Given an array of **n** positive integers. Initially we are at first position. We can jump to position y (1 <= x <= n) from position x (1 <= x <= n) if x divides y and x < y. The task is to print maximum sum path ending at every position x.
Note : Since first element is at position 1, we can jump to any position from here as 1 divides all other position numbers.
**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 the implementation of this approach:

## C++

// 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; }

## Java

// Java program to print maximum path // sum ending with each position x such // that all path step positions divide x. import java.util.*; class GFG { static void printMaxSum(int arr[], int n) { // Create an array such that dp[i] // stores maximum path sum ending with i. int dp[] = new int[n]; Arrays.fill(dp, 0); // 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 <= Math.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++) System.out.print(dp[i] + " "); } // Driver code public static void main(String[] args) { int arr[] = { 2, 3, 1, 4, 6, 5 }; int n = arr.length; // Function calling printMaxSum(arr, n); } } // This code is contributed by Anant Agarwal.

## Python3

# Python program to print maximum # path sum ending with each position # x such that all path step positions # divide x. def printMaxSum(arr, n): # Create an array such that dp[i] # stores maximum path sum ending with i. dp = [0 for i in range(n)] # Calculating maximum sum path # for each element. for i in range(n): dp[i] = arr[i] # Finding previous step for arr[i] # Moving from 1 to sqrt(i + 1) since all the # divisiors are present from sqrt(i + 1). maxi = 0 for j in range(1, int((i + 1) ** 0.5) + 1): # Checking if j is divisior of i + 1. if ((i + 1) % j == 0 and (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 and j != 1): maxi = dp[(i + 1) // j - 1] dp[i] += maxi # Printing the answer # (Maximum path sum ending # with every position i + 1). for i in range(n): print(dp[i], end = ' ') # Driver Program arr = [2, 3, 1, 4, 6, 5] n = len(arr) printMaxSum(arr, n) # This code is contributed by Soumen Ghosh.

## C#

// C# program to print maximum path // sum ending with each position x such // that all path step positions divide x. using System; class GFG { static void printMaxSum(int[] arr, int n) { // Create an array such that dp[i] // stores maximum path sum ending with i. int[] dp = new int[n]; // 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 <= Math.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++) Console.Write(dp[i] + " "); } // Driver code public static void Main() { int[] arr = { 2, 3, 1, 4, 6, 5 }; int n = arr.Length; // Function calling printMaxSum(arr, n); } } // This code is contributed by vt_m.

Output:

2 5 3 9 8 10

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

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