HDU5011

简单博弈。

Game

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 155 Accepted Submission(s): 120

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 ai(ai < 231) means there are ai 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

Source

2014 ACM/ICPC Asia Regional Xi’an Online

这题的常规做法就是，暴力打SG表。 找规律， 发现sg\[i\] = i; 然后就是所有数异或起来，为0就是输，否则赢。   当然如果你YY功力足够好，直接猜出异或，然后AC， 那也是可以的。

/* ***
Author :kuangbin
Created Time :2014/9/15 0:15:56
File Name :E:\2014ACM\2014网络赛\2014西安\HDU5011.cpp
************************************************ */

#include <stdio.h>

#include <string.h>

#include

#include <math.h>

#include <stdlib.h>

#include <time.h>
using namespace std;

int main()
{
//freopen(“in.txt”,”r”,stdin);
//freopen(“out.txt”,”w”,stdout);
int n;
while(scanf(“%d”,&n) == 1){
int a,sum = 0;
while(n–){
scanf(“%d”,&a);
sum ^= a;
}
if(sum)printf(“Win\n”);
else printf(“Lose\n”);
}
return 0;
}