Count the number of subarrays whose product is not divisible by k

Question

Given an array, I want to count the number of subarrays (adjacent) that, when executed, will not be divided by k.

Eg. Let A = [1, 2, 3, 4, 5, 6]and K = 2

Then the number of subarrays such that the product is not divisible by K is 2:

{1}
{3}
{5}

The rest of them are divided by 2.

{1, 2}
{1, 2, 3}
{1, 2, 3, 4}
{1, 2, 3, 4, 5}
{1, 2, 3, 4, 5, 6}
{2}
{2, 3}
{2, 3, 4}
{2, 3, 4, 5} etc..

I tried to calculate the total number of subarrays (n) (n + 1) / 2 first and then subtract the number of subarrays that are divisible by k using mod, but that does not work. How do I get around this?

This calculates (erroneously) the number of subarrays whose product is divided by K:

int curr = 1;
    int x = 0;
    for (int i = 0; i < a.length; i++) {
        curr = curr * a[i] % k;
        if (curr == 0) {
            curr = 1;
            if (x > 0)
                ans = ans.subtract(NumberUtils.nChooseR(x + 1, 2));
            if (a[i] % k != 0) {
                x = 1;
            } else {
                x = 0;
            }
        } else {
            x++;
        }
    }
    if (x != 0) {
        ans = ans.subtract(NumberUtils.nChooseR(x + 1, 2));
    }
    return ans;

Bit related question this , but it includes the addition, so it can not be applied here.

: 10 ^ 5, - 10 ^ 9. ,

Answer 1

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