PAT 甲级 1027 Colors in Mars (20 分) 进制转换问题我的代码+别人的代码-蒲公英云

PAT 甲级 1027 Colors in Mars (20 分) 进制转换问题我的代码+别人的代码

1027 Colors in Mars (20 分)

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

我的代码：

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
string convert(int n)
{
    string s="";
    while(n!=0)
    {
        if(n%13>9)
        s+=(n%13-10+65);
        else
        s+=(n%13+'0');
        n/=13;
    }
    return s;
}
void print(string t)
{
    if(t.length()==0)
    printf("00");
    else
    if(t.length()==1)
    printf("0%c",t[0]);
    else
    printf("%c%c",t[1],t[0]);
}
int main()
{
    int a,b,c,i,j,k;
    string t;
    cin>>a>>b>>c;
    printf("#");
    t=convert(a);
    print(t);
    t=convert(b);
    print(t);
    t=convert(c);
    print(t);
    printf("\n");
    return 0;
}

别人的代码：

差距还是有的！

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
char radix[13] ={
  '0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'A', 'B', 'C'
};
int main() {
    int r, g, b;
    scanf("%d%d%d", &r, &g, &b);
    printf("#");
    printf("%c%c", radix[r / 13], radix[r % 13]);
    printf("%c%c", radix[g / 13], radix[g % 13]);
    printf("%c%c", radix[b / 13], radix[b % 13]);
    return 0;
}