LA-5059(组合游戏)

题意：

有n堆石子,分别有a1,a2,...,an个,两个游戏者轮流操作,每次可以选一堆m拿走至少一个且不超过一半的石子,谁不能拿石子就算输;

思路：

a1太大打印sg表找规律,然后就是异或和了;

AC代码：

#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('
');
}

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;

LL get_sg(LL x)
{
    if(x%2==0)return x/2;
    return get_sg(x/2);
}
int main()
{       
        int t;
        read(t);
        while(t--)
        {
            int n;
            LL ans=0,a;
            read(n);
            For(i,1,n)
            {
                read(a);
                ans^=get_sg(a);
            }
            if(ans)printf("YES
");
            else printf("NO
");
            //printf("%s
",ans ? "YES":"NO");
        }
        
        return 0;
}