标签：ret lan end ica iso less .net https 概念
本题来自 Project Euler 第21题：https://projecteuler.net/problem=21
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Project Euler: Problem 21: Amicable numbers
Let d(n) be defined as the sum of proper divisors of n
(numbers less than n which divide evenly into n).
If d(a) = b and d(b) = a, where a ≠ b, then a and b are an amicable pair
and each of a and b are called amicable numbers.
For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110;
therefore d(220) = 284.
The proper divisors of 284 are 1, 2, 4, 71 and 142; so d(284) = 220.
Evaluate the sum of all the amicable numbers under 10000.
Answer: 31626
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def d(n): #计算数字n所有真因数之和
res = 0
for i in range(1, n//2+1):
if n/i == float(n//i):
res += i
return(res)
lst = [] #亲和数列表
for i in range(1, 10000):
a = d(i)
b = d(a)
if b == i and b != a:
lst.append(i)
res = 0 #所有亲和数之和
for i in range(len(lst)):
res += lst[i]
print(res)
首先需要明确两个数学概念：
概念弄清楚了，接下来就好办了。既是求10000以内所有亲和数之和，当然首先得找出所有亲和数（汇总为列表 lst）。而想找出亲和数，首先得定义出函数 d(n)，用来计算数字n的所有真因数之和。这之后，只要用之前所述的两个条件将10000以内所有数字遍历，就能找出所有亲和数了。
思路虽然清晰，但实际计算起来，还是费了些时间。想必是有更优的算法可以计算真因数之和吧，但……算了，我这数学渣，能算出来就不错了，不指望有啥好算法了……
Python练习题 048：Project Euler 021：10000以内所有亲和数之和
标签：ret lan end ica iso less .net https 概念
原文地址：http://www.cnblogs.com/iderek/p/6081732.html
