A1027 Colors in Mars (20分)

1027 Colors in Mars (20分)

首先,先贴柳神的博客

https://www.liuchuo.net/ 这是地址

想要刷好PTA,强烈推荐柳神的博客,和算法笔记

目录

• 1027 Colors in Mars (20分)
  • • Input Specification:
    • Output Specification:
    • Sample Input:
    • Sample Output:
  • 题目大意
  • 代码如下

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

题目大意

就是换进制,3个十进制的,转换为13进制 10,11,12 用A,B,C代替

每个数字都要输出两位,不足的用0补齐

代码如下

#include<iostream>
#include<cmath>
using namespace std;
int main(void) {
	int Red[3];
	char r[7] = {'0'};
	for (int i = 0; i < 7; i++) {
		r[i] = '0';
	}
	cin >> Red[0] >> Red[1] >> Red[2];
	for (int i = 0; i < 3; i++) {
		while (Red[i]) {
			if (Red[i] % 13 >= 10) {
				r[i * 2 + 1] = Red[i] % 13+55;
			}
			else {
				r[i * 2 + 1] = Red[i] % 13 + 48;
			}
			Red[i] /= 13;
			if (Red[i] % 13 >= 10) {
				r[i * 2] = Red[i] % 13 + 55;
			}
			else {
				r[i * 2] = Red[i] % 13 + 48;
			}
			Red[i] /= 13;

		}
	}
	printf("#");
	for (int i = 0; i < 6; i++) {
		printf("%c", r[i]);
	}
	return 0;
}