Count of arrays in which all adjacent elements are such that one of them divide the another

Given two positive integer n and n. The task is to find the number of arrays of size n that can be formed such that :

1. Each element is in range [1, m]
2. All adjacent element are such that one of them divide the another i.e element Ai divides Ai + 1 or Ai + 1 divides Ai + 2.

Examples:

```Input : n = 3, m = 3.
Output : 17
{1,1,1}, {1,1,2}, {1,1,3}, {1,2,1},
{1,2,2}, {1,3,1}, {1,3,3}, {2,1,1},
{2,1,2}, {2,1,3}, {2,2,1}, {2,2,2},
{3,1,1}, {3,1,2}, {3,1,3}, {3,3,1},
{3,3,3} are possible arrays.

Input : n = 1, m = 10.
Output : 10
```

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

We try to find number of possible values at each index of the array. First, at index 0 all values are possible from 1 to m. Now observe for each index, we can reach either to its multiple or its factor. So, precompute that and store it for all the elements. Hence for each position i, ending with integer x, we can go to next position i + 1, with the array ending in integer with multiple of x or factor of x. Also, multiple of x or factor of x must be less than m.
So, we define an 2D array dp[i][j], which is number of possible array (divisible adjacent element) of size i with j as its first index element.

```1 <= i <= m, dp[1][m] = 1.
for 1 <= j <= m and 2 <= i <= n
dp[i][j] = dp[i-1][j] + number of factor
of previous element less than m
+ number of multiples of previous
element less than m.
```

Below is the C++ implementation of this approach:

```// C++ program to count number of arrays
// of size n such that every element is
// in range [1, m] and adjacen are
// divisible
#include <bits/stdc++.h>
#define  MAX 1000
using namespace std;

int numofArray(int n, int m)
{
int dp[MAX][MAX];

// For storing factors.
vector<int> di[MAX];

// For storing multiples.
vector<int> mu[MAX];

memset(dp, 0, sizeof dp);
memset(di, 0, sizeof di);
memset(mu, 0, sizeof mu);

// calculating the factors and multiples
// of elements [1...m].
for (int i = 1; i <= m; i++)
{
for (int j = 2*i; j <= m; j += i)
{
di[j].push_back(i);
mu[i].push_back(j);
}
di[i].push_back(i);
}

// Initalising for size 1 array for
// each i <= m.
for (int i = 1; i <= m; i++)
dp[1][i] = 1;

// Calculating the number of array possible
// of size i and starting with j.
for (int i = 2; i <= n; i++)
{
for (int j = 1; j <= m; j++)
{
dp[i][j] = 0;

// For all previous possible values.
for (auto x:di[j])
dp[i][j] += dp[i-1][x];

for (auto x:mu[j])
dp[i][j] += dp[i-1][x];
}
}

// Calculating the total count of array
// which start from [1...m].
int ans = 0;
for (int i = 1; i <= m; i++)
{
ans += dp[n][i];
di[i].clear();
mu[i].clear();
}

return ans;
}

// Driven Program
int main()
{
int n = 3, m = 3;
cout << numofArray(n, m) << "\n";
return 0;
}
```

Output:

```17
```

Time Complexity: O(N*M).

This article is contributed by Anuj Chauhan. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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