LA5059 Playing With Stones

Playing With Stones

 大意：n堆石子，每堆a1,a2....an个，两人轮流操作，每次选一堆，减少至少一个至多石子数一般下取整个。不能拿的人输。给定n,a,问先手必胜/必败

打表求得一堆的sg函数：

0 1 0 2 1 3 0 4 2 5 1 6 3 7 0 8 4 9 2 10 5 11 1 12 6 13 3 14 7 15

不难发现其中有‘’1 2 3 5.....15"

即sg(2n) = n

考虑sg(2n + 1) 为0 1 0 2 1 3 0 4 2 5 1 6 3 7

恰好是原数列，即sg(2n+1) = sg(n) = sg([(2n+1)/2])

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <vector>
#include <cmath> 
#define min(a, b) ((a) < (b) ? (a) : (b))
#define max(a, b) ((a) > (b) ? (a) : (b))
#define abs(a) ((a) < 0 ? (-1 * (a)) : (a))
inline void swap(long long &a, long long &b)
{
    long long tmp = a;a = b;b = tmp;
}
inline void read(long long &x)
{
    x = 0;char ch = getchar(), c = ch;
    while(ch < '0' || ch > '9') c = ch, ch = getchar();
    while(ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
    if(c == '-') x = -x;
}

const long long INF = 0x3f3f3f3f;
const long long MAXN = 100 + 10;

long long t,n,a;

long long sg(long long x)
{
    if(x == 0) return 0;
    return (x & 1) == 0 ? x/2 : sg(x/2);
}

int main()
{
    read(t);
    for(;t;--t)
    {
        read(n);long long tmp = 0;
        for(register long long i = 1;i <= n;++ i) 
        {
            read(a);
            tmp ^= sg(a);
        }
        if(tmp) printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}