筹码分布预测赢家。

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Game

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 662 Accepted Submission(s): 454

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

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
typedef __int64 ll;
using namespace std;

int main()
{
    int n;
    while (scanf("%d", &n) != EOF)
    {
        ll ans = 0;
        ll tmp;
        for (int i = 0; i < n; i++)
        {
            scanf ("%I64d", &tmp);
            ans = ans^tmp;
        }
        if (ans == 0)
            printf("Lose\n");
        else printf("Win\n");
    }
}