Hdoj 5011 game

The game is exactly the same as the NIM game.

Game Time Limit: 2000/1000 MS (Java/others) memory limit: 65536/65536 K (Java/Others)
Total submission (s): 148 accepted submission (s): 116


Problem descriptionhere 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 coshould 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.
Inputthere 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 AI (AI <231) means there are ai beads in the I-th pile.
Outputfor each test case, if the first player can win the game, ouput "win" and if he can't, ouput "lose"
Sample Input

1121 131 2 3

 
Sample output

WinLoseLose

 
Source2014 ACM/ICPC Asia Regional Xi 'an online

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int main(){    int n;    while(scanf("%d",&n)!=EOF)    {        int x,ans=0;        for(int i=0;i<n;i++)        {            scanf("%d",&x);            ans=ans^x;        }        if(ans==0)            puts("Lose");        else puts("Win");    }    return 0;}

Hdoj 5011 game