本题来自 Project Euler 第21题:https://projecteuler.net/problem=21
‘‘‘ 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 ‘‘‘ 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:
- 真因数(proper divisor):除去数字本身的所有因数(不要求是素数)。比如:12的所有真因数是:1、2、3、4、6
- 亲和数(amicable number):先求出数字n所有真因数之和a,然后再求出数字a所有真因数之和b。如果 a != b 且 n == b,则 a、b、n都是亲和数
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……
时间: 2024-10-12 14:36:13