其所有元素有力数的最大子数组的长度

给定整数元素数组arr [] ，任务是找到arr []的最大子数组的长度，以使子数组中的所有元素都是Powerful number 。

A number n is said to be Powerful Number if, for every prime factor p of it, p² also divides it.

例子：

Input: arr[] = {1, 7, 36, 4, 6, 28, 4}
Output:2
Maximum length sub-array with all elements as Powerful number is {36, 4}

Input: arr[] = {25, 100, 2, 3, 9, 1}
Output: 2
Maximum length sub-array with all elements as Powerful number is {25, 100} or {9, 1}

方法：

• 从左到右遍历数组。用0初始化max_length和current_length变量。
• 如果当前元素是一个有力数，则增加current_length变量并继续。否则，将current_length设置为0 。
• 在每个步骤中，将max_length分配为max_length = max(current_length，max_length) 。
• 最后输出max_length的值。

C++

// C++ program to find the length of the
// largest sub-array of an array every
// element of whose is a powerful number
  
#include 
using namespace std;
  
// Function to check if the
// number is powerful
bool isPowerful(int n)
{
    // First divide the number repeatedly by 2
    while (n % 2 == 0) {
        int power = 0;
        while (n % 2 == 0) {
            n /= 2;
            power++;
        }
  
        // If only 2^1 divides n
        // (not higher powers),
        // then return false
        if (power == 1)
            return false;
    }
  
    // If n is not a power of 2
    // then this loop will execute
    // repeat above process
    for (int factor = 3;
         factor <= sqrt(n);
         factor += 2) {
  
        // Find highest power of "factor"
        // that divides n
        int power = 0;
        while (n % factor == 0) {
            n = n / factor;
            power++;
        }
  
        // If only factor^1 divides n
        // (not higher powers),
        // then return false
        if (power == 1)
            return false;
    }
  
    // n must be 1 now
    // if it is not a prime number.
    // Since prime numbers are not powerful,
    // we return false if n is not 1.
    return (n == 1);
}
  
// Function to return the length of the
// largest sub-array of an array every
// element of whose is a powerful number
int contiguousPowerfulNumber(int arr[], int n)
{
  
    int current_length = 0;
    int max_length = 0;
  
    for (int i = 0; i < n; i++) {
  
        // If arr[i] is
        // a Powerful number
        if (isPowerful(arr[i]))
            current_length++;
        else
            current_length = 0;
  
        max_length = max(max_length,
                         current_length);
    }
  
    return max_length;
}
  
// Driver code
int main()
{
    int arr[] = { 1, 7, 36, 4, 6, 28, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    cout << contiguousPowerfulNumber(arr, n);
  
    return 0;
}

Java

// Java program to find the length of the
// largest sub-array of an array every
// element of whose is a powerful number
import java.util.*;
  
class solution{
  
// Function to check if the
// number is powerful
static boolean isPowerful(int n)
{
    // First divide the number repeatedly by 2
    while (n % 2 == 0) {
        int power = 0;
        while (n % 2 == 0) {
            n /= 2;
            power++;
        }
  
        // If only 2^1 divides n
        // (not higher powers),
        // then return false
        if (power == 1)
            return false;
    }
  
    // If n is not a power of 2
    // then this loop will execute
    // repeat above process
    for (int factor = 3;
        factor <= Math.sqrt(n);
         factor += 2) {
  
        // Find highest power of "factor"
        // that divides n
        int power = 0;
        while (n % factor == 0) {
            n = n / factor;
            power++;
        }
  
        // If only factor^1 divides n
        // (not higher powers),
        // then return false
        if (power == 1)
            return false;
    }
  
    // n must be 1 now
    // if it is not a prime number.
    // Since prime numbers are not powerful,
    // we return false if n is not 1.
    return (n == 1);
}
  
// Function to return the length of the
// largest sub-array of an array every
// element of whose is a powerful number
static int contiguousPowerfulNumber(int[] arr, int n)
{
  
    int current_length = 0;
    int max_length = 0;
  
    for (int i = 0; i < n; i++) {
  
        // If arr[i] is
        // a Powerful number
        if (isPowerful(arr[i]))
            current_length++;
        else
            current_length = 0;
  
        max_length = Math.max(max_length,
                        current_length);
    }
  
    return max_length;
}
  
// Driver code
public static void main(String args[])
{
    int []arr = { 1, 7, 36, 4, 6, 28, 4 };
    int n = arr.length;
  
    System.out.println(contiguousPowerfulNumber(arr, n));
  
}
}
  
// This code is contributed by Bhupendra_Singh

Python3

# Python 3 program to find the length of 
# the largest sub-array of an array every 
# element of whose is a powerful number
  
import math
  
# function to check if 
# the number is powerful 
def isPowerful(n): 
    
    # First divide the number repeatedly by 2 
    while (n % 2 == 0): 
    
        power = 0
        while (n % 2 == 0): 
            
            n = n//2 
            power = power + 1
            
             
        # If only 2 ^ 1 divides 
        # n (not higher powers), 
        # then return false 
        if (power == 1): 
            return False
        
     
    # If n is not a power of 2 
    # then this loop will execute 
    # repeat above process 
    for factor in range(3, int(math.sqrt(n))+1, 2): 
        
        # Find highest power of 
        # "factor" that divides n 
        power = 0
        while (n % factor == 0): 
            
            n = n//factor 
            power = power + 1
            
     
        # If only factor ^ 1 divides 
        # n (not higher powers), 
        # then return false 
        if (power == 1): 
            return false 
        
     
     # n must be 1 now if it 
     # is not a prime numenr. 
     # Since prime numbers are 
     # not powerful, we return 
     # false if n is not 1. 
    return (n == 1) 
  
  
# Function to return the length of the 
# largest sub-array of an array every 
# element of whose is a powerful number
def contiguousPowerfulNumber(arr, n): 
    current_length = 0
    max_length = 0
  
    for i in range(0, n, 1): 
          
        # If arr[i] is a
        # Powerful number
        if (isPowerful(arr[i])): 
            current_length += 1
        else: 
            current_length = 0
  
        max_length = max(max_length, 
                        current_length) 
      
    return max_length 
  
# Driver code 
if __name__ == '__main__': 
    arr = [1, 7, 36, 4, 6, 28, 4] 
    n = len(arr) 
  
    print(contiguousPowerfulNumber(arr, n))

C#

// C# program to find the length of the
// largest sub-array of an array every
// element of whose is a powerful number
using System;
  
class solution{
  
// Function to check if the
// number is powerful
static bool isPowerful(int n)
{
  
    // First divide the number repeatedly by 2
    while (n % 2 == 0) {
        int power = 0;
        while (n % 2 == 0) {
            n /= 2;
            power++;
        }
  
        // If only 2^1 divides n
        // (not higher powers),
        // then return false
        if (power == 1)
            return false;
    }
  
    // If n is not a power of 2
    // then this loop will execute
    // repeat above process
    for (int factor = 3;
             factor <= Math.Sqrt(n);
             factor += 2) {
  
        // Find highest power of "factor"
        // that divides n
        int power = 0;
        while (n % factor == 0) {
            n = n / factor;
            power++;
        }
  
        // If only factor^1 divides n
        // (not higher powers),
        // then return false
        if (power == 1)
            return false;
    }
  
    // n must be 1 now
    // if it is not a prime number.
    // Since prime numbers are not powerful,
    // we return false if n is not 1.
    return (n == 1);
}
  
// Function to return the length of the
// largest sub-array of an array every
// element of whose is a powerful number
static int contiguousPowerfulNumber(int[] arr, int n)
{
    int current_length = 0;
    int max_length = 0;
    for (int i = 0; i < n; i++) {
  
        // If arr[i] is
        // a Powerful number
        if (isPowerful(arr[i]))
            current_length++;
        else
            current_length = 0;
  
        max_length = Math.Max(max_length,
                              current_length);
    }
    return max_length;
}
  
// Driver code
public static void Main(String []args)
{
    int []arr = { 1, 7, 36, 4, 6, 28, 4 };
    int n = arr.Length;
    Console.WriteLine(contiguousPowerfulNumber(arr, n));
}
}
  
// This code is contributed by gauravrajput1

输出：

2

时间复杂度： O(N×√N)
辅助空间复杂度： O(1)