PAT 1027. Colors in Mars (20)
2017-03-13 13:08
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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
Sample Output
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
/* 此题略水。。 */ #include "iostream" #include "algorithm" #include "stack" #include "cstring" using namespace std; int main() { int a, b, c; int ans[6]; cin >> a >> b >> c; if (a >= 13) { ans[1] = a % 13; a /= 13; ans[0] = a % 13; } else { ans[0] = 0; ans[1] = a%13; } if (b >= 13) { ans[3] = b % 13; b /= 13; ans[2] = b % 13; } else { ans[2] = 0; ans[3] = b % 13; } if (c >= 13) { ans[5] = c % 13; c /= 13; ans[4] = c % 13; } else { ans[4] = 0; ans[5] = c % 13; } cout << "#"; for (int i = 0; i < 6; i++) if (ans[i] >= 10) { cout << (char)(ans[i] - 10 + 'A'); } else cout << ans[i]; cout << endl; return 0; }
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