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入坑codewars第16天-Transformation of a Number Through Prime Factorization

題目:

Every natural number, n, may have a prime factorization like:

We define the geometric derivative of n, as a number with the following value:

For example: calculate the value of n* for n = 24500.

24500 = 2²5³7²
n* = (2*2) * (3*5²) * (2*7) = 4200

Make a function, f

 that can perform this calculation

f(n) ----> n*

Every prime number will have n* = 1.

Every number that does not have an exponent ki, higher than 1, will give a n* = 1, too

f(24500) == 4200

f(997) == 1

Do your best!

題意:

題意就是按照那些公式算;首先把一個數分解成質因數相乘;然後按照第二個公式計算即可。

思路:

思路就是先分解成質因數相乘,然後儲存在列表裡;
然後利用列表的計數函式,計算出每個質因子的個數就是kn;
按照公式求就ok了

第一次提交的程式碼如下超時了:

import math
from collections import Counter
def f(n):
    m=n
    nn=1
    list1=[]
    for i in range(2,m):
        while m%i==0:
            m=m/i
            list1.append(i)
    result = Counter(list1)
    for key in result:
        nn=nn*(result[key]*(key**(result[key]-1)))
    if m==n:return 1
    return nn

我之前寫過類似的知道超時原因是什麼? 就是這個m太大會導致超時,解決辦法就是處理一下m變成根號m:

import math
from collections import Counter
def f(n):
    m=n
    nn=1
    list1=[]
    for i in range(2,int(math.sqrt(m))+1):
        while m%i==0:
            m=m/i
            list1.append(i)
    result = Counter(list1)
    for key in result:
        nn=nn*(result[key]*(key**(result[key]-1)))
    if m==n:return 1
    return nn

然後就可以了。