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
火星上的人用和地球人相似的方式在他们的电脑中表示颜色。也就是说,颜色由6位数字表示,其中前2位数字表示红色,中间2位数字表示绿色,最后2位数字表示蓝色。唯一的区别是它们使用基数13(0-9和A-C)而不是16。现在给一个颜色三个十进制数字(每个数字在0到168之间),你应该输出他们的Mars RGB值。
输入规格:
每个输入文件包含一个测试用例,该测试用例占用一行,该行包含三个十进制颜色值。
输出规格:
对于每个测试用例,您应该以以下格式输出Mars RGB值:首先输出#,然后是一个6位数字,其中所有英文字符都必须大写。如果单色只有1位数字长,则必须在其左侧打印0。
样本输入:
15 43 71
样本输出:
#123456
进制转换
将一个10进制的数转换为13进制的数。10进制的数转换为其他进制,可以采用短除法的思想。
但是这道题给出的数,是有一个范围0-168(169=13*13),那么转换后的数一定是位数最多为两位数。所以可以映射即可。
♦如果输出的数为0,直接输出00
♦13进制之下,10-12表示为ABC
♦如果输出的数<13(之前傻傻得用10进制,以为是小于10),转换之后的数,在数字前面自动补齐为0.
#include
#include
#include
using namespace std;
vector<char>v;
int weight[13] = {
'0','1','2','3','4','5','6','7','8','9','A','B','C' };
void convert(int n)
{
v.clear();
while (n)
{
v.push_back(weight[n % 13]);
n = n / 13;
}
reverse(v.begin(), v.end());
}
void Print()
{
for (int i = 0; i < v.size(); i++) cout << v[i];
}
int main()
{
int n1, n2, n3;
cin >> n1 >> n2 >> n3;
if (n1 >= 0 && n1 <= 168 && n2 >= 0 && n2 <= 168 && n3 >= 0 && n3 <= 168)
{
cout << "#";
if (n1 == 0) cout << "00";
else
{
if (n1 <= 12)
{
cout << "0";
}
convert(n1);
Print();
}
if (n2 == 0) cout << "00";
else
{
if (n2 <=12)
{
cout << "0";
}
convert(n2);
Print();
}
if (n3 == 0) cout << "00";
else
{
if (n3 <=12)
{
cout << "0";
}
convert(n3);
Print();
}
}
return 0;
}
#include
using namespace std;
int main()
{
char weight[13] = {
'0','1','2','3','4','5','6','7','8','9','A','B','C' };
int n1, n2, n3;
cin >> n1 >> n2 >> n3;
cout << "#";
cout << weight[n1 / 13] << weight[n1 % 13] << weight[n2 / 13] << weight[n2 % 13] << weight[n3 / 13] << weight[n3 % 13];
return 0;
}