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📜  小于N的数字恰好是两个不同质数的乘积

📅  最后修改于: 2021-04-27 16:35:51             🧑  作者: Mango

给定一个数字N  。任务是找到所有小于N的数字,并且它们是两个截然不同的质数的乘积。
例如,33是两个不同质数的乘积,即11 * 3,而像60这样的数字则具有三个不同质数的乘积,即2 * 2 * 3 * 5。
注意:这些数字不能是一个完美的正方形。
例子:

算法

  1. 遍历到N并检查每个数字是否正好有两个素数。
  2. 现在,要避免出现这样的情况,例如49具有两个素数的7 * 7乘积,请检查该数字是否是一个完美的平方,以确保它具有两个不同的素数。
  3. 如果满足第1步和第2步的要求,则将其添加到引导程序列表中。
  4. 遍历矢量并打印其中的所有元素。

下面是上述方法的实现:

C++
// C++ program to find numbers that are product
// of exactly two distinct prime numbers
 
#include 
using namespace std;
 
// Function to check whether a number
// is a PerfectSquare or not
bool isPerfectSquare(long double x)
{
 
    long double sr = sqrt(x);
 
    return ((sr - floor(sr)) == 0);
}
 
// Function to check if a number is a
// product of exactly two distinct primes
bool isProduct(int num)
{
    int cnt = 0;
 
    for (int i = 2; cnt < 2 && i * i <= num; ++i) {
        while (num % i == 0) {
            num /= i;
            ++cnt;
        }
    }
 
    if (num > 1)
        ++cnt;
 
    return cnt == 2;
}
 
// Function to find numbers that are product
// of exactly two distinct prime numbers.
void findNumbers(int N)
{
    // Vector to store such numbers
    vector vec;
 
    for (int i = 1; i <= N; i++) {
        if (isProduct(i) && !isPerfectSquare(i)) {
 
            // insert in the vector
            vec.push_back(i);
        }
    }
 
    // Print all numers till n from the vector
    for (int i = 0; i < vec.size(); i++) {
        cout << vec[i] << " ";
    }
}
 
// Driver function
int main()
{
    int N = 30;
 
    findNumbers(N);
 
    return 0;
}


Java
// Java program to find numbers that are product
// of exactly two distinct prime numbers
import java.util.*; 
 
class GFG{
// Function to check whether a number
// is a PerfectSquare or not
static boolean isPerfectSquare(double x)
{
 
    double sr = Math.sqrt(x);
 
    return ((sr - Math.floor(sr)) == 0);
}
 
// Function to check if a number is a
// product of exactly two distinct primes
static boolean isProduct(int num)
{
    int cnt = 0;
 
    for (int i = 2; cnt < 2 && i * i <= num; ++i) {
        while (num % i == 0) {
            num /= i;
            ++cnt;
        }
    }
 
    if (num > 1)
        ++cnt;
 
    return cnt == 2;
}
 
// Function to find numbers that are product
// of exactly two distinct prime numbers.
static void findNumbers(int N)
{
    // Vector to store such numbers
    Vector vec = new Vector();
 
    for (int i = 1; i <= N; i++) {
        if (isProduct(i) && !isPerfectSquare(i)) {
 
            // insert in the vector
            vec.add(i);
        }
    }
 
    // Print all numers till n from the vector
    Iterator itr = vec.iterator(); 
            while(itr.hasNext()){ 
                 System.out.print(itr.next()+" "); 
            } 
}
 
// Driver function
public static void main(String[] args)
{
    int N = 30;
 
    findNumbers(N);
}
}
// This Code is Contributed by mits


Python 3
# Python 3 program to find numbers that are product
# of exactly two distinct prime numbers
 
import math
# Function to check whether a number
# is a PerfectSquare or not
def isPerfectSquare(x):
  
    sr = math.sqrt(x)
  
    return ((sr - math.floor(sr)) == 0)
 
# Function to check if a number is a
# product of exactly two distinct primes
def isProduct( num):
    cnt = 0
  
    i = 2
    while cnt < 2 and i * i <= num:
        while (num % i == 0) :
            num //= i
            cnt += 1
        i += 1
  
    if (num > 1):
        cnt += 1
  
    return cnt == 2
  
# Function to find numbers that are product
# of exactly two distinct prime numbers.
def findNumbers(N):
    # Vector to store such numbers
    vec = []
  
    for i in range(1,N+1) :
        if (isProduct(i) and not isPerfectSquare(i)) :
  
            # insert in the vector
            vec.append(i)
  
    # Print all numers till n from the vector
    for i in range(len( vec)):
        print(vec[i] ,end= " ")
  
# Driver function
if __name__=="__main__":
     
    N = 30
    findNumbers(N)


C#
// C# program to find numbers that are product
// of exactly two distinct prime numbers
using System;
using System.Collections.Generic;
 
class GFG
{
    // Function to check whether a number
    // is a PerfectSquare or not
    static bool isPerfectSquare(double x)
    {
 
        double sr = Math.Sqrt(x);
 
        return ((sr - Math.Floor(sr)) == 0);
    }
 
    // Function to check if a number is a
    // product of exactly two distinct primes
    static bool isProduct(int num)
    {
        int cnt = 0;
 
        for (int i = 2; cnt < 2 && i * i <= num; ++i)
        {
            while (num % i == 0)
            {
                num /= i;
                ++cnt;
            }
        }
 
        if (num > 1)
            ++cnt;
 
        return cnt == 2;
    }
 
    // Function to find numbers that are product
    // of exactly two distinct prime numbers.
    static void findNumbers(int N)
    {
        // Vector to store such numbers
        List vec = new List();
 
        for (int i = 1; i <= N; i++)
        {
            if (isProduct(i) && !isPerfectSquare(i))
            {
 
                // insert in the vector
                vec.Add(i);
            }
        }
 
        // Print all numers till n from the vector
        foreach(var a in vec)
                    Console.Write(a + " ");
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int N = 30;
 
        findNumbers(N);
    }
}
 
// This code has been contributed by 29AjayKumar


PHP
 1)
        ++$cnt;
 
    return $cnt == 2;
}
 
// Function to find numbers that are product
// of exactly two distinct prime numbers.
function findNumbers($N)
{
    // Vector to store such numbers
    $vec = array();
 
    for ($i = 1; $i <= $N; $i++)
    {
        if (isProduct($i) &&
           !isPerfectSquare($i))
        {
 
            // insert in the vector
            array_push($vec, $i);
        }
    }
 
    // Print all numers till n from the vector
    for ($i = 0; $i < sizeof($vec); $i++)
    {
        echo $vec[$i] . " ";
    }
}
 
// Driver Code
$N = 30;
 
findNumbers($N);
 
// This code is contributed by ita_c


Javascript


输出:
6 10 14 15 21 22 26

时间复杂度: O( n  * $\sqrt{n}$  )