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310进制数字(0-168),输出一个"#"号,把他们都转成13进制(0-9A-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-->A11-->B12-->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;
}

Q.E.D.


Think twice, code once.