【题解】POJ-1740 A New Stone Game

A New Stone Game(POJ-1740)

题面

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.

输入

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.

输出

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

样例输入

3
2 1 3
2
1 1
0

样例输出

1
0

提示

无

思路

当石子堆数为偶数，且石子数相等的堆也是偶数个时，后手可以模仿先手操作获得胜利，此时为必败态。其它情况时，先手都可以把最大堆的石子取走一个，并将剩下的石子分配给别的堆使得局面为必败态。

代码

using namespace std;
const int mxn = 1e5 + 5;
int a[mxn];

int main()
{
    int n;
    while(~scanf("%d", &n) && n)
    {
        for(int i=0; i<n; i++)
            scanf("%d", &a[i]);

        if(n&1){
            printf("1\n");
            continue;
        }
        sort(a, a+n);

        int f = 1;
        for(int i=0; i<n; i+=2){
            if(a[i+1] != a[i]){
                printf("1\n");
                f = 0;
                break;
            }
        }
        if(f)
            printf("0\n");
    }
    return 0;
}