Game

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 493    Accepted Submission(s): 380

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 < 10⁵) means there are n piles of beads. The next line contains n postive integer, the i-th postive integer a_i(a_i < 2³¹) 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

1
1
2
1 1
3
1 2 3

Sample Output

Win
Lose
Lose

分析：尼姆博弈题，虽然稍微有点不同，但可以忽略

这里引用一下关于尼姆博弈的简单说法：

尼姆博弈：

指的是一个游戏，目前有任意堆石子，每堆的石子都是任意的，双方轮流从中取石子，有如下规则：

1，每一步只能从某一堆取走至少一枚石子；

2，谁取到最后一次石子，谁胜利。

结论：假设有n堆物品，每堆物品有a[i]件，若a[0]^a[1]^a[2]^........a[n-1]!=0 先手必胜，否则必败。

#include <cstdio>
#include <iostream>
using namespace std;
typedef long long ll;
int c[1000010];
int main()
{
	int n,ans;
	while(scanf("%d",&n)!=EOF)
	{
		ans=0;
		for(int i=0;i<n;i++)
		{
		  scanf("%d",&c[i]);
		  ans=ans^c[i];
	    }
		if(ans!=0)
		puts("Win");
		else
		puts("Lose");
	}
	return 0;
}