poj 3975&&hdu 1850 (nim)

//赢得了上风
//从n几年移除堆叠一堆石头，有多少可取的石头堆
# include <stdio.h>
# include <string.h>
# include <algorithm>
using namespace std;
int main()
{
    int n,sum,i,j,a[1010],cot,vis[1010];
    while(~scanf("%d",&n),n)
    {
        sum=0;

        for(i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
        }
        memset(vis,0,sizeof(vis));
        cot=0;
        for(i=0;i<n;i++)
        {
            sum=0;
           vis[i]=1;
            for(j=0;j<n;j++)
            {
                if(!vis[j])
                  sum^=a[j];
            }
            vis[i]=0;
            if(a[i]>sum)//要异或到零的状态。由于要取出几个石头，全部a[i]要大于sum，同样的两个数异或为零
                cot++;
        }
        printf("%d\n",cot);
    }
    return 0;
}