问题描述:
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.
解决方法:
找到一个数(a)的因子的和(b),然后这再找到这个和的因子的和(c)。
找到符合a==c的数。
public static int count_divisors_sum(long number) { //找到一个数的因子的 和 int count = 1; for(int i=2; i*i<=number; i++){ if(number%i==0){ count+=i; if(i*i<number){ count += number/i; } } } return count; }