python当中没有scanf这种东西，所以我用find()找'['位置，再切片

from math import sqrt,ceil,gcd,log;re=lambda:map(int,input().strip().split())

a = {
    "B":0,
    "KB":1,
    "MB":2,
    "GB":3,
    "TB":4,
    "PB":5,
    "EB":6,
    "ZB":7,
    "YB":8
}

n, = re()
for i in range(1, n + 1):
    s = input()
    idx = s.find('[')
    s = s[idx + 1 : -1]
    print(f"Case #{i}: {100 - 100 * pow(1000 / 1024, a[s]):.2f}%")

由于python强处理字符串和其解包特性，也可以这样写：

from math import sqrt,ceil,gcd,log;re=lambda:map(int,input().strip().split())

a = {
    "B":0,
    "KB":1,
    "MB":2,
    "GB":3,
    "TB":4,
    "PB":5,
    "EB":6,
    "ZB":7,
    "YB":8
}

n, = re()
for i in range(1, n + 1):
    *_, s = input().strip(']').split('[')
    print(f"Case #{i}: {100 - 100 * pow(1000 / 1024, a[s]):.2f}%")