由网友(风云再起ゞ)分享简介：从项目欧拉问题500 120除数的数是16。事实上120具有16的除数的最小数字。The number of divisors of 120 is 16. In fact 120 is the smallest number having 16 divisors.查找最小的数与2 ** 500500除数。给你的答案...

从项目欧拉问题500

120除数的数是16。事实上120具有16的除数的最小数字。

The number of divisors of 120 is 16. In fact 120 is the smallest number having 16 divisors.

查找最小的数与2 ** 500500除数。给你的答案模500500507。

Find the smallest number with 2**500500 divisors. Give your answer modulo 500500507.

这是很简单的计算n的约数，例如。在Python LEN（[我为我的range（1，N + 1）如果n％我== 0]）。这是O（n）。

It's simple enough to count the divisors of n, eg. in Python len([i for i in range(1,n+1) if n % i == 0]). This is O(n).

我试图蛮力搜索，发现最小的数字与32除数是840，但它的多为上述问题的速度太慢。从不平等 count_divisors（N）其中​​。= N ，这个数字将是巨大的。

I tried brute force search and found the smallest number with 32 divisors is 840, but it's much too slow for the problem above. From the inequality count_divisors(n) <= n, the number is going to be massive.

你怎么看？有任何想法吗？如何计算与一定数量的约数最小的数字？

What do you think? Any ideas? How to calculate smallest number with certain number of divisors?

修改：我不认为这是一个重复。这个问题是比较具体的，它涉及某一类的多的较大的数字。 href="http://stackoverflow.com/questions/8861994">其他问题一般询问 = 0。现在的问题是 - 为了获得 X ，是它更好地提高权力2（即增加一个 1 ），或使用7（即取 4 = 1，而不是 4 = 0） ？那么，它的简单检查，2 * 3 * 5 * 7> 2 3 * 3 * 5 = 120，这就是为什么120的答案在这种情况下。

where ai >= 0. The question is now - in order to get minimum value of x, is it better to increase the powers of 2 (i.e., increment a1), or to use 7 (i.e. take a4=1 instead of a4=0)? Well, it's simple to check, 2*3*5*7 > 23 * 3 * 5 = 120, that's why the 120 is answer in this case.

如何推广这种做法？您应该创建分堆，你就会把素数的权力，照顾的分频器的数量达到指定值。如果16​​，这分堆将包含数字2，3，5，7，2 2 3 2 2 4 等。 为什么？由于16 = 2 4 ，所以每个（一个我 +1）有16分，也就是说，它必须是2，当您添加新的电源功率每次，就应该增加左侧（即变量 D（X））由2的幂，因为你的最终目标是找到最小的数与2 500500 约数。

How to generalize this approach? You should create min-heap where you'll put the powers of primes, taking care that the number of divisors reaches the specified value. In case of 16, this min-heap would contain numbers 2, 3, 5, 7, 22, 32, 24 etc. Why? Because 16 = 24, so each of (ai+1) has to divide 16, i.e. it has to be power of 2. Every time when you add new power, it should increase the left hand side (i.e., the variable d(x)) by power of 2, since your final goal is to find the smallest number with 2500500 divisors.

HTH。