10进制数字向0~3999的罗马数字的转换

31 class Solution {
32 public:
33     string intToRoman(int num) {
34         char* c[4][10] = {
35             { "","I","II","III","IV","V","VI","VII","VIII","IX" },
36             { "","X","XX","XXX","XL","L","LX","LXX","LXXX","XC" },
37             { "","C","CC","CCC","CD","D","DC","DCC","DCCC","CM" },
38             { "","M","MM","MMM" }
39         };
40         string roman;
41         roman.append(c[3][num / 1000 % 10]);
42         roman.append(c[2][num / 100 % 10]);
43         roman.append(c[1][num / 10 % 10]);
44         roman.append(c[0][num % 10]);
45 
46         return roman;
47     }
48 };
49 
50 int main() {
51     int n;
52     cin >> n;
53     Solution s;
54     cout << s.intToRoman(n) << endl;
55     return 0;
56 }
原文地址:https://www.cnblogs.com/baocong/p/6756500.html