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题目链接:
Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 65536/65536 K (Java/Others)
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
//#include <bits/stdc++.h>
#include <stack>
using namespace std;
#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
typedef long long LL;
template<class T> void read(T&num) {
char CH; bool F=false;
for(CH=getchar();CH<‘0‘||CH>‘9‘;F= CH==‘-‘,CH=getchar());
for(num=0;CH>=‘0‘&&CH<=‘9‘;num=num*10+CH-‘0‘,CH=getchar());
F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
if(!p) { puts("0"); return; }
while(p) stk[++ tp] = p%10, p/=10;
while(tp) putchar(stk[tp--] + ‘0‘);
putchar(‘\n‘);
}
const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=2e6+10;
const int maxn=500+10;
const double eps=1e-8;
int sg[N],vis[25];
inline int get_sg(int x)
{
mst(vis,0);
//cout<<x<<endl;
for(int i=19;i>=0;i--)
{
if(x&(1<<i))
{
int temp=x;
for(int j=i-1;j>=0;j--)
{
if(!(x&(1<<j)))
{
temp^=(1<<i)^(1<<j);
//cout<<temp<<endl;
vis[sg[temp]]=1;
break;
}
}
}
}
for(int i=0;i<=19;i++)if(!vis[i])return i;
return 0;
}
inline void Init()
{
For(i,0,(1<<20)-1)sg[i]=get_sg(i);
}
int main()
{
Init();
int t;
read(t);
while(t--)
{
int n,a,m,ans=0;
read(n);
while(n--)
{
int sum=0;
read(m);
For(i,1,m)
{
read(a);
sum|=(1<<(20-a));
}
ans^=sg[sum];
}
if(ans)printf("YES\n");
else printf("NO\n");
}
return 0;
}
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原文地址:http://www.cnblogs.com/zhangchengc919/p/5701384.html