仅由素数组成的子数组的计数

给定一个长度为N的数组A[] ，任务是找到仅由素数组成的子数组的数量。

例子：

Input: arr[] = {2, 3, 4, 5, 7} 
Output: 6 
Explanation: 
All possible subarrays made up of only prime numbers are {{2}, {3}, {2, 3}, {5}, {7}, {5, 7}}
Input: arr[] = {2, 3, 5, 6, 7, 11, 3, 5, 9, 3} 
Output: 17

朴素方法：解决问题的最简单方法是从给定数组生成所有可能的子数组，并检查它是否仅由素数组成。
时间复杂度： O(N ³ * √max(array))，其中√M是检查一个数是否为素数所需的时间，这个M 的范围是[min(arr), max(arr)]
辅助空间： O(1)
高效的方法：需要进行以下观察来优化上述方法：

Count of subarrays from an array of length M is equal to M * (M + 1) / 2.

因此，从给定的数组中，仅由素数组成的长度为M的连续子数组将生成M * (M + 1) / 2个长度的子数组。
请按照以下步骤解决问题：

• 遍历数组并检查每个元素是否为素数。
• 对于找到的每个素数，不断增加count 。
• 对于每个非素数元素，通过添加count * (count + 1) / 2并将count重置为0来更新所需的答案。
• 最后，打印所需的子数组。

下面是上述方法的实现：

C++

// C++ Program to implement
// the above approach
#include 
using namespace std;
 
// Function to check if a number
// is prime or not.
bool is_prime(int n)
{
    if (n <= 1)
        return 0;
 
    for (int i = 2; i * i <= n; i++) {
 
        // If n has any factor other than 1,
        // then n is non-prime.
        if (n % i == 0)
            return 0;
    }
 
    return 1;
}
 
// Function to return the count of
// subarrays made up of prime numbers only
int count_prime_subarrays(int ar[], int n)
{
 
    // Stores the answer
    int ans = 0;
 
    // Stores the count of continous
    // prime numbers in an array
    int count = 0;
 
    for (int i = 0; i < n; i++) {
 
        // If the current array
        // element is prime
        if (is_prime(ar[i]))
 
            // Increase the count
            count++;
        else {
 
            if (count) {
 
                // Update count of subarrays
                ans += count * (count + 1)
                       / 2;
                count = 0;
            }
        }
    }
 
    // If the array ended with a
    // continous prime sequence
    if (count)
        ans += count * (count + 1) / 2;
 
    return ans;
}
 
// Driver Code
int main()
{
    int N = 10;
    int ar[] = { 2, 3, 5, 6, 7,
                 11, 3, 5, 9, 3 };
    cout << count_prime_subarrays(ar, N);
}

Java

// Java Program to implement
// the above approach
import java.util.*;
class GFG{
 
// Function to check if a number
// is prime or not.
static boolean is_prime(int n)
{
    if (n <= 1)
         return false;
 
    for (int i = 2; i * i <= n; i++)
    {
 
        // If n has any factor other than 1,
        // then n is non-prime.
        if (n % i == 0)
             return false; 
    }
    return true;
}
 
// Function to return the count of
// subarrays made up of prime numbers only
static int count_prime_subarrays(int ar[], int n)
{
 
    // Stores the answer
    int ans = 0;
 
    // Stores the count of continous
    // prime numbers in an array
    int count = 0;
 
    for (int i = 0; i < n; i++)
    {
 
        // If the current array
        // element is prime
        if (is_prime(ar[i]))
 
            // Increase the count
            count++;
        else
        {
            if (count != 0)
            {
 
                // Update count of subarrays
                ans += count * (count + 1) / 2;
                count = 0;
            }
        }
    }
 
    // If the array ended with a
    // continous prime sequence
    if (count != 0)
        ans += count * (count + 1) / 2;
 
    return ans;
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 10;
    int []ar = { 2, 3, 5, 6, 7,
                11, 3, 5, 9, 3 };
    System.out.print(count_prime_subarrays(ar, N));
}
}
 
// This code is contributed by PrinciRaj1992

Python3

# Python3 program to implement
# the above approach
 
# Function to check if a number
# is prime or not.
def is_prime(n):
 
    if(n <= 1):
        return 0
 
    i = 2
    while(i * i <= n):
 
        # If n has any factor other than 1,
        # then n is non-prime.
        if(n % i == 0):
            return 0
 
        i += 1
 
    return 1
 
# Function to return the count of
# subarrays made up of prime numbers only
def count_prime_subarrays(ar, n):
 
    # Stores the answer
    ans = 0
 
    # Stores the count of continous
    # prime numbers in an array
    count = 0
 
    for i in range(n):
 
        # If the current array
        # element is prime
        if(is_prime(ar[i])):
 
            # Increase the count
            count += 1
 
        else:
            if(count):
 
                # Update count of subarrays
                ans += count * (count + 1) // 2
                count = 0
 
    # If the array ended with a
    # continous prime sequence
    if(count):
        ans += count * (count + 1) // 2
 
    return ans
 
# Driver Code
N = 10
ar = [ 2, 3, 5, 6, 7,
       11, 3, 5, 9, 3 ]
 
# Function call
print(count_prime_subarrays(ar, N))
 
# This code is contributed by Shivam Singh

C#

// C# Program to implement
// the above approach
using System;
class GFG{
 
// Function to check if a number
// is prime or not.
static bool is_prime(int n)
{
    if (n <= 1)
         return false;
 
    for (int i = 2; i * i <= n; i++)
    {
 
        // If n has any factor other than 1,
        // then n is non-prime.
        if (n % i == 0)
             return false; 
    }
    return true;
}
 
// Function to return the count of
// subarrays made up of prime numbers only
static int count_prime_subarrays(int []ar, int n)
{
 
    // Stores the answer
    int ans = 0;
 
    // Stores the count of continous
    // prime numbers in an array
    int count = 0;
 
    for (int i = 0; i < n; i++)
    {
 
        // If the current array
        // element is prime
        if (is_prime(ar[i]))
 
            // Increase the count
            count++;
        else
        {
            if (count != 0)
            {
 
                // Update count of subarrays
                ans += count * (count + 1) / 2;
                count = 0;
            }
        }
    }
 
    // If the array ended with a
    // continous prime sequence
    if (count != 0)
        ans += count * (count + 1) / 2;
 
    return ans;
}
 
// Driver Code
public static void Main(String[] args)
{
    int N = 10;
    int []ar = { 2, 3, 5, 6, 7,
                11, 3, 5, 9, 3 };
    Console.Write(count_prime_subarrays(ar, N));
}
}
 
// This code is contributed by Rajput-Ji

Javascript

 
      // JavaScript Program to implement
      // the above approach
 
      // Function to check if a number
      // is prime or not.
      function is_prime(n) {
        if (n <= 1) return 0;
 
        for (var i = 2; i * i <= n; i++) {
           
          // If n has any factor other than 1,
          // then n is non-prime.
          if (n % i == 0) return 0;
        }
 
        return 1;
      }
 
      // Function to return the count of
      // subarrays made up of prime numbers only
      function count_prime_subarrays(ar, n) {
        // Stores the answer
        var ans = 0;
 
        // Stores the count of continous
        // prime numbers in an array
        var count = 0;
 
        for (var i = 0; i < n; i++) {
          // If the current array
          // element is prime
          if (is_prime(ar[i]))
            // Increase the count
            count++;
          else {
            if (count) {
              // Update count of subarrays
              ans += (count * (count + 1)) / 2;
              count = 0;
            }
          }
        }
 
        // If the array ended with a
        // continous prime sequence
        if (count) ans += (count * (count + 1)) / 2;
 
        return ans;
      }
 
      // Driver Code
 
      var N = 10;
      var ar = [2, 3, 5, 6, 7, 11, 3, 5, 9, 3];
      document.write(count_prime_subarrays(ar, N));
       

输出：

17

时间复杂度： O(N * √max(arr))，其中√M是检查一个数是否为素数所需的时间，这个M 的范围是[min(arr), max(arr)]
辅助空间： O(1)

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