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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 Input15 43 71Sample 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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原文地址：http://www.cnblogs.com/mengfanrong/p/4802106.html
