1027. Colors in Mars (20)

主题如以下：

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

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

这个题目的本质是考察除k取余法，方法为用十进制数去除以基数k，每次把商数作为下一次的值，余数写在旁边，一直做到商数小于基数，结束运算。然后从下到上。从最后一个商数到第一个余数的路径上全部的数构成了结果，比如将十进制数38转化为13进制数（0-9,A-C）

从下到上，各自是2、12，这个数是2C。故38的13进制形式为2C，依照这种方法设计程序就可以。

由于13进制涉及到了字母，因此使用string来存储这个数字。每次在string头部插入字符。

#include <iostream>
#include <vector>
#include <string>
#include <string.h>

using namespace std;

char int2char(int n){
    if(n <= 9){
        return '0' + n;
    }else{
        return 'A' + (n - 10);
    }
}

string convertMars(int value){
    string temp;
    int shang,yu;
    int radix = 13;
    while(1){
        shang = value / radix;
        yu = value % radix;
        temp.insert(temp.begin(),int2char(yu));
        value /= radix;
        if(value < radix){
            temp.insert(temp.begin(),int2char(value));
            break;
        }
    }
    return temp;
}

int main()
{
    string mR,mG,mB;
    int R,G,B;
    cin >> R >> G >> B;
    mR = convertMars(R);
    mG = convertMars(G);
    mB = convertMars(B);
    cout << "#" << mR << mG << mB;
    return 0;
}

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