hdu 2014西安赛区网络赛 5011Game

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一做网络赛就感到压力十足挤进前90 压力山大啊。。。

Game

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 72 Accepted Submission(s): 56

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

Sample Output

  Win
Lose
Lose

说是跟SG函数有关的东西将所有的数取异或值就可以了判断是否大于0 大于0就是Win

至于说为什么我也不造。。

http://baike.baidu.com/view/2855458.htm?fr=aladdin SG函数

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string.h>
#include <string>
#include <queue>#define eps 1e-8
#define op operator
#define MOD  10009
#define MAXN  100100#define FOR(i,a,b)  for(int i=a;i<=b;i++)
#define FOV(i,a,b)  for(int i=a;i>=b;i--)
#define REP(i,a,b)  for(int i=a;i<b;i++)
#define REV(i,a,b)  for(int i=a-1;i>=b;i--)
#define MEM(a,x)    memset(a,x,sizeof a)
#define ll __int64using namespace std;//int a[100010];
//priority_queue<int,vector<int>,less<int> > Q;int main()
{
//freopen("ceshi.txt","r",stdin);int n;while(scanf("%d",&n)!=EOF){int res=0;for(int i=0;i<n;i++){int x;scanf("%d",&x);res=res^x;}if(res>0)  puts("Win");else puts("Lose");}return 0;
}

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