Given an array of **n** positive integers. Initially we are at first position. We can jump position y to position x if y divides x and y < x. The task is to print maximum sum path ending with every position x where 1 <= x <= n.

Here position is index plus one assuming indexes begin with 0.

Examples:

Input :arr[] = {2, 3, 1, 4, 6, 5}Output :2 5 3 9 8 10 Maximum sum path ending with position 1 is 2. For position 1, last position to visit is 1 only. So maximum sum for position 1 = 2. Maximum sum path ending with position 2 is 5. For position 2, path can be jump from position 1 to 2 as 1 divides 2. So maximum sum for position 2 = 2 + 3 = 5. For position 3, path can be jump from position 1 to 3 as 1 divides 3. So maximum sum for position 3 = 2 + 3 = 5. For position 4, path can be jump from position 1 to 2 and 2 to 4. So maximum sum for position 4 = 2 + 3 + 4 = 9. For position 5, path can be jump from position 1 to 5. So maximum sum for position 5 = 2 + 6 = 8. For position 6, path can be jump from position 1 to 2 and 2 to 6 or 1 to 3 and 3 to 6. But path 1 -> 2 -> 6 gives maximum sum for position 6 = 2 + 3 + 5 = 10.

The idea is to use Dynamic Programming to solve this problem.

Create an 1-D array dp[] where each element dp[i] stores maximum sum path ending at index i (or position x where x = i+1) with divisible jumps. The recurrence relation for dp[i] can be defined as:dp[i] = max(dp[i], dp[divisor of i+1] + arr[i])To find all the divisor of i+1, move from 1 divisor to sqrt(i+1).

Below is C++ implementation of this approach:

// C++ program to print maximum path sum ending with // each position x such that all path step positions // divide x. #include<bits/stdc++.h> using namespace std; void printMaxSum(int arr[], int n) { // Create an array such that dp[i] stores maximum // path sum ending with i. int dp[n]; memset(dp, 0, sizeof dp); // Calculating maximum sum path for each element. for (int i = 0; i < n; i++) { dp[i] = arr[i]; // Finding previous step for arr[i] // Moving from 1 to sqrt(i+1) since all the // divisors are present from sqrt(i+1). int maxi = 0; for (int j=1; j<=sqrt(i+1); j++) { // Checking if j is divisor of i+1. if (((i + 1)%j == 0) && (i + 1) != j) { // Checking which divisor will provide // greater value. if (dp[j-1] > maxi) maxi = dp[j-1]; if (dp[(i+1)/j - 1] > maxi && j != 1) maxi = dp[(i+1)/j - 1]; } } dp[i] += maxi; } // Printing the answer (Maximum path sum ending // with every position i+1. for (int i = 0; i < n; i++) cout << dp[i] << " "; } // Driven Program int main() { int arr[] = { 2, 3, 1, 4, 6, 5 }; int n = sizeof(arr)/sizeof(arr[0]); printMaxSum(arr, n); return 0; }

Output:

2 5 3 9 8 10

**Time Complexity: **O(n*sqrt(n)).

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