codecs 1264 芳香数

1264 芳香数

题目描述 Description

This question involves calculating the value of aromatic numbers which are a combination of Arabic digits and Roman numerals.

本题是关于计算芳香数数值的问题，芳香数是阿拉伯数字和罗马数字的组合。

An aromatic number is of the form ARARAR...AR, where each A is an Arabic digit, and each R is a Roman numeral. Each pair AR contributes a value described below, and by adding or subtracting these values together we get the value of the entire aromatic number.

芳香数的格式是ARARAR..ARA，其中A代表阿拉伯数字，R代表罗马数字。每一对AR按照下面的计算方式计算一个值，通过把这些数值加减起来，就得到了整个芳香数的数值。

An Arabic digit A can be 0, 1, 2, 3, 4, 5, 6, 7, 8 or 9. A Roman numeral R is one of the seven letters I, V, X, L, C, D, or M. Each Roman numeral has a base value:

阿拉伯数字是0,1,2..9，罗马数字是I,V,X,L,C,D,M。

Symbol I V X L C D M Base value 1 5 10 50 100 500 1000

符号I V X L C D M的值是1 5 10 50 100 500 1000。

The value of a pair AR is A times the base value of R. Normally, you add up the values of the pairs to get the overall value. However, wherever there are consecutive symbols ARA0R0 with R0 having a strictly bigger base value than R, the value of pair AR must be subtracted from the total, instead of being added.

一对AR的值计算为A乘以R。一般的，我们把所有的AR的值加起来就得到了芳香数的值。但是如果存在连续的两个数对ARA0R0，其中R0严格大于R的话，则需要减去AR的值，而不是加上。

For example, the number 3M1D2C has the value 3∗1000+1∗500+2∗100 = 3700 and 3X2I4X has the value 3 ∗ 10 − 2 ∗ 1 + 4 ∗ 10 = 68.

举个例子，3M1D2C 的值为3*1000+1*500+2*100=3700，而3X2I4X的值为3*10-2*1+4*10=68

Write a program that computes the values of aromatic numbers.

你的任务是写一个程序来计算一个给定的芳香数的值。

输入描述 Input Description

 The input is a valid aromatic number consisting of between 2 and 20 symbols.

输入是一个合法的芳香数，包含了2-20个字符。

输出描述 Output Description

 The output is the decimal value of the given aromatic number.

输出是一个十进制的整数代表这个芳香数的值。

样例输入 Sample Input

样例输入 1： 3M1D2C

样例输入 2： 2I3I2X9V1X

样例输出 Sample Output

样例输出 1： 3700

样例输出 2： -16

题目挺吓人，但其实就那么回事，耐心读就好了

#include<cstdio>
#include<cmath>
#include<cstring>
#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
int tot,x,flag,y;
int a[100000];
int b[100000];
string n;
int main()
{
    ios::sync_with_stdio(false);
    cin>>n;
    for(int i=0;i<n.size();i+=2)
    {
        a[x]=n[i]-48;
        switch(n[i+1])
        {
            case 'I':
                b[x++]=1;
                break;
            case 'V':
                b[x++]=5;
                break;
            case 'X':
                b[x++]=10;
                break;
            case 'L':
                b[x++]=50;
                break;
            case 'C':
                b[x++]=100;
                break;
            case 'D':
                b[x++]=500;
                break;
            case 'M':
                b[x++]=1000; 
        }
    }
    for(int i=0;i<x;i++)
    {
        if(b[i+1]>b[i])
            tot-=a[i]*b[i];
        else
            tot+=a[i]*b[i];
    }
    cout<<tot;
    return 0;
}