下一个元素的索引为 arr[arr[i] + i] 的最长子序列

给定一个数组arr[] ，任务是从数组中找到满足以下条件的最大长度子序列：
可以选择任何元素作为子序列的第一个元素，但下一个元素的索引将由arr[arr[i] + i]确定，其中i是序列中前一个元素的索引。
例子：

Input: arr[] = {1, 2, 3, 4, 5}
Output: 1 2 4
arr[0] = 1, arr[1 + 0] = arr[1] = 2, arr[2 + 1] = arr[3] = 4
Other possible sub-sequences are {2, 4}, {3}, {4} and {5}
Input: arr[] = {1, 6, 3, 1, 12, 1, 4}
Output: 3 1 4

编程需要懂一点英语

方法：

利用两个数组temp和print 。
temp数组将存储当前正在考虑的数组元素，而print数组将存储要作为最终输出打印的数组元素。
从0迭代到n – 1并将当前元素视为序列的第一个元素。
将当前序列的所有元素存储到临时数组中。
如果临时数组的大小变得大于打印数组，则将临时数组的所有内容复制到打印数组。
考虑完所有序列后，打印打印数组的内容。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to print the maximum length sub-sequence
void maxLengthSubSeq(int a[], int n)
{
    // Arrays to store the values to be printed
    int temp[n], print[n];
    int y = 0;
 
    for (int i = 0; i < n; i++) {
        int j = 0;
        int x = 0;
 
        // Store the first value into the temp array
        temp[j++] = a[x];
 
        // Index of the next element
        x = a[x] + x;
 
        // Iterate till index is in range of the array
        while (x < n) {
            temp[j++] = a[x];
            x = a[x] + x;
        }
 
        // If the length (temp) > the length (print) then
        // copy the contents of the temp array into
        // the print array
        if (y < j) {
            for (int k = 0; k < j; k++) {
                print[k] = temp[k];
                y = j;
            }
        }
    }
 
    // Print the contents of the array
    for (int i = 0; i < y; i++)
        cout << print[i] << " ";
}
 
// Driver code
int main()
{
    int a[] = { 1, 2, 3, 4, 5 };
    int n = sizeof(a) / sizeof(a[0]);
    maxLengthSubSeq(a, n);
    return 0;
}

Java

//Java  implementation of the approach/
 
import java.io.*;
 
class GFG {
     
// Function to print the maximum length sub-sequence
static void maxLengthSubSeq(int a[], int n)
{
    // Arrays to store the values to be printed
    int temp[]=new int[n];
    int print[]=new int[n];
    int y = 0;
 
    for (int i = 0; i < n; i++) {
        int j = 0;
        int x = 0;
 
        // Store the first value into the temp array
        temp[j++] = a[x];
 
        // Index of the next element
        x = a[x] + x;
 
        // Iterate till index is in range of the array
        while (x < n) {
            temp[j++] = a[x];
            x = a[x] + x;
        }
 
        // If the length (temp) > the length (print) then
        // copy the contents of the temp array into
        // the print array
        if (y < j) {
            for (int k = 0; k < j; k++) {
                print[k] = temp[k];
                y = j;
            }
        }
    }
 
    // Print the contents of the array
    for (int i = 0; i < y; i++)
            System.out.print(print[i] + " ");
}
 
// Driver code
    public static void main (String[] args) {
 
    int a[] = { 1, 2, 3, 4, 5 };
    int n = a.length;
    maxLengthSubSeq(a, n);
    }
//This code is contributed by @Tushil.   
}

Python3

# Python 3 implementation of the approach
 
# Function to print the maximum length
# sub-sequence
def maxLengthSubSeq(a, n):
     
    # Arrays to store the values to be printed
    temp = [0 for i in range(n)]
    print1 = [0 for i in range(n)]
    y = 0
 
    for i in range(0, n, 1):
        j = 0
        x = 0
 
        # Store the first value into
        # the temp array
        temp[j] = a[x]
        j += 1
 
        # Index of the next element
        x = a[x] + x
 
        # Iterate till index is in range
        # of the array
        while (x < n):
            temp[j] = a[x]
            j += 1
            x = a[x] + x
         
        # If the length (temp) > the length
        # (print) then copy the contents of
        # the temp array into the print array
        if (y < j):
            for k in range(0, j, 1):
                print1[k] = temp[k]
                y = j
             
    # Print the contents of the array
    for i in range(0, y, 1):
        print(print1[i], end = " ")
 
# Driver code
if __name__ == '__main__':
    a = [1, 2, 3, 4, 5]
    n = len(a)
    maxLengthSubSeq(a, n)
 
# This code is contributed by
# Surendra_Gangwar

C#

//C# implementation of the approach/
 
using System;
 
public class GFG{
         
// Function to print the maximum length sub-sequence
static void maxLengthSubSeq(int []a, int n)
{
    // Arrays to store the values to be printed
    int []temp=new int[n];
    int []print=new int[n];
    int y = 0;
 
    for (int i = 0; i < n; i++) {
        int j = 0;
        int x = 0;
 
        // Store the first value into the temp array
        temp[j++] = a[x];
 
        // Index of the next element
        x = a[x] + x;
 
        // Iterate till index is in range of the array
        while (x < n) {
            temp[j++] = a[x];
            x = a[x] + x;
        }
 
        // If the length (temp) > the length (print) then
        // copy the contents of the temp array into
        // the print array
        if (y < j) {
            for (int k = 0; k < j; k++) {
                print[k] = temp[k];
                y = j;
            }
        }
    }
 
    // Print the contents of the array
    for (int i = 0; i < y; i++)
            Console.Write(print[i] + " ");
}
 
// Driver code
    static public void Main (){
         
    int []a = { 1, 2, 3, 4, 5 };
    int n = a.Length;
    maxLengthSubSeq(a, n);
    }
//This code is contributed by ajit.
}

PHP

 the length
        // (print) then copy the contents of
        // the temp array into the print array
        if ($y < $j)
        {
            for ($k = 0; $k < $j; $k++)
            {
                $print[$k] = $temp[$k];
                $y = $j;
            }
        }
    }
 
    // Print the contents of the array
    for ($i = 0; $i < $y; $i++)
        echo $print[$i] . " ";
}
 
// Driver code
$a = array(1, 2, 3, 4, 5);
$n = sizeof($a);
maxLengthSubSeq($a, $n);
 
// This code is contributed
// by Akanksha Rai

Javascript

输出：

1 2 4