打印数组中所有最长除法子序列

给定N个整数的数组arr [] ，任务是打印由给定数组形成的最长除法子序列的所有元素。如果我们有多个序列且最大长度，则打印所有序列。

例子：

Input: arr[] = { 2, 11, 16, 12, 36, 60, 71 }
Output:
2 12 36
2 12 60
Explanation:
There are two subsequence with maximum length 3.
1. 2, 12, 36
2. 2, 12, 60

Input: arr[] = { 2, 4, 16, 32, 64, 60, 12 };
Output:
2 4 16 32 64
Explanation:
There is only one subsequence with maximum length 5.
1. 2 4 16 32 64

方法：

1. 声明一个二维数组LDS [] [] ，该数组存储最长除法子序列。
2. 使用本文讨论的方法遍历给定的数组arr []并找到直到当前元素为止的最长划分子序列。
3. 遍历给定的LDS [] []数组，并以最大长度打印序列中的所有元素。

下面是上述方法的实现：

C++

// C++ program for the above approach
  
#include "bits/stdc++.h"
using namespace std;
  
// Function to print LDS[i] element
void printLDS(vector& Max)
{
  
    // Traverse the Max[]
    for (auto& it : Max) {
        cout << it << ' ';
    }
}
  
// Function to construct and print Longest
// Dividing Subsequence
void LongestDividingSeq(int arr[], int N)
{
    // 2D vector for storing sequences
    vector > LDS(N);
  
    // Push the first element to LDS[][]
    LDS[0].push_back(arr[0]);
  
    // Interate over all element
    for (int i = 1; i < N; i++) {
  
        // Loop for every index till i
        for (int j = 0; j < i; j++) {
  
            // if current elements divides
            // arr[i] and length is greater
            // than the previous index, then
            // insert the current element
            // to the sequences LDS[i]
            if ((arr[i] % arr[j] == 0)
                && (LDS[i].size() < LDS[j].size() + 1))
                LDS[i] = LDS[j];
        }
  
        // L[i] ends with arr[i]
        LDS[i].push_back(arr[i]);
    }
  
    int maxLength = 0;
  
    // LDS stores the sequences till
    // each element of arr[]
    // Traverse the LDS[][] to find the
    // the maximum length
    for (int i = 0; i < N; i++) {
        int x = LDS[i].size();
        maxLength = max(maxLength, x);
    }
  
    // Print all LDS with maximum length
    for (int i = 0; i < N; i++) {
  
        // Find size
        int size = LDS[i].size();
  
        // If size = maxLength
        if (size == maxLength) {
  
            // Print LDS
            printLDS(LDS[i]);
            cout << '\n';
        }
    }
}
  
// Driver Code
int main()
{
    int arr[] = { 2, 11, 16, 12, 36, 60, 71 };
    int N = sizeof(arr) / sizeof(arr[0]);
  
    // Function Call
    LongestDividingSeq(arr, N);
  
    return 0;
}

Python3

# Python3 program for the above approach
  
# Function to print LDS[i] element
def printLDS(Max):
      
    # Traverse the Max[]
    for it in Max:
        print(it, end = " ")
  
# Function to construct and print 
# Longest Dividing Subsequence
def LongestDividingSeq(arr, N):
      
    # 2D vector for storing sequences
    LDS = [[] for i in range(N)]
  
    # Push the first element to LDS[][]
    LDS[0].append(arr[0])
  
    # Interate over all element
    for i in range(1, N):
          
        # Loop for every index till i
        for j in range(i):
  
            # If current elements divides
            # arr[i] and length is greater
            # than the previous index, then
            # insert the current element
            # to the sequences LDS[i]
            if ((arr[i] % arr[j] == 0) and
                (len(LDS[i]) < len(LDS[j]) + 1)):
                LDS[i] = LDS[j].copy()
  
        # L[i] ends with arr[i]
        LDS[i].append(arr[i])
  
    maxLength = 0
  
    # LDS stores the sequences till
    # each element of arr[]
    # Traverse the LDS[][] to find 
    # the maximum length
    for i in range(N):
        x = len(LDS[i])
        maxLength = max(maxLength, x)
  
    # Print all LDS with maximum length
    for i in range(N):
  
        # Find size
        size = len(LDS[i])
  
        # If size = maxLength
        if (size == maxLength):
  
            # Print LDS
            printLDS(LDS[i])
            print()
  
# Driver Code
arr = [ 2, 11, 16, 12, 36, 60, 71 ]
N = len(arr)
  
# Function call
LongestDividingSeq(arr, N);
  
# This code is contributed by ANKITKUMAR34

输出：

2 12 36 
2 12 60

时间复杂度： O(N ² )
辅助空间： O(N ² )