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

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<algorithm>
 4 #include<cmath>
 5 #include<cstring>
 6 #include<vector>
 7  
 8 using namespace std;
 9 
10 const int MAXN = 2333;
11 int n, _;
12 int f[MAXN];
13 int sg[MAXN];
14 bool tf[MAXN];
15 
16 int get_() {
17     f[0] = 1;
18     int i;
19     for (i = 1; f[i - 1] <= 1100 ; ++i) {
20         f[i] = f[i - 1] * 2;
21     }
22     return i - 1;
23 }
24 
25 void get_sg() {
26     sg[0] = 0;
27     for (int i = 1; i <= 1100; ++i) {
28         for (int j = 0; j <= 1100; ++j) tf[j] = 1;
29         for (int j = 0; j <= 1100 && f[j] <= i; ++j) tf[sg[i - f[j]]] = 0;
30         int j = 0;
31         while (!tf[j]) ++j;
32         sg[i] = j;
33     }
34 }
35 
36 int main() {
37     _ = get_();
38     get_sg();
39     while (scanf("%d", &n) != EOF) {
40         if (sg[n] > 0) cout << "Kiki\n"; else cout << "Cici\n";
41     }
42     return 0;
43 }

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posted on 2018-03-17 11:34 ^m 阅读(111) 评论(0) 编辑收藏举报