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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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