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

Sample Input:

15 43 71

Sample Output:

#123456

題目大意：顏色由 R G B 三個字元表示，每個字元對應一個十三進位制數。現給出 RGB 三個十進位制數，將他們轉換位十三進位制數

輸出以#字元開始

思路：1.由於最後輸出的有“#”字元，並且13進位制涉數大於9的都用字母表示，因此選擇用char 或string型別來儲存。

           2.因為int型別轉char型別比較麻煩，可將單個的char型別當作string類，所以選擇string型別

           3.十進位制轉十三進製為連續除13取餘並反向輸出，為了方便可以用棧臨時儲存

注意事項： If a single color is only 1-digit long, you must print a 0 to its left.  如果轉換的13進位制數只有一位，高位用0補齊

#include<iostream>
#include<stack>
#include<vector>
#include<string>
using namespace std;
//將十進位制數轉換為13進位制數
void changeNum(vector<string>&result, int n) {
	stack<int> s;//儲存轉換的13進位制數
    //如果十進位制數為0則將0壓入棧中
	if (n == 0) {
		s.push(0);
	}
    //連續取餘進位制轉換
	while (n != 0) {
		int temp = n % 13;
		s.push(temp);
		n /= 13;
	}
    //如果13進位制數只有一位，則用零補齊
	if (s.size() == 1) {
		s.push(0);
	}
    //將13進位制數轉換為字串
	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個十進位制數，邊輸入邊轉換
	for (int i = 0;i < 3;i++) {
		int n;
		cin >> n;
		changeNum(result, n);
	}
    //輸出
	cout << "#";
	for (int i = 0;i < result.size();i++) {
		cout << result[i];
	}
	cout << endl;
	return 0;
}