HDUOJ----(4788)Hard Disk Drive

最新推荐文章于 2025-08-06 16:23:50 发布

weixin_30549175

最新推荐文章于 2025-08-06 16:23:50 发布

阅读量53

点赞数

CC 4.0 BY-SA版权

文章标签： java

原文链接：http://www.cnblogs.com/gongxijun/p/3440590.html

本文介绍了一个计算硬盘容量缺失比例的简单算法，通过输入硬盘大小和单位，输出缺失部分的比例。算法覆盖了从字节到Yottabyte的所有单位，并支持不同制造商和操作系统的容量测量差异。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 252 Accepted Submission(s): 144

Problem Description

　　Yesterday your dear cousin Coach Pang gave you a new 100MB hard disk drive (HDD) as a gift because you will get married next year. 　　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

　　The first line contains an integer T, which indicates the number of test cases. 　　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

　　For each test case, output one line “Case #x: y”, where x is the case number (starting from 1) and y is the percentage of the “missing part”. The answer should be rounded to two digits after the decimal point.

Sample Input

   2 100[MB] 1[B]
  

Sample Output

   Case #1: 4.63%
  

   Case #2: 0.00% 
   
     Hint

Source

2013 Asia Chengdu Regional Contest

简单的题：

 1 #include<stdio.h>
 2 #include<string.h>
 3 #include<stdlib.h>
 4 int main()
 5 {
 6     char str[20];
 7     int t,i,cnt=1;
 8     double a;
 9     scanf("%d",&t);
10     while(t--)
11     {
12         scanf("%s",str);
13         for(i=0;str[i]!='[';i++);
14         switch(str[++i])
15         {
16         case'B': a=0.00;  break;
17         case'K': a=2.34;  break;
18         case'M': a=4.63;  break;
19         case'G': a=6.87;  break;
20         case'T': a=9.05;  break;
21         case'P': a=11.18; break;
22         case'E': a=13.26; break;
23         case'Z': a=15.30; break;
24         case'Y': a=17.28; break;
25         }
26         printf("Case #%d: %.2lf%%\n",cnt++,a);
27     }
28     return 0;
29 }