HDU 4788 Hard Disk Drive

最新推荐文章于 2025-09-10 22:21:01 发布

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最新推荐文章于 2025-09-10 22:21:01 发布

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文章标签： java

原文链接：http://www.cnblogs.com/GODLIKEING/p/3432542.html

本文介绍了一个简单的程序，用于计算硬盘制造商所标示的容量与操作系统实际显示容量之间的百分比差异。通过解析输入字符串中的数值与单位，程序可以处理从字节到尧字节等不同单位，并输出每种情况下的“缺失部分”百分比。

Hard Disk Drive

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 125 Accepted Submission(s): 75

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

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思路：水题

代码：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <queue>
using namespace std;
int t;
int the_number;
double sum;
int the_flag;
double the_result;
char map[10];
char str;
int main()
{
    scanf("%d",&t);
    for(int k = 1;k <= t;k ++)
    {
        memset(map,0,sizeof(map));
        the_flag = 0;
        the_number = 0;
        sum = 1;
        the_result = 0;
        scanf("%s",map);
        int m = 1;
        for(int i = 0;i < 10;i ++)
        {
            if(map[i] == '[')
            {
                str = map[i + 1];
                break ;
            }
            the_number = the_number * m + map[i] - '0';
            m = 10;
        }
        printf("Case #%d: ",k);
        if(str == 'B')
           the_flag = 0;
        if(str == 'K')
           the_flag = 1;
        if(str == 'M')
           the_flag = 2;
        if(str == 'G')
           the_flag = 3;
        if(str == 'T')
           the_flag = 4;
        if(str == 'P')
           the_flag = 5;
        if(str == 'E')
           the_flag = 6;
        if(str == 'Z')
           the_flag = 7;
        if(str == 'Y')
           the_flag = 8;
        if(the_flag == 0)
        {
            printf("0.00");
            printf("%%\n");
        }
        else
        {
            while(the_flag --)
            {
                sum = sum * (1000.0 / 1024.0);
            }
            the_result = (1 - sum) * 100;;
            printf("%.2lf",the_result);
            printf("%%\n");
        }
    }
    return 0;
}