PTA 1027 Colors in Mars （20 分）燚

版权声明：哈哈 你能采纳 我很开心 https://blog.csdn.net/m0_38090681/article/details/84144182

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：颜色由 R G B 三个字符表示，每个字符对应一个十三进制数。现给出 RGB 三个十进制数，将他们转换位十三进制数

输出以#字符开始

思路：1.由于最后输出的有“#”字符，并且13进制涉数大于9的都用字母表示，因此选择用char 或string类型来存储。

           2.因为int类型转char类型比较麻烦，可将单个的char类型当作string类，所以选择string类型

           3.十进制转十三进制为连续除13取余并反向输出，为了方便可以用栈临时存储

注意事项： If a single color is only 1-digit long, you must print a 0 to its left.  如果转换的13进制数只有一位，高位用0补齐

#include<iostream>
#include<stack>
#include<vector>
#include<string>
using namespace std;
//将十进制数转换为13进制数
void changeNum(vector<string>&result, int n) {
	stack<int> s;//存储转换的13进制数
    //如果十进制数为0则将0压入栈中
	if (n == 0) {
		s.push(0);
	}
    //连续取余进制转换
	while (n != 0) {
		int temp = n % 13;
		s.push(temp);
		n /= 13;
	}
    //如果13进制数只有一位，则用零补齐
	if (s.size() == 1) {
		s.push(0);
	}
    //将13进制数转换为字符串
	while (!s.empty()) {
		if (s.top() > 9) {
			switch (s.top()) {
			case 10:result.push_back("A");break;
			case 11:result.push_back("B");break;
			case 12:result.push_back("C");break;
			}
			s.pop();
		}
		else {
			result.push_back(to_string(s.top()));
			s.pop();
		}
	}
}
int main() {
	vector<string>result;
    //3个十进制数，边输入边转换
	for (int i = 0;i < 3;i++) {
		int n;
		cin >> n;
		changeNum(result, n);
	}
    //输出
	cout << "#";
	for (int i = 0;i < result.size();i++) {
		cout << result[i];
	}
	cout << endl;
	return 0;
}