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。

 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。

 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 1： 3M1D2C

样例输入 2： 2I3I2X9V1X

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;
}