欧拉计划21

昨天困得不得了，忘记做了，还是发现用python写这种东西太方便了，再来试一下。

原题如下:

Amicable numbers

Problem 21

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.

代码如下:

# -*- coding: UTF-8 -*-
def sum1(n):
sum = 0
for i in range(1,n/2 + 1):
if n%i == 0 :
#print i
sum+=i
#print sum
return sum

if __name__=="__main__":
sum = 0;
for i in range(1,10000 + 1):
test1 = sum1(i)
if( i == sum1(test1) and i != test1):
sum += i
print sum

结果为:  31626 !!