【kata Daily 190905】What's a Perfect Power anyway?（完美幂）

原题：

A perfect power is a classification of positive integers:

In mathematics, a perfect power is a positive integer that can be expressed as an integer power of another positive integer. More formally, n is a perfect power if there exist natural numbers m > 1, and k > 1 such that m^k = n.

Your task is to check wheter a given integer is a perfect power. If it is a perfect power, return a pair m and k with m^k = n as a proof. Otherwise return Nothing, Nil, null, NULL, None or your language's equivalent.

Note: For a perfect power, there might be several pairs. For example 81 = 3^4 = 9^2, so (3,4) and (9,2) are valid solutions. However, the tests take care of this, so if a number is a perfect power, return any pair that proves it.

Examples

　　isPP(4) => [2,2]

　　isPP(9) => [3,2]

　　isPP(5) => None

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题目大意：给定一个数n，判断这个数是否是完美幂，即：有一个数m的k次数等于n。

解题思路：

　　我自己的解题思路很粗暴，但是并不能过审，这里也说一下我的思路

def isPP(n):
    # your code here
    for i in range(n):
        for j in range(n):
            if i**j == n:
                return [i, j]
    return None

没错。。。很笨而且很好资源的办法，，，ヽ(ー_ー)ノ

　　看一下其他网友的办法：

def isPP(n):
    #your code here
    from math import sqrt
    m=int(sqrt(n))
    for i in range(2,m+1):
        k=0
        while i**k < n:
            k+=1
        if i**k==n:
            return [i,k]
    return None

解读：先对n进行开根号，得到最大的m值，然后根据逐步逼近的办法来确定k的值。很好理解，是个好办法。

　　看一下最多人推荐的：

from math import ceil, log, sqrt

def isPP(n):
    for b in xrange(2, int(sqrt(n)) + 1):
        e = int(round(log(n, b)))
        if b ** e == n:
            return [b, e]
    return None

困惑：e 的根据是什么不太懂，，，

知识点：

1、数的幂次方运算，**两个星号表示幂运算：2**3=8

2、数的对数运算，math.log(x[, base])：base默认为e

3、数的开根号，math.sqrt(n)

4、逐步逼近的算法，根据判断的结果取最后的值用于运算