POJ 1740 A New Stone Game

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本文深入探讨了两人参与的石子博弈游戏规则，玩家通过取走石头并自由分配到其他堆，目标是使对手无法行动。分析了游戏胜负的关键在于堆数的奇偶性，提出了一种判断策略，确保在奇数堆情况下必胜。

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A New Stone Game

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 5310		Accepted: 2912

Description

Alice and Bob decide to play a new stone game.At the beginning of the game they pick n(1<=n<=10) piles of stones in a line. Alice and Bob move the stones in turn.
At each step of the game,the player choose a pile,remove at least one stones,then freely move stones from this pile to any other pile that still has stones.
For example:n=4 and the piles have (3,1,4,2) stones.If the player chose the first pile and remove one.Then it can reach the follow states.
2 1 4 2
1 2 4 2（move one stone to Pile 2）
1 1 5 2（move one stone to Pile 3）
1 1 4 3（move one stone to Pile 4）
0 2 5 2（move one stone to Pile 2 and another one to Pile 3）
0 2 4 3（move one stone to Pile 2 and another one to Pile 4）
0 1 5 3（move one stone to Pile 3 and another one to Pile 4）
0 3 4 2（move two stones to Pile 2）
0 1 6 2（move two stones to Pile 3）
0 1 4 4（move two stones to Pile 4）
Alice always moves first. Suppose that both Alice and Bob do their best in the game.
You are to write a program to determine who will finally win the game.

Input

The input contains several test cases. The first line of each test case contains an integer number n, denoting the number of piles. The following n integers describe the number of stones in each pile at the beginning of the game, you may assume the number of stones in each pile will not exceed 100.
The last test case is followed by one zero.

Output

For each test case, if Alice win the game,output 1,otherwise output 0.

Sample Input

Sample Output

1
0

题意：Alice和Bob玩取石头的游戏，轮流做两种操作，先从某一堆石头中取走至少1个，之后从一堆石头中取走任意个（含0个）石头分配到其他仍然有石头的堆，如果最后没有了则输。

题解：如果为奇数堆石头的话，那么Alice可以通过以上两种操作使得一堆变为0，而剩余的偶数堆达到两两配对（即两堆石头数目相等）。那么Bob操作之后，Alice又可以变换成类似上一次的状态（配对状态）。那么Alice则必胜。

如果为偶数堆石头的话，若不是配对状态，则Alice可以变换成配对状态，则Alice必胜，若已经是配对状态则Alice必输。

#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=110;
int main()
{
    int n,num,cnt[maxn];
    while(scanf("%d",&n)&&n)
    {
        memset(cnt,0,sizeof(cnt));
        int flag=0;
        for(int i=0;i<n;i++)
        {
            scanf("%d",&num);
            cnt[num]++;
        }
        for(int i=0;i<maxn;i++)
        {
            if(cnt[i]&1)
            {
                flag=1;
            }
        }
        if(flag) printf("1\n");
        else printf("0\n");
    }
    return 0;
}