hdu 4788 Hard Disk Drive (水题)

题意：

[align=left]Input[/align]
　　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.

[align=left]Output[/align]
　　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.

[align=left]Sample Input[/align]

2 100[MB] 1[B]

[align=left]Sample Output[/align]

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

代码：

int T;
char s[100];

ll Pow(int a,int x){
ll res = 1;
rep(i,1,x) res *= (ll)a;
return res;
}
int main(){
cin >> T;
rep(t,1,T){
scanf("%s",s);
int l = strlen(s);
char x = s[l-3];
int a;
switch(x){
case '[': a=0; break;
case 'K': a=1; break;
case 'M': a=2; break;
case 'G': a=3; break;
case 'T': a=4; break;
case 'P': a=5; break;
case 'E': a=6; break;
case 'Z': a=7; break;
case 'Y': a=8; break;
}
ll A = Pow(125,a), B = Pow(128,a);
//printf("%I64d %I64d\n",A,B);
double ans = (double)100 - ((double)A*100/(double)B);
printf("Case #%d: %.2lf%%\n",t,ans);
}
}