PAT 1116 Come on! Let's C [简单]
1116 Come on! Let's C (20 分)
"Let's C" is a popular and fun programming contest hosted by the College of Computer Science and Technology, Zhejiang University. Since the idea of the contest is for fun, the award rules are funny as the following:
- 0、 The Champion will receive a "Mystery Award" (such as a BIG collection of students' research papers...).
- 1、 Those who ranked as a prime number will receive the best award -- the Minions (小黄人)!
- 2、 Everyone else will receive chocolates.
Given the final ranklist and a sequence of contestant ID's, you are supposed to tell the corresponding awards.
Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N (≤104), the total number of contestants. Then N lines of the ranklist follow, each in order gives a contestant's ID (a 4-digit number). After the ranklist, there is a positive integer K followed by K query ID's.
Output Specification:
For each query, print in a line ID: award where the award is Mystery Award, or Minion, or Chocolate. If the ID is not in the ranklist, print Are you kidding? instead. If the ID has been checked before, print ID: Checked.
Sample Input:
6
1111
6666
8888
1234
5555
0001
6
8888
0001
1111
2222
8888
2222
Sample Output:
8888: Minion
0001: Chocolate
1111: Mystery Award
2222: Are you kidding?
8888: Checked
2222: Are you kidding?
题目大意:给出n个排名的id,序号代表排名,如果是第一名输出Mystery Award,素数名输出Minion,其他人输出Chocolate,重复查询提示。
水水的AC:
#include <iostream>
#include <algorithm>
#include<cstdio>
#include <map>
#include<cmath>
using namespace std; map<int,string> award;
map<int,int> query;
bool prime(int a){
int q=sqrt(a);
for(int i=;i<=q;i++){
if(a%i==){
return false;
}
}
return true;
}
int main()
{
int n,no;
scanf("%d",&n);
for(int i=;i<n;i++){
scanf("%d",&no);
if(i==){
award[no]="Mystery Award";
}else{
if(prime(i+)){
award[no]="Minion";
}else{
award[no]="Chocolate";
}
}
}
int m;
scanf("%d",&m);
for(int i=;i<m;i++){
scanf("%d",&no);
if(award.count(no)==){
printf("%04d: Are you kidding?\n",no);
}else {
if(query.count(no)==){
printf("%04d: Checked\n",no);
}else{//使用printf怎么输出String啊啊?好像没办法。。。
printf("%04d: ",no);
cout<<award[no]<<"\n";
query[no]=;
}
}
}
return ;
}
//有一个槽点,就是在写最后出现了个问题,代码中有注释,所以就混合使用printf和cout输出了,罪过。
1.查了一下发现是需要将其转换成char*类型,直接使用:
printf("%04d: %s\n",no,award[no].c_str());
即可!c_str()就是将string转换为char*的函数!!!
PAT 1116 Come on! Let's C [简单]的更多相关文章
- PAT 1002. A+B for Polynomials (25) 简单模拟
1002. A+B for Polynomials (25) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue T ...
- PAT 1116 Come on! Let's C
1116 Come on! Let's C (20 分) "Let's C" is a popular and fun programming contest hosted b ...
- PAT Sign In and Sign Out[非常简单]
1006 Sign In and Sign Out (25)(25 分) At the beginning of every day, the first person who signs in th ...
- PAT 1069 The Black Hole of Numbers[简单]
1069 The Black Hole of Numbers(20 分) For any 4-digit integer except the ones with all the digits bei ...
- pat 1116 Come on! Let's C(20 分)
1116 Come on! Let's C(20 分) "Let's C" is a popular and fun programming contest hosted by t ...
- PAT 甲级 1077 Kuchiguse (20 分)(简单,找最大相同后缀)
1077 Kuchiguse (20 分) The Japanese language is notorious for its sentence ending particles. Person ...
- PAT 甲级 1035 Password (20 分)(简单题)
1035 Password (20 分) To prepare for PAT, the judge sometimes has to generate random passwords for ...
- PAT 数列求和-加强版 (20分)(简单模拟)
给定某数字A(1≤A≤9)以及非负整数N(0≤N≤100000),求数列之和S=A+AA+AAA+⋯+AA⋯A(N个A).例如A=1, N=3时,S=1+11+111=123 输入格式: 输入数字A与 ...
- PAT 1102 Invert a Binary Tree[比较简单]
1102 Invert a Binary Tree(25 分) The following is from Max Howell @twitter: Google: 90% of our engine ...
随机推荐
- hive组件和执行过程
转自http://blog.csdn.net/lifuxiangcaohui/article/details/40262021 对Hive的基本组成进行了总结: 1.组件: 元存储(Metastore ...
- 第二百七十四节,同源策略和跨域Ajax
同源策略和跨域Ajax 什么是同源策略 尽管浏览器的安全措施多种多样,但是要想黑掉一个Web应用,只要在浏览器的多种安全措施中找到某种措施的一个漏洞或者绕过一种安全措施的方法即可.浏览器的各种保安措 ...
- 学习:100个高质量Java开发者博客
谷歌关键字搜索:100个高质量Java开发者博客. Java开发牛人十大必备网站.
- 在ChemDraw中如何使用ChemACX
ChemACX是一款功能强大的化学品比价数据库,可与E-Notebook和ChemDraw整合使用,极大地方便生化科学家们采购化学品.那么很多用户就会开始疑惑该如何在ChemDraw化学绘图软件调用C ...
- uva 707(记忆化搜索)
题目链接:http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=21261 思路:此题需要记忆化搜索,dp[x][y][t]表示当前状 ...
- 正则表达式Regex
1.概念 正则表达式,又称规则表达式.(英语:Regular Expression,在代码中常简写为regex.regexp或RE),计算机科学的一个概念.正则表通常被用来检索.替换那些符合某个模式( ...
- layui多选框
多选下拉框:http://sun.faysunshine.com/layui/formSelects-v4/example/example_v4.html 1.下载formSelects-v4.1 2 ...
- WPF进阶之接口(1):IValueConverter,IMultiValueConverter
看一个例子,FontFamily="Trebuchet MS, GlobalSansSerif.CompositeFont" .这样一条简单的语句,熟悉WPF的人在xaml中可能经 ...
- Securi-Pi:使用树莓派作为安全跳板
导读 像很多 LinuxJournal 的读者一样,我也过上了当今非常普遍的“科技游牧”生活,在网络之间,从一个接入点到另一个接入点,我们身处现实世界的不同地方却始终保持连接到互联网和日常使用的其它网 ...
- poj_2823 线段树
题目大意 给定一行数,共N个.有一个长度为K的窗口从左向右滑动,窗口中始终有K个数字,窗口每次滑动一个数字.求各个时刻窗口中的最大值和最小值. 题目分析 直接搜索,复杂度为O(n^2).本题可以看做是 ...