PAT A1027 Colors in Mars(20)[火星颜色，进制]

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

题意

R，G，B 三个十进制数转化为十三进制输出

思路1

1. 单纯的进制转换，需要注意补齐两位 0，最后反转结果

代码1

#include <bits/stdc++.h>
using namespace std;

const int radix = 13;

void change(int x)
{
    vector<int> ans;
    while(x)
    {
        int t = x % radix;
        char c;
        if(t > 9)
            c = t - 10 + 'A';
        else 
            c = t + '0';
        ans.push_back(c);
        x /= radix;
    }
    while(ans.size() != 2) ans.push_back('0');
    reverse(ans.begin(), ans.end());
    printf("%c%c", ans[0], ans[1]);
}


int main()
{
    int R, G, B;
    scanf("%d%d%d", &R, &G, &B);
    printf("#");
    change(R);
    change(G);
    change(B);
    return 0;
}

思路2

1. 0-168 转化的结果只有两位，即低位为 a[i] % 13 和 高位 a[i] / 13，将这两位转化为十三进制字符即可。

代码2

#include <iostream>
using namespace std;
const int N = 3;

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

int main()
{
    int a[N];
    for(int i = 0; i < N; i++)
        scanf("%d", &a[i]);
    printf("#");
    for(int i = 0; i < N; i++)
        printf("%c%c", get(a[i] / 13), get(a[i] % 13));
    puts("");
    return 0;
}