POJ 1740 A New Stone Game

A New Stone Game

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 4431		Accepted: 2388

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

3
2 1 3
2
1 1
0

Sample Output

1
0
题目大意：对于n堆石子，每堆若干个，两人轮流操作，每次操作分两步，第一步从某堆中去掉至少一个，第二步（可省略）把该堆剩余石子的一部分分给其它的某些堆。最后谁无子可取即输。
解题方法：

当组数为偶数时候，如果2N组可分为N组两两相同的组，就是先行者必败。关键在于找必败点

分析：

　　　　　　　只有一堆时先手必胜。
                   有两堆时若两堆相等则后手只用和先手一样决策即可保证胜利，后手必胜。若不同则先手可以使其变成相等的两堆，先手必胜。
                   有三堆时先手只用一次决策即可将其变成两堆相等的局面，先手必胜。
                   有四堆时由于三堆必胜，无论先手后手都想逼对方取完其中一堆，而只有在四堆都为一颗时才会有人取完其中一堆，联系前面的结论可以发现，只有当四堆可以分成    两两相等的两对时先手才会失败。
                   

从而得到3条性质

 首先，末状态为必败态，第一条性质符合。
 其次，可以证明任何一个胜态都有策略变成必败态（分奇数堆和偶数堆两种情况讨论）。
 最后，证明任何一个必败态都无法变成另一个必败态（比较简单）。

#include <stdio.h>
#include <iostream>
#include <string.h>
using namespace std;

int main()
{
    int Count[105];
    int n, x;

    while(scanf("%d", &n) != EOF && n != 0)
    {    
        bool flag = true;
        memset(Count, 0, sizeof(Count));
        for (int i = 0; i < n; i++)
        {
            scanf("%d", &x);
            Count[x]++;
        }
        if (n % 2 == 1)
        {
            printf("1
");
        }
        else
        {
            for (int i = 0; i <= 100; i++)
            {
                if (Count[i] % 2 == 1)
                {
                    flag = false;
                    break;
                }
            }
            if (!flag)
            {
                printf("1
");
            }
            else
            {
                printf("0
");
            }
        }
    }
    return 0;
}