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📜  给定数组中最长递增素数子序列的长度

📅  最后修改于: 2021-09-17 06:54:10             🧑  作者: Mango

给定一个由N 个正整数组成的数组arr[] ,任务是找到给定数组中由素数组成的最长递增子序列的长度。

例子:

朴素方法:最简单的方法是生成给定数组的所有可能子序列,并按递增顺序打印由素数组成的最长子序列的长度。
时间复杂度: O(2 N )
辅助空间: O(N)

Efficient Approach:思路是使用Dynamic Programming的方式来优化上面的方式。此问题是最长递增子序列 (LIS) 问题的基本变体。以下是步骤:

  • 初始化一个大小为N的辅助数组dp[]使得dp[i]将存储以索引i结尾的素数的 LIS 的长度。
  • 下面是寻找最长递增素数的递推关系:
  • 使用埃拉托色尼筛法将所有质数存储到10 5
  • 在给定数组上迭代两个嵌套循环,并根据上述递推关系更新数组dp[]
  • 经过以上所有步骤,数组dp[]中的最大元素就是给定数组中素数最长递增子序列的长度。

下面是上述方法的实现:

C++
// C++ program for the above approach
 
#include 
using namespace std;
#define N 100005
 
// Function to find the prime numbers
// till 10^5 using Sieve of Eratosthenes
void SieveOfEratosthenes(bool prime[],
                         int p_size)
{
    // False here indicates
    // that it is not prime
    prime[0] = false;
    prime[1] = false;
 
    for (int p = 2; p * p <= p_size; p++) {
 
        // If prime[p] is not changed,
        // then it is a prime
        if (prime[p]) {
 
            // Update all multiples of p,
            // set them to non-prime
            for (int i = p * 2;
                 i <= p_size; i += p)
                prime[i] = false;
        }
    }
}
 
// Function which computes the length
// of the LIS of Prime Numbers
int LISPrime(int arr[], int n)
{
    // Create an array of size n
    int lisp[n];
 
    // Create boolean array to
    // mark prime numbers
    bool prime[N + 1];
 
    // Initialize all values to true
    memset(prime, true, sizeof(prime));
 
    // Precompute N primes
    SieveOfEratosthenes(prime, N);
 
    lisp[0] = prime[arr[0]] ? 1 : 0;
 
    // Compute optimized LIS having
    // prime numbers in bottom up manner
    for (int i = 1; i < n; i++) {
        if (!prime[arr[i]]) {
            lisp[i] = 0;
            continue;
        }
 
        lisp[i] = 1;
        for (int j = 0; j < i; j++) {
 
            // Check for LIS and prime
            if (prime[arr[j]]
                && arr[i] > arr[j]
                && lisp[i] < lisp[j] + 1) {
                lisp[i] = lisp[j] + 1;
            }
        }
    }
 
    // Return maximum value in lis[]
    return *max_element(lisp, lisp + n);
}
 
// Driver Code
int main()
{
    // Given array
    int arr[] = { 1, 2, 5, 3, 2, 5, 1, 7 };
 
    // Size of array
    int M = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    cout << LISPrime(arr, M);
 
    return 0;
}


Java
// Java program for the above approach
import java.util.*;
 
class GFG{
     
static final int N = 100005;
 
// Function to find the prime numbers
// till 10^5 using Sieve of Eratosthenes
static void SieveOfEratosthenes(boolean prime[],
                                int p_size)
{
     
    // False here indicates
    // that it is not prime
    prime[0] = false;
    prime[1] = false;
 
    for(int p = 2; p * p <= p_size; p++)
    {
 
        // If prime[p] is not changed,
        // then it is a prime
        if (prime[p])
        {
 
            // Update all multiples of p,
            // set them to non-prime
            for(int i = p * 2;
                    i <= p_size;
                    i += p)
                prime[i] = false;
        }
    }
}
 
// Function which computes the length
// of the LIS of Prime Numbers
static int LISPrime(int arr[], int n)
{
     
    // Create an array of size n
    int []lisp = new int[n];
 
    // Create boolean array to
    // mark prime numbers
    boolean []prime = new boolean[N + 1];
 
    // Initialize all values to true
    for(int i = 0; i < prime.length; i++)
        prime[i] = true;
 
    // Precompute N primes
    SieveOfEratosthenes(prime, N);
 
    lisp[0] = prime[arr[0]] ? 1 : 0;
 
    // Compute optimized LIS having
    // prime numbers in bottom up manner
    for(int i = 1; i < n; i++)
    {
        if (!prime[arr[i]])
        {
            lisp[i] = 0;
            continue;
        }
 
        lisp[i] = 1;
        for(int j = 0; j < i; j++)
        {
             
            // Check for LIS and prime
            if (prime[arr[j]] &&
                arr[i] > arr[j] &&
               lisp[i] < lisp[j] + 1)
            {
                lisp[i] = lisp[j] + 1;
            }
        }
    }
 
    // Return maximum value in lis[]
    return Arrays.stream(lisp).max().getAsInt();
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given array
    int arr[] = { 1, 2, 5, 3, 2, 5, 1, 7 };
 
    // Size of array
    int M = arr.length;
 
    // Function call
    System.out.print(LISPrime(arr, M));
}
}
 
// This code is contributed by Amit Katiyar


Python3
# Python3 program for
# the above approach
N = 100005
  
# Function to find the prime numbers
# till 10^5 using Sieve of Eratosthenes
def SieveOfEratosthenes(prime, p_size):
 
  # False here indicates
  # that it is not prime
  prime[0] = False
  prime[1] = False
 
  p = 2
  while p * p <= p_size:
 
    # If prime[p] is not changed,
    # then it is a prime
    if (prime[p]):
 
      # Update all multiples of p,
      # set them to non-prime
      for i in range (p * 2,
                      p_size + 1, p):
        prime[i] = False
 
        p += 1
 
# Function which computes the length
# of the LIS of Prime Numbers
def LISPrime(arr, n):
 
  # Create an array of size n
  lisp = [0] * n
 
  # Create boolean array to
  # mark prime numbers
  prime = [True] * (N + 1)
 
  # Precompute N primes
  SieveOfEratosthenes(prime, N)
 
  if prime[arr[0]]:
    lisp[0] = 1
    else:
      lisp[0] = 0
 
      # Compute optimized LIS having
      # prime numbers in bottom up manner
      for i in range (1, n):
        if (not prime[arr[i]]):
          lisp[i] = 0
          continue
 
          lisp[i] = 1
          for j in range (i):
            # check for LIS and prime
            if (prime[arr[j]] and
                arr[i] > arr[j] and
                lisp[i] < lisp[j] + 1):
              lisp[i] = lisp[j] + 1
 
              # Return maximum value in lis[]
              return max(lisp)
 
# Driver Code
if __name__ == "__main__":
 
  # Given array
  arr = [1, 2, 5, 3,
         2, 5, 1, 7]
 
  # Size of array
  M = len(arr)
 
  # Function Call
  print (LISPrime(arr, M))
 
# This code is contributed by Chitranayal


C#
// C# program for the above approach
using System;
using System.Linq;
 
class GFG{
     
static readonly int N = 100005;
 
// Function to find the prime numbers
// till 10^5 using Sieve of Eratosthenes
static void SieveOfEratosthenes(bool []prime,
                                int p_size)
{
     
    // False here indicates
    // that it is not prime
    prime[0] = false;
    prime[1] = false;
 
    for(int p = 2; p * p <= p_size; p++)
    {
 
        // If prime[p] is not changed,
        // then it is a prime
        if (prime[p])
        {
 
            // Update all multiples of p,
            // set them to non-prime
            for(int i = p * 2;
                    i <= p_size;
                    i += p)
                prime[i] = false;
        }
    }
}
 
// Function which computes the length
// of the LIS of Prime Numbers
static int LISPrime(int []arr, int n)
{
     
    // Create an array of size n
    int []lisp = new int[n];
 
    // Create bool array to
    // mark prime numbers
    bool []prime = new bool[N + 1];
 
    // Initialize all values to true
    for(int i = 0; i < prime.Length; i++)
        prime[i] = true;
 
    // Precompute N primes
    SieveOfEratosthenes(prime, N);
 
    lisp[0] = prime[arr[0]] ? 1 : 0;
 
    // Compute optimized LIS having
    // prime numbers in bottom up manner
    for(int i = 1; i < n; i++)
    {
        if (!prime[arr[i]])
        {
            lisp[i] = 0;
            continue;
        }
 
        lisp[i] = 1;
        for(int j = 0; j < i; j++)
        {
             
            // Check for LIS and prime
            if (prime[arr[j]] &&
                arr[i] > arr[j] &&
               lisp[i] < lisp[j] + 1)
            {
                lisp[i] = lisp[j] + 1;
            }
        }
    }
 
    // Return maximum value in lis[]
    return lisp.Max();
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given array
    int []arr = { 1, 2, 5, 3, 2, 5, 1, 7 };
 
    // Size of array
    int M = arr.Length;
 
    // Function call
    Console.Write(LISPrime(arr, M));
}
}
 
// This code is contributed by Amit Katiyar


Javascript


输出:
4

时间复杂度: O(N 2 )
辅助空间: O(N)

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