HDOJ/HDU 2352 Verdis Quo(罗马数字与10进制数的转换)
Problem Description
The Romans used letters from their Latin alphabet to represent each of the seven numerals in their number system. The list below shows which
letters they used and what numeric value each of those letters represents:
I = 1
V = 5
X = 10
L = 50
C = 100
D = 500
M = 1000
Using these seven numerals, any desired number can be formed by following the two basic additive and subtractive rules. To form a number using
the additive rule the Roman numerals are simply written from left to right in descending order, and the value of each roman numeral is added
together. For example, the number MMCLVII has the value 1000 + 1000 + 100 + 50 + 5 + 1 + 1 = 2157. Using the addition rule alone could lead to
very long strings of letters, so the subtraction rule was invented as a result. Using this rule, a smaller Roman numeral to the left of a larger one is
subtracted from the total. In other words, the number MCMXIV is interpreted as 1000 - 100 + 1000 + 10 - 1 + 5 = 1914.
Over time the Roman number writing system became more standardized and several additional rules were developed. The additional rules used today
are:
- The I, X, or C Roman numerals may only be repeated up to three times in succession. In other words, the number 4 must be represented as IV
and not as IIII. - The V, L, or D numerals may never be repeated in succession, and the M numeral may be repeated as many 2. times as necessary.
- Only one smaller numeral can be placed to the left of another. For example, the number 18 is represented as XVIII but not as XIIX.
- Only the I, X, or C can be used as subtractive numerals.
- A subtractive I can only be used to the left of a V or X. Likewise a X can only appear to the left of a L or C, and a C can only be used to the
left of a D or M. For example, 49 must be written as XLIX and not as IL.
Your goal is to write a program which converts Roman numbers to base 10 integers.
Input
The input to this problem will consist of the following:
A line with a single integer “N” (1 ≤ N ≤ 1000), where N indicates how many Roman numbers are to be converted.
A series of N lines of input with each line containing one Roman number. Each Roman number will be in the range of 1 to 10,000 (inclusive)
and will obey all of the rules laid out in the problem’s introduction.
Output
For each of the N Roman numbers, print the equivalent base 10 integer, one per line.
Sample Input
3
IX
MMDCII
DXII
Sample Output
9
2602
512
罗马数字共有7个,即I(1)、V(5)、X(10)、L(50)、C(100)、D(500)和M(1000)。
1、重复数次:一个罗马数字重复几次,就表示这个数的几倍。
2、右加左减:
2.1 在较大的罗马数字的右边记上较小的罗马数字,表示大数字加小数字。
2.2 在较大的罗马数字的左边记上较小的罗马数字,表示大数字减小数字。
2.3 左减的数字有限制,仅限于I、X、C。比如45不可以写成VL,只能是XLV
2.4 但是,左减时不可跨越一个位数。比如,99不可以用IC(100 - 1)表示,而是用XCIX([100 - 10] + [10 - 1])表示。(等同于阿拉伯数字每位数字分别表示。)
2.5 左减数字必须为一位,比如8写成VIII,而非IIX。
注意的就是:I只能在V,X的左边。X只能在L,C的左边。C只能在D,M的左边。
知道这些就可以AC了。
import java.util.Scanner;
/**
* @author 陈浩翔
* 2016-6-5
*/
public class Main{
static char[] chS={'I','V','X','L','C','D','M'};
static int[] chN={1,5,10,50,100,500,1000};
static String[] strS={"IV","IX","XL","XC","CD","CM"};
static int[] strN={2,2,20,20,200,200};
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t=sc.nextInt();
while(t-->0){
String str=sc.next();
int num=0;
for(int i=0;i<str.length();i++){
for(int j=0;j<chS.length;j++){
if(str.charAt(i)==chS[j]){
num+=chN[j];
break;
}
}
}
String s="";
for(int i=1;i<str.length();i++){
s=""+str.charAt(i-1)+str.charAt(i);
for(int j=0;j<strS.length;j++){
if(s.equals(strS[j])){
num-=strN[j];
break;
}
}
}
System.out.println(num);
}
}
}
HDOJ/HDU 2352 Verdis Quo(罗马数字与10进制数的转换)的更多相关文章
- HDOJ(HDU) 1877 又一版 A+B(进制、、)
Problem Description 输入两个不超过整型定义的非负10进制整数A和B(<=231-1),输出A+B的m (1 < m <10)进制数. Input 输入格式:测试输 ...
- [转]as3 算法实例【输出1 到最大的N 位数 题目:输入数字n,按顺序输出从1 最大的n 位10 进制数。比如输入3,则输出1、2、3 一直到最大的3 位数即999。】
思路:如果我们在数字前面补0的话,就会发现n位所有10进制数其实就是n个从0到9的全排列.也就是说,我们把数字的每一位都从0到9排列一遍,就得到了所有的10进制数. /** *ch 存放数字 *n n ...
- c++描述将一个2进制数转化成10进制数(用到初始化栈,进栈,入栈)
/* c++描述将2进制数转化成10进制数 问题,1.初始化栈后,用new,不知道delete是否要再写一个函数释放内存, 还是在哪里可以加上delete 2.如果栈满了,我要分配多点空间,我想的办法 ...
- c语言将2进制数转化为10进制数(栈的初始化,进栈,出栈)
//c语言描述 将2进制转化为10进制 #include <stdio.h> #include <stdlib.h> #include <math.h> #defi ...
- python中的2、8、16、10进制之间的转换
python除法的坑 众所周知,python除法有两个运算符,一个是/,还有一个是//,那么这两个有什么不同之处呢? 从图片可以得知,使用//返回一个float类型,而使用/返回一个int类型.我们总 ...
- ORACLE 36进制和10进制,互相转换函数
第一部分 --36转10进制 create or replace function f_36to10 (str varchar) return int is returnValue int; s ...
- HDU 2352 Verdis Quo
罗马数字转化为十进制的值 题目非常的长 提取有效信息 并且介绍很多规则 但是事实上有用的信息就是如何加 什么时候减 当当前字母小于下一个字母时 减去当前字母的值 #include <iostre ...
- 【LeetCode】将罗马数字转换成10进制数
Roman to Integer Given a roman numeral, convert it to an integer. 首先介绍罗马数字 罗马数字共有七个,即I(1),V(5),X(10) ...
- 已知从BUF开始存放了10个字类型有符号数据,编程求出这10个数中的最大数和最小数(将最大数存入MAX字单元、最小数存入MIN字单元),并将其以10进制数的形式在屏幕上显示出来。
data segment pmax db 0dh,0ah , 'MAX : ','$' pmin db 0dh,0ah , 'MIN : ','$' buf ...
随机推荐
- Java RandomAccessFile的使用(转载的文章,出处http://www.2cto.com/kf/201208/149816.html)
Java的RandomAccessFile提供对文件的读写功能,与普通的输入输出流不一样的是RamdomAccessFile可以任意的访问文件的任何地方.这就是“Random”的意义所在. Rando ...
- 九度OJ 1079 手机键盘
题目地址:http://ac.jobdu.com/problem.php?pid=1079 题目描述: 按照手机键盘输入字母的方式,计算所花费的时间 如:a,b,c都在“1”键上,输入a只需要按一次, ...
- Linux网络设置高级指南
from:http://www.oschina.net/question/23734_117144 Linux网络设置高级指南 本文面向的是被Linux复杂的有线无线网络架构弄得头昏脑胀:或者被网上半 ...
- 清理SQL多余登录信息
服务器列表.登陆帐户.密码等信息都记录在 %AppData%\Microsoft\Microsoft SQL Server\100\Tools\Shell\SqlStudio.bin (2008)%A ...
- Spring-Boot-XML-Restful-Service
http://docs.spring.io/spring-boot/docs/current/reference/htmlsingle/#howto-write-an-xml-rest-service ...
- scrapy使用爬取多个页面
scrapy是个好玩的爬虫框架,基本用法就是:输入起始的一堆url,让爬虫去get这些网页,然后parse页面,获取自己喜欢的东西.. 用上去有django的感觉,有settings,有field.还 ...
- USB Key插入和移除监控
近期在做USB Key插入和移除监控,已经做到了插入和移除USB Key时,程序能够及时感应到. 如下为源代码: private void Form1_Load(object sender, Even ...
- 在ubuntu中获得root权限
在终端中输入:(1)sudo passwd rootEnter new UNIX password: (在这输入你的密码)Retype new UNIX password: (确定你输入的密码)pas ...
- Entity Framework Code First 映射继承关系
转载 http://www.th7.cn/Program/net/201301/122153.shtml Code First如何处理类之间的继承关系.Entity Framework Code Fi ...
- Js处理数据——前端分页工具
这几天有小伙伴讨论起了分页的相关问题,这里我也简单讲下前端如何简单便捷的利用Js(库)写出优雅,好用的分页工具. 分页是个很简单又超多接触的技术点,粗略来讲分如下两种: 真分页--每次根据页码.页大小 ...