HDU - 1847 Good Luck in CET-4 Everybody!(sg函数，水题)

题目链接：点击查看

题目大意：给出n张牌，两个人轮流摸牌，每次只能摸2的幂次，问先手必胜还是必败

题目分析：简单sg打表题，先预处理出sg表，然后O(1)查询即可

#include<iostream>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<climits>
#include<cmath>
#include<cctype>
#include<stack>
#include<queue>
#include<list>
#include<vector>
#include<set>
#include<map>
#include<sstream>
#define Pi acos(-1.0)
using namespace std;typedef long long LL;const int inf=0x3f3f3f3f;const int N=1e3+100; int n;int a[110];int sg[N];bool mex[N];void get_SG()
{sg[0]=0;for(int i=1;i<N;i++){memset(mex,false,sizeof(mex));for(int j=1;j<=n;j++){if(a[j]>i)break;mex[sg[i-a[j]]]=true;}for(int j=0;j<N;j++){if(!mex[j]){sg[i]=j;break;}}}
}void init()
{n=0;for(int i=1;i<=1000;i<<=1)a[++n]=i;get_SG();
}int main()
{
//  freopen("input.txt","r",stdin);init();int n;while(scanf("%d",&n)!=EOF){if(sg[n])printf("Kiki\n");elseprintf("Cici\n");}return 0;
}

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