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 beforeV
(5) andX
(10) to make 4 and 9.X
can be placed beforeL
(50) andC
(100) to make 40 and 90.C
can be placed beforeD
(500) andM
(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(); }