Given an integer, convert it to a roman numeral.

Input is guaranteed to be within the range from 1 to 3999.

 

没认真想，看到一种方法，应该算是最简单的方法了吧

基础知识：

罗马数字是由字符I,V,X,L,C,D,M等等表示的，其中

I = 1;

V = 5;

X = 10;

L = 50;

C = 100;

D = 500;

M = 1000;

接下来应该是V开始的重复，但是上面要加一个横线，表示对应数字的1000倍。

个位应该是：I,II,III,IV,V,VI,VII,VIII,IX

十位应该是：X,XX,XXX,XL,L,LX,LXX,LXXX,XC

百位应该是：C,CC,CCC,CD,D,DC,DCC,DCCC,CM

class Solution {public:    string intToRoman(int num) {         char c[10][10][10]={
   {
   "0","I","II","III","IV","V","VI","VII","VIII","IX"},{
   "0","X","XX","XXX","XL","L","LX"        ,"LXX","LXXX","XC"},{
   "0","C","CC","CCC","CD","D",              "DC","DCC","DCCC","CM"},{
   "0","M","MM","MMM"}};        int t=1;        int tmp=num;        string st;        if(tmp/1000!=0) st+=c[3][tmp/1000];        if((tmp%1000)/100!=0) st+=c[2][(tmp%1000)/100];        if((tmp%100)/10!=0) st+=c[1][(tmp%100)/10];        if(tmp%10!=0) st+=c[0][tmp%10];        return st;            }};