HDU - 1847 Good Luck in CET-4 Everybody!(sg函数)
2017-09-07 23:01
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Good Luck in CET-4 Everybody!
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 10838 Accepted Submission(s): 7038
Problem Description
大学英语四级考试就要来临了,你是不是在紧张的复习?也许紧张得连短学期的ACM都没工夫练习了,反正我知道的Kiki和Cici都是如此。当然,作为在考场浸润了十几载的当代大学生,Kiki和Cici更懂得考前的放松,所谓“张弛有道”就是这个意思。这不,Kiki和Cici在每天晚上休息之前都要玩一会儿扑克牌以放松神经。
“升级”?“双扣”?“红五”?还是“斗地主”?
当然都不是!那多俗啊~
作为计算机学院的学生,Kiki和Cici打牌的时候可没忘记专业,她们打牌的规则是这样的:
1、 总共n张牌;
2、 双方轮流抓牌;
3、 每人每次抓牌的个数只能是2的幂次(即:1,2,4,8,16…)
4、 抓完牌,胜负结果也出来了:最后抓完牌的人为胜者;
假设Kiki和Cici都是足够聪明(其实不用假设,哪有不聪明的学生~),并且每次都是Kiki先抓牌,请问谁能赢呢?
当然,打牌无论谁赢都问题不大,重要的是马上到来的CET-4能有好的状态。
Good luck in CET-4 everybody!
Input
输入数据包含多个测试用例,每个测试用例占一行,包含一个整数n(1<=n<=1000)。
Output
如果Kiki能赢的话,请输出“Kiki”,否则请输出“Cici”,每个实例的输出占一行。
Sample Input
1
3
Sample Output
Kiki
Cici
Author
lcy
Source
ACM Short Term Exam_2007/12/13
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lcy | We have carefully selected several similar problems for you: 1849 1846 2147 2149 2188
#include <bits/stdc++.h> using namespace std; const int N = 1e4 + 10; int n; int sg , vis , f[] = {0, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024}; void Get_sg(){ memset(sg, 0 ,sizeof(sg)); for(int i = 1; i <= 1000; i++){ memset(vis, 0 , sizeof(vis)); for(int j = 1; f[j] <= i; j++){ vis[sg[i-f[j]]] = 1; } for(int j = 0; j <= 1000; j++){ if(!vis[j]){ sg[i] = j; //printf("sg[%d] = %d\n", i, sg[i]); break; } } } } int main(){ Get_sg(); while(scanf("%d", &n) == 1){ if(sg ) printf("Kiki\n"); else printf("Cici\n"); } }
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