PAT1027 Colors in Mars (20分) 10进制转13进制

题目

People in Mars represent the colors in their computers in a similar way as the Earth people. That is, a color is represented by a 6-digit number, where the first 2 digits are for Red, the middle 2 digits for Green, and the last 2 digits for Blue. The only difference is that they use radix 13 (0-9 and A-C) instead of 16. Now given a color in three decimal numbers (each between 0 and 168), you are supposed to output their Mars RGB values.

Input Specification:

Each input file contains one test case which occupies a line containing the three decimal color values.

Output Specification:

For each test case you should output the Mars RGB value in the following format: first output #, then followed by a 6-digit number where all the English characters must be upper-cased. If a single color is only 1-digit long, you must print a 0 to its left.

Sample Input:

15 43 71

Sample Output:

123456

题目解读

简单说，就是给你3个10进制数字（0-168），输出一个"#"号，把他们都转成13进制（0-9，A-C）并输出，中间不要有空格，168也就是 CC，所以转换结果最多也就是 CC，宽度为2，但是要求转换结果只有1位的时候要前面补0，以2位的格式输出，并且字母只能是大写。（比如输出 #12A3BB）

思路

最核心的肯定就是把这个10进制的数（num）转成13进制，但是它最多只有两位，所以高位就是 num / 13，低位就是 num % 13，这不就是两个位置凑齐了？？

还有个问题是，10-->A，11-->B，12-->C，所以用一个字符数组作为映射表就可以了。

比如 char c[14] = {"0123456789ABC"}， 然后把原本的输出 cout << num / 13 << num % 13 变成 cout << c[num / 13] << c[num % 13] 就搞定

代码

#include <iostream>
using namespace std;

int main() {
    // 作为映射表
    char c[14] = {"0123456789ABC"};
    // cout << "#";
    printf("#");
    for(int i = 0; i < 3; ++i) {
        int num;
        // cin >> num;
        scanf("%d", &num);
        // 转成13进制，两位，高位是 / 13，地位是 % 13
        cout << c[num / 13] << c[num % 13];
    }
    return 0;
}