本文迁移自老博客，原始链接为 https://seven.blog.ustc.edu.cn/usacopreface-numbering/
传统罗马数字用单个字母表示特定的数值，给定罗马数字的组成规则，统计从1~n的数字转化为罗马数字以后，各个字母出现的总次数 解法：先预处理整数的各个位用罗马数字的表示方式，并存入数组，然后将整数转换成罗马数字并进行统计。
#include<iostream>
#include<algorithm>
#include<fstream>
#include<vector>
#include<map>
using namespace std;
int n;
string thousand[4] = {"","M","MM","MMM"};
string hundred[10] = {"","C","CC","CCC","CD","D","DC","DCC","DCCC","CM"};
string decade[10] = {"","X","XX","XXX","XL","L","LX","LXX","LXXX","XC"};
string unit[10] = {"","I","II","III","IV","V","VI","VII","VIII","IX"};
string preface(int i)
{
string str;
if(i / 1000 != 0)
{
str += thousand[i / 1000];
i %= 1000;
}
if(i / 100 != 0)
{
str += hundred[i / 100];
i %= 100;
}
if(i / 10 != 0)
{
str += decade[i / 10];
i %= 10;
}
if(i / 1 != 0)
str += unit[i / 1];
return str;
}
map<char , int> ans;
void cal(string str)
{
for(int i = 0 ; i < str.size(); i++)
ans[str[i]]++;
}
int main()
{
ofstream fout ("preface.out");
ifstream fin ("preface.in");
fin>>n;
char cnt[8] = "IVXLCDM";
for(int i = 1 ; i <= n ; i++)
cal(preface(i));
for(int i = 0 ; i < 7 ; i++)
{
if(ans[cnt[i]])
fout<<cnt[i]<<" "<<ans[cnt[i]]<<endl;
}
return 0;
}
