PAT basic 1059
来源:互联网 发布:诺亚方舟 知乎 编辑:程序博客网 时间:2024/06/18 18:10
#include <cstdio>#include <set>#include <cmath>using namespace std;int ran[10000];bool isprime(int a) { if(a <= 1) return false; int Sqrt = sqrt((double)a); for(int i = 2; i <= Sqrt; i++) { if(a % i == 0) return false; } return true;}int main() { int n, k; scanf("%d", &n); for(int i = 0; i < n; i++) { int id; scanf("%d", &id); ran[id] = i + 1; } scanf("%d", &k); set<int> ss; for(int i = 0; i < k; i++) { int id; scanf("%d", &id); printf("%04d: ", id); if(ran[id] == 0) { printf("Are you kidding?\n"); continue; } if(ss.find(id) == ss.end()) { ss.insert(id); } else { printf("Checked\n"); continue; } if(ran[id] == 1) { printf("Mystery Award\n"); }else if(isprime(ran[id])) { printf("Minion\n"); }else { printf("Chocolate\n"); } } return 0;}/*ran数组标记每个id对应的排名,集合ss存储所有已经询问过的id,如果发现当前id已经出现在ss中,则输出“Checked”,如果ran[id] == 0说明当前id不在排名列表中,所以输出“Are you kidding?”,如果ran[id]为1则输出“Minion”,如果ran[id]为素数则输出“Mystery Award”,否则输出“Chocolate”*/
阅读全文
0 0
- PAT basic 1059
- PAT Basic
- PAT (Basic Level) Practise
- PAT Basic 1001
- PAT Basic 1002
- PAT Basic 1005
- PAT Basic 1006
- PAT Basic 1007
- PAT Basic 1008
- PAT Basic 1009
- PAT Basic 1010
- pat basic level 1016
- pat basic level 1018
- pat basic level 1019
- PAT(basic level)题解
- PAT basic 1004 : 成绩排名
- PAT (Basic) 1001~1005
- PAT (Basic) 1006~1010
- Zemax中高斯光束设置的相关问题
- sqlmap教程2
- PAT basic 1058
- linux邮件系统
- 正则表达式-限定符_转义字符
- PAT basic 1059
- HDU-6154 CaoHaha's staff (找规律+二分)
- Centos7.2环境RPM 安装MySQL5.6.24
- SQL语言简介之最详细解读、SQL和PL/SQL区别
- https的单向和双向
- hdu 6168 Numbers (STL)
- 学习日记20
- CodeForces
- 图结构练习——BFS——从起始点到目标点的最短步数