H - Hard Disk Drive解题报告
一、原题
Hard Disk Drive
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1484 Accepted Submission(s): 813
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%
二、题目大意：
计算以1000而不是1024作为硬盘容量所造成的误差（以1024为标准）。
三、思路：
这个题目可以发现其实它跟前面的数字没有关系，只跟后面的“B”, “KB”, “MB”, “GB”, “TB”, “PB”, “EB”, “ZB”, “YB”有关。
可推出公式：ans=1-(125/128)^t
t=0 arr[i+1]=’B’
t=1 arr[i+1]=’K’
t=2 arr[i+1]=’M’
……
四、实现：预处理一下init
#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
using namespace std;
char arr[20];
double p[20];
void init()
{
p[0]=1;
for(int i=1;i<=10;i++)
{
p[i]=p[i-1]*(125.0/128.0);
}
p[0]=0;
for(int i=1;i<=10;i++)
{
p[i]=100*(1-p[i]);
}
}
int main()
{
// freopen("in.txt","r",stdin);
int T,ca=1;
scanf("%d",&T);
init();
cin.get();
while(T--)
{
gets(arr);
int lenth=strlen(arr);
int i=0;
double ans=0;
for(i=0;i<lenth;i++)
{
if(arr[i]=='[')
break;
}
switch(arr[i+1])
{
case 'B':ans=p[0];break;
case 'K':ans=p[1];break;
case 'M':ans=p[2];break;
case 'G':ans=p[3];break;
case 'T':ans=p[4];break;
case 'P':ans=p[5];break;
case 'E':ans=p[6];break;
case 'Z':ans=p[7];break;
case 'Y':ans=p[8];break;
}
printf("Case #%d: %.2lf",ca++,ans);
cout<<"%\n";
}
return 0;
}
最后附上链接：
http://acm.hdu.edu.cn/showproblem.php?pid=4788
Ps：这题还有点纪念意义的啊，毕竟是我人生中第一次1Aregion题，虽然它只是一个签到题。。。。But人家没有看题解自己推出公式的啊。。。