给定一个由n 个正整数组成的数组。最初我们处于第一位置。如果 x 除 y 且 x < y，我们可以从位置 x (1 <= x <= n) 跳转到位置 y (1 <= y <= n)。任务是打印在每个位置 x 结束的最大和路径。
注意：由于第一个元素在位置 1，我们可以从这里跳转到任何位置，因为 1 可以除掉所有其他位置编号。
例子 ：
Input : arr[] = {2, 3, 1, 4, 6, 5}
Output : 2 5 3 9 8 10
Explanation:
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 + 1 = 3.
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.方法：
这个想法是使用动态规划来解决这个问题。
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).下面是这个方法的实现：
C++
// C++ program to print maximum
// path sum ending with
// each position x such that all
// path step positions
// divide x.
#include
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
# Python3 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.PHP
$maxi)
$maxi = $dp[$j - 1];
if ($dp[($i + 1) / $j - 1] > $maxi && $j != 1)
$maxi = $dp[($i + 1) / $j - 1];
}
}
$dp[$i] += $maxi;
}
// Printing the answer (Maximum path sum ending
// with every position i+1.
for ($i = 0; $i < $n; $i++)
echo $dp[$i] , " " ;
}
// Driven Program
$arr = array(2, 3, 1, 4, 6, 5 );
$n = sizeof($arr);
printMaxSum($arr, $n);
// This code is contributed by Ryuga
?>Javascript
输出：
2 5 3 9 8 10如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。