在家无聊到开始随便找算法题做，嘛，既然都做了，那就写在笔记本上好了～

Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.

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

For example, two is written as II in Roman numeral, just two one's added together. Twelve is written as, XII, which is simply X + II. The number twenty seven is written as XXVII, which is XX + V + II.

Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not IIII. Instead, the number four is written as IV. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX. There are six instances where subtraction is used:

• I can be placed before V (5) and X (10) to make 4 and 9. 
• X can be placed before L (50) and C (100) to make 40 and 90. 
• C can be placed before D (500) and M (1000) to make 400 and 900.

Given an integer, convert it to a roman numeral. Input is guaranteed to be within the range from 1 to 3999.

Example 1:

Input: 3
Output: "III"

Example 2:

Input: 4
Output: "IV"

Example 3:

Input: 9
Output: "IX"

Example 4:

Input: 58
Output: "LVIII"
Explanation: L = 50, V = 5, III = 3.

Example 5:

Input: 1994
Output: "MCMXCIV"
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

Solution

#include <iostream>
#include <string>
#include <vector>
#include <sstream>

using namespace std;

string intToRoman(int num) {
    // all possible basic blocks
    std::vector<int> entry = {1, 4, 5, 9, 10, 40, 50, 90, 100, 400, 500, 900, 1000};
    // and their corresponding Roman numbers
    std::vector<string> roman = {"I", "IV", "V", "IX", "X", "XL", "L", "XC", "C", "CD", "D", "CM", "M"};
    
    // build result string
    std::ostringstream result;
    // iterate reversely
    unsigned long index = entry.size() - 1;
    // number remained to be translate
    int remain = num;
    // translate to Roman numbers until remaining value gets 0
    while (remain != 0) {
        // calculate number of repeats of current Roman number
        int repeat = remain / entry[index];
        // repeat current Roman number / translation
        for (int i = 0;i < repeat;i++) {
            result << roman[index];
        }
        // substract value represented in this round
        remain -= repeat * entry[index];
        // move to next Roman number
        index--;
    }
    
    return result.str();
}