PAT (Advanced Level)-1027. Colors in Mars (20)

PAT (Advanced Level)-1027. Colors in Mars (20)

1027. Colors in Mars (20)

时间限制
400 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序 Standard
作者 CHEN, Yue


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

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

Output

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 the left.

Sample Input

15 43 71

Sample Output

#123456

解读

单纯的一道十进制转换十三进制的题目。而且十三进制下不超过两位数。
进制转换的方式不做累述,PAT (Advanced Level)-1019. General Palindromic Number (20)的解读部分。

由于只有两位数的十三进制数,且有ABC这样的字符出现,所以最终我选择使用string字符串处理所有的火星数,0-9,A-C全部存入mars_base中直接做返回调用。可以减少此题中的额外处理。

完整代码

#include<iostream>
#include<string>
#include<cstring>

std::string mars_base[] = { "0","1","2","3","4","5","6","7","8","9","A","B","C" };

std::string transform_to_mars(int n)
{
	std::string RGB_mars;
	int base = 0, higher = 0;
	base = n % 13;
	higher = n / 13;

	RGB_mars = mars_base[higher] + mars_base[base];

	return RGB_mars;
}

int main()
{
	int R, G, B;
	std::cin >> R >> G >> B;

	std::string mR, mG, mB;
	mR = transform_to_mars(R);
	mG = transform_to_mars(G);
	mB = transform_to_mars(B);

	std::cout << "#" << mR << mG << mB << std::endl;

	return 0;
}

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