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**The Problem About the Sequence of Numbers**

In the given sequence of integers, we need to find the longest subsequence of integers of which each element of the subsequence can be divided by the previous with no remainders.

**Solution**

To solve the problem the elements of the array should be sorted by the absolute value (in non-decreasing order). If the array has elements equal to 0, then one of them will be the last element of a sequence, so we can ignore them. Let K(i) denote the maximum number of items that can be located in a certain sequence with the desired property, the last element of which is the i-th element.

Obviously, it is the element with the smallest absolute value, and nothing can’t be preceded by any element, so K(R)=0, where R is the index of the first nonzero element.

We need to determine the maximum number of elements that may precede the given element for each next element with the number j, j=R+1,…,N, taking into account the required properties. It is clear that this number is the maximum value of K(p1), (p2),…, K(pm), where the elements with the numbers p1,p2,…,pm, R<=p1<p2<…<pm<j are divisors of the element with number j. So we have the recurrence formula to compute K(j):

K(j) = max (K(p1) K(p2),…, K(pm)).

The value of K(N) calculated according to the above rule determines the maximum number of items that can be located in a certain sequence with the desired property (excluding a possible zero at the end of the sequence).

In order to find which elements form a maximal sequence, it is sufficient to remember the number of p1,p2,…,pm, which ensures a maximum of the numbers K(p1) K(p2),…, (pm) for each number j. These numbers can be defined during the calculation of the value of K(j) in a certain array, for example, ANCESTOR. Using this information, it is easy to determine the numbers of elements in the sequence, passing from element i with the maximum value of K(i) to the item that precedes (ANCESTOR(i)), until we come to the first element of the sequence f (PARENT(f)=0).

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