给定一个由N 个整数组成的数组arr[] ，任务是找到arr[]的最大子集，使得在该子集中的每一对数字中，一个数字是另一个的除数。
例子：
Input: arr[] = {1, 2, 3, 4, 5}
Output: 4 2 1
All possible pairs of the subsequence are:
(4, 2) -> 4 % 2 = 0
(4, 1) -> 4 % 1 = 0
and (2, 1) -> 2 % 1 = 0
Input: arr[] = {1, 3, 4, 9}
Output: 1 3 9
方法：这里的任务是找到最大的子集，其中在每对数字中，一个可以被另一个整除，即对于序列a 1 , a 2 , a 3 … a k其中a 1 ≤ a 2 ≤ … ≤ a k并且a i+1 % a i = 0对于每个i 。这个序列可以使用动态规划找到(类似于最长递增子序列)。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to find the required subsequence
void findSubSeq(int arr[], int n)
{
// Sort the array
sort(arr, arr + n);
// Keep a count of the length of the
// subsequence and the previous element
int count[n] = { 1 };
int prev[n] = { -1 };
// Set the initial values
memset(count, 1, sizeof(count));
memset(prev, -1, sizeof(prev));
// Maximum length of the subsequence and
// the last element
int max = 0;
int maxprev = -1;
// Run a loop for every element
for (int i = 0; i < n; i++) {
// Check for all the divisors
for (int j = i - 1; j >= 0; j--) {
// If the element is a divisor and the length
// of subsequence will increase by adding
// j as previous element of i
if (arr[i] % arr[j] == 0
&& count[j] + 1 > count[i]) {
// Increase the count
count[i] = count[j] + 1;
prev[i] = j;
}
}
// Update the max count
if (max < count[i]) {
max = count[i];
maxprev = i;
}
}
// Get the last index of the subsequence
int i = maxprev;
while (i >= 0) {
// Print the element
if (arr[i] != -1)
cout << arr[i] << " ";
// Move the index to the previous element
i = prev[i];
}
}
// Driver code
int main()
{
int arr[] = { 1, 2, 3, 4, 5 };
int n = sizeof(arr) / sizeof(int);
findSubSeq(arr, n);
return 0;
} Java
// Java implementation of the approach
import java.util.*;
class GFG
{
// Function to find the required subsequence
static void findSubSeq(int arr[], int n)
{
// Sort the array
Arrays.sort(arr);
// Keep a count of the length of the
// subsequence and the previous element
int count[] = new int[n];
int prev[] = new int[n];
int i, j;
// Set the initial values
for(i = 0 ; i < n; i++)
count[i] = 1;
for(j = 0; j < n; j++)
prev[j] = -1;
// Maximum length of the subsequence and
// the last element
int max = 0;
int maxprev = -1;
// Run a loop for every element
for ( i = 0; i < n; i++)
{
// Check for all the divisors
for ( j = i - 1; j >= 0; j--)
{
// If the element is a divisor and
// the length of subsequence will
// increase by adding j as
// previous element of i
if (arr[i] % arr[j] == 0 &&
count[j] + 1 > count[i])
{
// Increase the count
count[i] = count[j] + 1;
prev[i] = j;
}
}
// Update the max count
if (max < count[i])
{
max = count[i];
maxprev = i;
}
}
// Get the last index of the subsequence
i = maxprev;
while (i >= 0)
{
// Print the element
if (arr[i] != -1)
System.out.print(arr[i] + " ");
// Move the index to the previous element
i = prev[i];
}
}
// Driver code
public static void main (String[] args)
{
int arr[] = { 1, 2, 3, 4, 5 };
int n = arr.length;
findSubSeq(arr, n);
}
}
// This code is contributed by AnkitRai01Python3
# Python3 implementation of the approach
# Function to find the required subsequence
def findSubSeq(arr, n) :
# Sort the array
arr.sort();
# Keep a count of the length of the
# subsequence and the previous element
# Set the initial values
count = [1] * n;
prev = [-1] * n;
# Maximum length of the subsequence and
# the last element
max = 0;
maxprev = -1;
# Run a loop for every element
for i in range(n) :
# Check for all the divisors
for j in range(i - 1, -1, -1) :
# If the element is a divisor and the length
# of subsequence will increase by adding
# j as previous element of i
if (arr[i] % arr[j] == 0 and
count[j] + 1 > count[i]) :
# Increase the count
count[i] = count[j] + 1;
prev[i] = j;
# Update the max count
if (max < count[i]) :
max = count[i];
maxprev = i;
# Get the last index of the subsequence
i = maxprev;
while (i >= 0) :
# Print the element
if (arr[i] != -1) :
print(arr[i], end = " ");
# Move the index to the previous element
i = prev[i];
# Driver code
if __name__ == "__main__" :
arr = [ 1, 2, 3, 4, 5 ];
n = len(arr);
findSubSeq(arr, n);
# This code is contributed by AnkitRai01C#
// C# implementation of the approach
using System;
using System.Collections;
class GFG
{
// Function to find the required subsequence
static void findSubSeq(int []arr, int n)
{
// Sort the array
Array.Sort(arr);
// Keep a count of the length of the
// subsequence and the previous element
int []count = new int[n];
int []prev = new int[n];
int i, j;
// Set the initial values
for(i = 0; i < n; i++)
count[i] = 1;
for(j = 0; j < n; j++)
prev[j] = -1;
// Maximum length of the subsequence
// and the last element
int max = 0;
int maxprev = -1;
// Run a loop for every element
for ( i = 0; i < n; i++)
{
// Check for all the divisors
for ( j = i - 1; j >= 0; j--)
{
// If the element is a divisor and
// the length of subsequence will
// increase by adding j as
// previous element of i
if (arr[i] % arr[j] == 0 &&
count[j] + 1 > count[i])
{
// Increase the count
count[i] = count[j] + 1;
prev[i] = j;
}
}
// Update the max count
if (max < count[i])
{
max = count[i];
maxprev = i;
}
}
// Get the last index of the subsequence
i = maxprev;
while (i >= 0)
{
// Print the element
if (arr[i] != -1)
Console.Write(arr[i] + " ");
// Move the index to the previous element
i = prev[i];
}
}
// Driver code
public static void Main ()
{
int []arr = { 1, 2, 3, 4, 5 };
int n = arr.Length;
findSubSeq(arr, n);
}
}
// This code is contributed by AnkitRai01Javascript
输出：
4 2 1时间复杂度： O(N 2 )
辅助空间： O(1)
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