Hdoj Hard Disk Drive 4788 "2013 area game H-Water"

Hard Disk DriveTime limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 1887 Accepted Submission (s): 1042


Problem Description yesterday your dear cousin Coach Pang gave you a new 100MB hard disk drive (HDD) as a gift because yo U'll get married next year.
But your turned on your computer and the operating system (OS) told you the HDD are about 95MB. The 5MB of space is missing. It's known that the HDD manufacturers has a different capacity measurement. The manufacturers think 1 "kilo" is and the OS thinks that's 1024. There is several descriptions of the size of an HDD. They is 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 your 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, +] and unit is the description of size which could are "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 Percen Tage of the "missing part". The answer should is rounded to both digits after the decimal point.
Sample Input

2100[MB]1[B]

Sample Output

Case #1:4.63%case #2:0.00%Hint

Source2013 Asia Chengdu Regional Contest
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Water ~

#include <iostream> #include <stdio.h> #include <string.h> #include <algorithm>using namespace Std;int Main () {int t;scanf ("%d", &t), int xp=1;while (t--) {double X=0;char s[10];scanf ("%s", s); int L=strlen (s); char Str[5];int i;for (i=0;; i++) {if (s[i]== ' [') break;x=x*10+ (s[i]-' 0 ');} int k=0;for (int j=i+1;j<l-1;j++) {str[k++]=s[j];} str[k]= ';d ouble sum=x;int xn,dn;if (str[0]== ' B ') {sum=x;} else if (str[0]== ' K ') {sum=sum*1000.0/1024.0;} else if (str[0]== ' M ') {dn=xn=2;while (xn--) {sum*=1000.0;} while (dn--) {sum/=1024.0;}} else if (str[0]== ' G ') {dn=xn=3;while (xn--) {sum*=1000.0;} while (dn--) {sum/=1024.0;}} else if (str[0]== ' T ') {dn=xn=4;while (xn--) {sum*=1000.0;} while (dn--) {sum/=1024.0;}} else if (str[0]== ' P ') {dn=xn=5;while (xn--) {sum*=1000.0;} while (dn--) {sum/=1024.0;}} else if (str[0]== ' E ') {dn=xn=6;while (xn--) {sum*=1000.0;} while (dn--) {sum/=1024.0;}} else if (str[0]== ' Z ') {dn=xn=7;while (xn--) {sum*=1000.0;} while (dn--) {sum/=1024.0;}} else if (str[0]== ' Y ') {dn=xn=8;while (xn--) {sum*=1000.0;} while (dn--) {sum/=1024.0;}}    Double res= (x-sum) *100.0/x;printf ("Case #%d:%.2lf%%\n", xp++,res);} return 0;}

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Hdoj Hard Disk Drive 4788 "2013 area game H-Water"