HDU1847 Good Luck in CET-4 Everybody 博弈 SG函数

题意：给定n张牌，两个人轮流摸牌，每次摸牌张数为2的幂次，问先手胜还是后手胜  n≤1000

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1847

 #include<iostream>
 #include<cstdio>
 #include<algorithm>
 #include<cmath>
 #include<cstring>
 #include<vector>

 using namespace std;

 const int MAXN = ;
 int n, _;
 int f[MAXN];
 int sg[MAXN];
 bool tf[MAXN];

 int get_() {
     f[] = ;
     int i;
     for (i = ; f[i - ] <=  ; ++i) {
         f[i] = f[i - ] * ;
     }
     return i - ;
 }

 void get_sg() {
     sg[] = ;
     for (int i = ; i <= ; ++i) {
         for (int j = ; j <= ; ++j) tf[j] = ;
         for (int j = ; j <=  && f[j] <= i; ++j) tf[sg[i - f[j]]] = ;
         int j = ;
         while (!tf[j]) ++j;
         sg[i] = j;
     }
 }

 int main() {
     _ = get_();
     get_sg();
     while (scanf("%d", &n) != EOF) {
         if (sg[n] > ) cout << "Kiki\n"; else cout << "Cici\n";
     }
     return ;
 }