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;
}