标签：
Game
Time Limit:1000MS Memory Limit:65536KB
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
1
1
2
1 1
3
1 2 3
Sample Output
Win
Lose
Lose
题解：
这道题是典型的Nim游戏，与Nim游戏不同的是该题中除了至少拿走一颗珠子以外，还可以将该堆剩下的珠子分为两堆，其实这对游戏的胜负判断是没有影响的。因为至少拿走了一颗珠子，那么xor求和，之前为0，拿走珠子以后，无论是否将剩下的该堆珠子分为两堆，此时的xor和必不为零，反之亦然。
关于Nim游戏，已经有大量的解释，在此略过。
?以下是代码：
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#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <algorithm>#include <cctype>#include <sstream>#include <queue>using namespace std;#define F(i,s,e) for(int i = s;i<e;i++)#define ss(x) scanf("%d",&x)#define write() freopen("1.in","r",stdin)#define W(x) while(x)int main(){ //write(); int n,t,sum; W(ss(n)!=EOF){ sum=0; W(n--){ ss(t); sum^=t;//对所有的输入xor求和 } if(sum)printf("Win\n"); else printf("Lose\n");//和为零，则先手必输 }} |
Spring-1-E Game（HDU 5011）解题报告及测试数据
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原文地址：http://www.cnblogs.com/gzdaijie/p/4311730.html
