ACM The Famous Clock
The Famous Clock
- 描述
-
Mr. B, Mr. G and Mr. M are now in Warsaw, Poland, for the 2012’s ACM-ICPC World Finals Contest. They’ve decided to take a 5 hours training every day before the contest. Also, they plan to start training at 10:00 each day since the World Final Contest will do so. The scenery in Warsaw is so attractive that Mr. B would always like to take a walk outside for a while after breakfast. However, Mr. B have to go back before training starts, otherwise his teammates will be annoyed. Here is a problem: Mr. B does not have a watch. In order to know the exact time, he has bought a new watch in Warsaw, but all the numbers on that watch are represented in Roman Numerals. Mr. B cannot understand such kind of numbers. Can you translate for him?
- 输入
- Each test case contains a single line indicating a Roman Numerals that to be translated. All the numbers can be found on clocks. That is, each number in the input represents an integer between 1 and 12. Roman Numerals are expressed by strings consisting of uppercase ‘I’, ‘V’ and ‘X’. See the sample input for further information.
- 输出
- For each test case, display a single line containing a decimal number corresponding to the given Roman Numerals.
- 样例输入
-
I
II
III
IV
V
VI
VII
VIII
IX
X
XI
XII - 样例输出
-
Case 1: 1
Case 2: 2
Case 3: 3
Case 4: 4
Case 5: 5
Case 6: 6
Case 7: 7
Case 8: 8
Case 9: 9
Case 10: 10
Case 11: 11
Case 12: 12#include <iostream>
#include <map>
#include <string>
using namespace std; int romanToInt(string s){
int length = s.length();
if (length < ) return ;
map<int,int> romanMap;
romanMap['I'] = ;romanMap['V'] = ;romanMap['X'] = ;
int index = length - , sum = romanMap[s[index]];
while (--index >= ) {
if (romanMap[s[index+]] >romanMap[s[index]]) sum -= romanMap[s[index]];
else sum += romanMap[s[index]];
}
return sum;
} int main(){
int i = ;
string str;
while(cin >> str) cout<<"Case "<<++i<<": "<<romanToInt(str)<<endl;
return ;
}
ACM The Famous Clock的更多相关文章
- HDU 4256 The Famous Clock
The Famous Clock Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) ...
- nyoj 484-The Famous Clock
484-The Famous Clock 内存限制:64MB 时间限制:1000ms 特判: No 通过数:2 提交数:2 难度:1 题目描述: Mr. B, Mr. G and Mr. M are ...
- The Famous Clock
描述 Mr. B, Mr. G and Mr. M are now in Warsaw, Poland, for the 2012’s ACM-ICPC World Finals Contest. T ...
- 【南阳OJ分类之语言入门】80题题目+AC代码汇总
小技巧:本文之前由csdn自动生成了一个目录,不必下拉一个一个去找,可通过目录标题直接定位. 本文转载自本人的csdn博客,复制过来的,排版就不弄了,欢迎转载. 声明: 题目部分皆为南阳OJ题目. 代 ...
- JavaScript美术馆进化史
内容选自<<JavaScript DOM 编程艺术>>第4-6章,跟着作者一起见证美术馆的进化吧. 先放效果图,然后一步步做出每个效果.每个效果都有它实用的地方,且知道过程可以 ...
- javascript DOM(2) 一个网页上切换显示不同的图片或文本
摘自: javascript DOM 编程艺术 1. 在一个网页上切换显示不同的图片 <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Tran ...
- 《DOM Scripting》学习笔记-——第五章、第六章 案列改进
第四章的案例代码可以得到更好的改进.例如:预留退路.向后兼容性和分离js. 原html代码: <!DOCTYPE html> <html lang="en"> ...
- 《DOM Scripting》学习笔记-——第四章 案列分析 JS美术馆(点击链接到图片)
实现效果:点击图片链接,可以在当前网页显示图片,并且显示图片标题. Html代码 <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.1//EN&qu ...
- HDU 4256 翻译罗马数字
参考自:https://www.cnblogs.com/ECJTUACM-873284962/p/6414173.html The Famous Clock Time Limit: 2000/1000 ...
随机推荐
- jquery学习笔记----元素筛选
1.eq() 筛选指定索引号的元素2.first() 筛选出第一个匹配的元素3.last() 筛选出最后一个匹配的元素4.hasClass() 检查匹配的元素是否含有指定的类5.filter() 筛 ...
- sprintf_s的使用
int sprintf_s(char *restrict buffer, rsize_t bufsz, const char *restrict format, ...); ...
- MyEclipse2014配置Tomcat开发JavaWeb程序JSP以及Servlet(转载)
转载地址:http://blog.csdn.net/21aspnet/article/details/21867241 1.安装准备 1).下载安装MyEclipse2014,这已经是最新版本. 2) ...
- ASP.NET Web Api 服务器端变了,客户端该如何修改请求(转载)
转载地址:http://www.cnblogs.com/fzrain/p/3558765.html 前言 一旦我们将API发布之后,消费者就会开始使用并和其他的一些数据混在一起.然而,当新的需求出现时 ...
- android:id="@id/resid" , andorid:id="@+id/resid" 的区别
的区别?android:id="@id/resid" // 引用现有的资源idandorid:id="@+id/resid" // 新增一个资源id i ...
- HDU2546 饭卡(背包)
开始写成01背包的形式,求m元可买物品价值的最大值 dp[j] = max(dp[j], dp[j - pri[i]] + pri[i]) 结果为m - dp[m] 但后来发现是有问题的, 比如这组过 ...
- 使用git进行团队合作开发
1.git 和 svn 的差异 git和svn 最大的差异在于git是分布式的管理方式而svn是集中式的管理方式.如果不习惯用代码管理工具,可能比较难理解分布式管理和集中式管理的概念.下面介绍两种工具 ...
- 攻城狮在路上(肆)How tomcat works(一) 简单的web服务器
该节总共三个类:Request\Response\HttpServer---user.dir 该节的目的是实现简单web服务器对静态文件的访问.需要对请求头.请求体的格式有所了解,不然就没有 ...
- Educational Codeforces Round 3 E. Minimum spanning tree for each edge LCA/(树链剖分+数据结构) + MST
E. Minimum spanning tree for each edge Connected undirected weighted graph without self-loops and ...
- connot resolve symbol R
出现这个原因大都是layout里的xml文件出错,就不会自动生成R文件