计算给定范围内具有确切5个不同因子的数字

给定两个整数L和R ，任务是从正好具有5个不同正因子的范围[L，R]中计算数量。

例子：

Input: L = 1, R= 100 
Output: 2 
Explanation: The only two numbers in the range [1, 100] having exactly 5 prime factors are 16 and 81. 
Factors of 16 are {1, 2, 4, 8, 16}. 
Factors of 8 are {1, 3, 9, 27, 81}.

Input: L = 1, R= 100 
Output: 2 
 

天真的方法：解决此问题的最简单方法是遍历范围[L，R]，并为每个数字计算其因素。如果因子计数等于5 ，则将计数增加1 。
时间复杂度： (R – L)×√N
辅助空间： O(1)
高效方法：为了优化上述方法，需要对具有5个因子的数字进行以下观察。

Let the prime factorization of a number be p₁^a₁×p₂^a₂ × … ×p_n^a_n. 
Therefore, the count of factors of this number can be written as (a₁ + 1)×(a₂ + 1)× … ×(a_n + 1). 
Since this product must be equal to 5 (which is a prime number), only one term greater than 1 must exist in the product. That term must be equal to 5. 
Therefore, if a_i + 1 = 5 
=> a_i = 4

请按照以下步骤解决问题：

• 所需的计数是包含p ⁴作为因子的范围内的数字计数，其中p是质数。
• 为了在较大范围( [ 1，10 ¹⁸ ] )上有效地计算p ⁴ ，想法是使用Eratosthenes的Sieve存储最多10 ^4.5的所有素数。

下面是上述方法的实现：

C++14

// C++ Program to implement
// the above approach
 
#include 
using namespace std;
 
const int N = 2e5;
 
// Stores all prime numbers
// up to 2 * 10^5
vector prime;
 
// Function to generate all prime
// numbers up to 2 * 10 ^ 5 using
// Sieve of Eratosthenes
void Sieve()
{
    prime.clear();
    vector p(N + 1, true);
 
    // Mark 0 and 1 as non-prime
    p[0] = p[1] = false;
 
    for (int i = 2; i * i <= N; i++) {
 
        // If i is prime
        if (p[i] == true) {
 
            // Mark all its factors as non-prime
            for (int j = i * i; j <= N; j += i) {
                p[j] = false;
            }
        }
    }
 
    for (int i = 1; i < N; i++) {
 
        // If current number is prime
        if (p[i]) {
 
            // Store the prime
            prime.push_back(1LL * pow(i, 4));
        }
    }
}
 
// Function to count numbers in the
// range [L, R] having exactly 5 factors
void countNumbers(long long int L,
                  long long int R)
{
 
    // Stores the required count
    int Count = 0;
 
    for (int p : prime) {
 
        if (p >= L && p <= R) {
            Count++;
        }
    }
    cout << Count << endl;
}
 
// Driver Code
int main()
{
    long long L = 16, R = 85000;
 
    Sieve();
 
    countNumbers(L, R);
 
    return 0;
}

Java

// Java Program to implement
// the above approach
import java.util.*;
class GFG
{
  static int N = 200000;
 
  // Stores all prime numbers
  // up to 2 * 10^5
  static int prime[] = new int [20000];
  static int index = 0;
 
  // Function to generate all prime
  // numbers up to 2 * 10 ^ 5 using
  // Sieve of Eratosthenes
  static void Sieve()
  {
    index = 0;
    int p[] = new int [N + 1];
    for(int i = 0; i <= N; i++)
    {
      p[i] = 1;
    }
 
    // Mark 0 and 1 as non-prime
    p[0] = p[1] = 0;
    for (int i = 2; i * i <= N; i++)
    {
 
      // If i is prime
      if (p[i] == 1)
      {
 
        // Mark all its factors as non-prime
        for (int j = i * i; j <= N; j += i)
        {
          p[j] = 0;
        }
      }
    }
    for (int i = 1; i < N; i++)
    {
 
      // If current number is prime
      if (p[i] == 1)
      {
 
        // Store the prime
        prime[index++] = (int)(Math.pow(i, 4));
      }
    }
  }
 
  // Function to count numbers in the
  // range [L, R] having exactly 5 factors
  static void countNumbers(int L,int R)
  {
 
    // Stores the required count
    int Count = 0;
    for(int i = 0; i < index; i++)
    {
      int p = prime[i];
      if (p >= L && p <= R)
      {
        Count++;
      }
    }
    System.out.println(Count);
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    int L = 16, R = 85000;
    Sieve();
    countNumbers(L, R);
  }
}
 
// This code is contributed by amreshkumar3.

Python3

# Python3 implementation of
# the above approach
N = 2 * 100000
 
# Stores all prime numbers
# up to 2 * 10^5
prime = [0] * N
 
# Function to generate all prime
# numbers up to 2 * 10 ^ 5 using
# Sieve of Eratosthenes
def Sieve() :
    p = [True] * (N + 1)
 
    # Mark 0 and 1 as non-prime
    p[0] = p[1] = False
    i = 2
    while(i * i <= N) :
 
        # If i is prime
        if (p[i] == True) :
 
            # Mark all its factors as non-prime
            for j in range(i * i, N, i):
                p[j] = False    
        i += 1
    for i in range(N):
 
        # If current number is prime
        if (p[i] != False) :
 
            # Store the prime
            prime.append(pow(i, 4))
 
# Function to count numbers in the
# range [L, R] having exactly 5 factors
def countNumbers(L, R) :
 
    # Stores the required count
    Count = 0
    for p in prime :
        if (p >= L and p <= R) :
            Count += 1
    print(Count)
 
# Driver Code
L = 16
R = 85000
Sieve()
countNumbers(L, R)
 
# This code is contributed by code_hunt.

C#

// C# Program to implement
// the above approach
using System;
class GFG
{
  static int N = 200000;
 
  // Stores all prime numbers
  // up to 2 * 10^5
  static int []prime = new int [20000];
  static int index = 0;
 
  // Function to generate all prime
  // numbers up to 2 * 10 ^ 5 using
  // Sieve of Eratosthenes
  static void Sieve()
  {
    index = 0;
    int []p = new int [N + 1];
    for(int i = 0; i <= N; i++)
    {
      p[i] = 1;
    }
 
    // Mark 0 and 1 as non-prime
    p[0] = p[1] = 0;
    for (int i = 2; i * i <= N; i++)
    {
 
      // If i is prime
      if (p[i] == 1)
      {
 
        // Mark all its factors as non-prime
        for (int j = i * i; j <= N; j += i)
        {
          p[j] = 0;
        }
      }
    }
    for (int i = 1; i < N; i++)
    {
 
      // If current number is prime
      if (p[i] == 1)
      {
 
        // Store the prime
        prime[index++] = (int)(Math.Pow(i, 4));
      }
    }
  }
 
  // Function to count numbers in the
  // range [L, R] having exactly 5 factors
  static void countNumbers(int L,int R)
  {
 
    // Stores the required count
    int Count = 0;
    for(int i = 0; i < index; i++)
    {
      int p = prime[i];
      if (p >= L && p <= R)
      {
        Count++;
      }
    }
    Console.WriteLine(Count);
  }
 
  // Driver Code
  public static void Main(String[] args)
  {
    int L = 16, R = 85000;
    Sieve();
    countNumbers(L, R);
  }
}
 
// This code is contributed by shikhasingrajput

输出：

7

时间复杂度： O(N * log(log(N)))
辅助空间： O(N)