Question

我创建了一个将字母转换为数字的函数（a = 1，b = 2等等）然后我对它们做了这个等式：

b^(a / (b/2))
b = letter being translated
a = previous letter (or if b is the first letter, a = len(string)

例如，以下是翻译字母'e'的方法：

'Hello' = [34, 5, 12, 12, 15]
5^(34 / (5/2)) = 3206210698.73506

以下是我用来执行此操作的代码：

alphabet = {
'H' : 34,
'e' : 5,
'l' : 12,
'o' : 15,
}   
entry = "Hello"
password = []
hashed = []
firstNumber = True
lengthPower = len(entry)
counter = 0

for letter in entry:

    key = alphabet[letter]
    password.append(key)

for number in password:
    if firstNumber:
        powered = number ** lengthPower
        lastNumber = number
        firstNumber = False
    else:
        powered = number ** (password[counter]/(number/2))
        lastNumber = number
        counter += 1
    hashed.append(powered)

但我怎样才能扭转这种行为呢？ '3206210698.73506'回到5（'e'）？

b^(34 / (b/2)) = 3206210698.73506

Answer 1

y = x ^ n，x = y ^ (1/n)。请参阅this或谷歌'逆功率'。所以在你的情况下，这将指导你。

n = 34/(5/2)
y = pow(5, n)  # 3206210698.7350655
x = int(pow(y, 1/n))  # 5

求解python中的指数方程

1 个答案: