HDU - 4788 Hard Disk Drive (成都邀请赛H 水题)
HDU - 4788
Time Limit:1000MS | Memory Limit:32768KB | 64bit IO Format:%I64d & %I64u |
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Description
But you turned on your computer and the operating system (OS) told you the HDD is about 95MB. The 5MB of space is missing. It is known that the HDD manufacturers have a different capacity measurement. The manufacturers think 1 “kilo” is 1000 but the OS thinks
that is 1024. There are several descriptions of the size of an HDD. They are byte, kilobyte, megabyte, gigabyte, terabyte, petabyte, exabyte, zetabyte and yottabyte. Each one equals a “kilo” of the previous one. For example 1 gigabyte is 1 “kilo” megabytes.
Now you know the size of a hard disk represented by manufacturers and you want to calculate the percentage of the “missing part”.
Input
For each test case, there is one line contains a string in format “number[unit]” where number is a positive integer within [1, 1000] and unit is the description of size which could be “B”, “KB”, “MB”, “GB”, “TB”, “PB”, “EB”, “ZB”, “YB” in short respectively.
Output
Sample Input
100[MB]
1[B]
Sample Output
Case #2: 0.00%
Hint
思路:基本的区别是在第一位
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
int n;
char u[8];
int main()
{
while (scanf("%d", &n) != EOF) {
for (int cas = 1; cas <= n; cas++) {
scanf("%*d%s", u);
printf("Case #%d: ", cas);
if (u[1] == 'B')
printf("0.00");
else if (u[1] == 'K')
printf("2.34");
else if (u[1] == 'M')
printf("4.63");
else if (u[1] == 'G')
printf("6.87");
else if (u[1] == 'T')
printf("9.05");
else if (u[1] == 'P')
printf("11.18");
else if (u[1] == 'E')
printf("13.26");
else if (u[1] == 'Z')
printf("15.30");
else if (u[1] == 'Y')
printf("17.28");
printf("%%\n");
}
}
return 0;
}
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