51Nod

每个数的SG值之和他有多少个1相关

打表复杂度:找K个有序的<n的非负数的复杂度为n^k/(k!)

则这题的SG打表复杂度为64⁸/7! 为1e10左右

void dfs(int cur, int yu, int ans, int next)
{
    if(yu==0)
    {
        vis[ans]=1;
        return ;
    }
    for(int i=next; i<cur; i++)
        dfs(cur, yu-1, ans^sg[i], i);
}
void init()
{
    sg[0]=0;
    for(int i=1;i<=64; i++)
    {
        memset(vis,0,sizeof(vis));
        dfs(i, 7, 0, 0);
        for(int j=0;;j++)
            if(!vis[j])
            {
                sg[i]=j;
                break;
            }
        printf("%d
", sg[i]);
    }
}

#include <bits/stdc++.h>
using namespace std;
typedef unsigned long long uLL;

int sg[] = {0, 1, 2, 4, 8, 16, 32, 64, 128, 255, 256, 512,
            1024, 2048, 3855, 4096, 8192, 13107, 16384, 21845,
            27306, 32768, 38506, 65536, 71576, 92115, 101470,
            131072, 138406, 172589, 240014, 262144, 272069,
            380556, 524288, 536169, 679601, 847140, 1048576,
            1072054, 1258879, 1397519, 2005450, 2097152, 2121415,
            2496892, 2738813, 3993667, 4194304, 4241896, 4617503,
            5821704, 7559873, 8388608, 8439273, 8861366, 11119275,
            11973252, 13280789, 16777216, 16844349, 17102035,
            19984054, 21979742, 23734709
           };
           
int cal(uLL x)
{
    int ret=0;
    for(int i=0;i<64;i++)
        if((x>>i)&1) ret++;
    return ret;
}
int main()
{
    int n;
    while(cin>>n)
    {
        int ans=0;
        while(n--)
        {
            uLL t;
            cin>>t;
            ans ^= sg[cal(t)];
        }
        if(ans) puts("B");
        else puts("L");
    }
}