# [Question] Count Multiples of Array

### Question

N是一个很大的正整数——可能到1015次方，

A 是一个array，里面存着一些正整数，up to 1000个

### Solution

It’s a very difficult question.

We can’t do it like a Sieve of Eratosthenes, cuz N is too large. The best solution is at this post, level 9:

Consider the simplest case: A={2}, then any odd number below N is OK, so the result would be (N - N/2). Then consider A={2, 3}, any number below N that is not mutiply of 2 or 3 is OK, so the result would be (N - N/2 - N/3 + N/6). Then consider A={2, 3, 5}, the result would be (N - N/2 - N/3 - N/5 + N/6 + N/10 + N/15 - N/30).

So there is a general rule:

for A={a1, a2, …, aN}, if ai is not dividable by aj for any i != j, then we could:

1. for i from 1 to N, calc r1 = N - SUM(N/ai);
2. for i, j from 1 to N, i != j, calc r2 = r1 + SUM(N/(ai*aj));
3. for i, j, k from 1 to N, i != j != k, calc r3 = r2 - SUM(N/(aiajak));
4. until all numbers in A are chosen.
5. then the final rN is the result.

So for the problem, first we preprocess A to eliminate any multiplies in A. For example, A={2, 4, 5}, we can eliminate 4 because it is a mutiply of 2 which is also in A. So A'={2, 5}, then we calc:

r1 = 10 - 10/2 - 10/5 = 10 - 5 - 2 = 3; r2 = r1 + 10/10 = 3 + 1 = 4;

then the final result is 4.

Refer to [Question] Multiples of 3 and 5.

not written