hdu1850Being a Good Boy in Spring Festival NiM博弈

//当该点时p-position时，先手输

//当该点为n-position时，即a1^a2......^an = k时;

//先手的第一步是使得a1^a2^.......^an^k = 0;

//那么只要ai^k < ai就行

#include<iostream>

#include<cstdio>

#include<cstring>

using namespace std ;

const int maxn = 10000010 ;

int num[maxn] ;

int main()

{

// freopen("in.txt","r",stdin);

int  n ;

while(scanf("%d",&n)&&n)

{

int sum = 0 ;

for(int i = 1;i <= n ;i++)

{

scanf("%d" ,&num[i]) ;

sum^=num[i] ;

}

if(sum == 0)

{

printf("0\n") ;

continue ;

}

int ans = 0 ;

for(int i = 1;i <= n ;i++)

{

int t = num[i]^sum ;

if(t < num[i])

ans++ ;

}

printf("%d\n",ans );

}

return 0;

}