题意：
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%
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 ",A,B); double ans = (double)100 - ((double)A*100/(double)B); printf("Case #%d: %.2lf%% ",t,ans); } }