最小的数只有 4 个除数，最多有两个差 D

给定数字D ，找到最小的数字N ，使其恰好有四个除数，并且其中任何两个除数之间的差大于或等于D 。

例子：

Input: 1
Output: 6
Explanation: 6 has four divisors 1, 2, 3, and 6. 
Difference between any two of them is always greater or equal to 1.

Input: 2
Output: 15
Explanation: 15 has four divisors 1, 3, 5 and 15. 
Difference between any two of them is always greater or equal to 2

Input: 3
Output: 55
Explanation: 55 has four divisors 1, 5, 11 and 55. 
Difference between any two of them is always greater than 3.

方法：很明显 '1' 和数字本身是N的除数。所以恰好有 4 个除数的数的除数分别为1、a、b、a*b 。对于它恰好有 4 个除数的条件， a和b都必须是素数。对于任意两者之差至少为D的条件，从1+d开始寻找a并检查它是否为素数，如果不是素数，我们将找到它旁边的素数。同样，从a + d开始查找b并检查它是否为素数，并执行与查找a相同的操作。

下面是上述方法的实现。

C++

// C++ program for the above approach
#include 
using namespace std;
 
int next_prime(int x)
{
    // Edge case
    if (x == 1 || x == 2) {
        return 2;
    }
 
    // Checking if x is prime
    bool is_prime = false;
 
    // Loop until next prime is found
    while (!is_prime) {
        is_prime = true;
        for (int i = 2; i <= sqrt(x); i++) {
            if (x % i == 0) {
                is_prime = false;
            }
        }
        x++;
    }
    return x - 1;
}
 
// Function to find the number
int findN(int D)
{
    // Assuming 1+D as first required
    // divisor because it is
    // at distance D from 1
    int a = 1 + D;
 
    // Checking whether 1+D is prime or not
    // otherwise it will return prime number
    // just next to it
    a = next_prime(a);
 
    // Checking the same for next divisor
    int b = a + D;
    b = next_prime(b);
 
    int N = a * b;
    return N;
}
 
// Driver Code
int main()
{
    int D = 2;
 
    int ans = findN(D);
    cout << ans;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG {
 
  static int next_prime(int x)
  {
    // Edge case
    if (x == 1 || x == 2) {
      return 2;
    }
 
    // Checking if x is prime
    Boolean is_prime = false;
 
    // Loop until next prime is found
    while (!is_prime) {
      is_prime = true;
      for (int i = 2; i * i <= x; i++) {
        if (x % i == 0) {
          is_prime = false;
        }
      }
      x++;
    }
    return x - 1;
  }
 
  // Function to find the number
  static int findN(int D)
  {
    // Assuming 1+D as first required
    // divisor because it is
    // at distance D from 1
    int a = 1 + D;
 
    // Checking whether 1+D is prime or not
    // otherwise it will return prime number
    // just next to it
    a = next_prime(a);
 
    // Checking the same for next divisor
    int b = a + D;
    b = next_prime(b);
 
    int N = a * b;
    return N;
  }
 
  // Driver Code
  public static void main (String[] args) {
    int D = 2;
 
    int ans = findN(D);
    System.out.println(ans);
  }
}
 
// This code is contributed by hrithikgarg03188.

Python3

# Python code for the above approach
def next_prime(x):
   
    # Edge case
    if (x == 1 or x == 2):
        return 2
 
    # Checking if x is prime
    is_prime = False
 
    # Loop until next prime is found
    while (not is_prime):
        is_prime = True
        for i in range(2, int(x ** 0.5) + 1):
            if (x % i == 0):
                is_prime = False
 
        x += 1
 
    return x - 1
 
# Function to find the number
def findN(D):
 
    # Assuming 1+D as first required
    # divisor because it is
    # at distance D from 1
    a = 1 + D
 
    # Checking whether 1+D is prime or not
    # otherwise it will return prime number
    # just next to it
    a = next_prime(a)
 
    # Checking the same for next divisor
    b = a + D
    b = next_prime(b)
 
    N = a * b
    return N
 
# Driver Code
D = 2
 
ans = findN(D)
print(ans)
 
# This code is contributed by Saurabh Jaiswal

C#

// C# program for the above approach
using System;
class GFG {
 
  static int next_prime(int x)
  {
     
    // Edge case
    if (x == 1 || x == 2) {
      return 2;
    }
 
    // Checking if x is prime
    bool is_prime = false;
 
    // Loop until next prime is found
    while (!is_prime) {
      is_prime = true;
      for (int i = 2; i * i <= x; i++) {
        if (x % i == 0) {
          is_prime = false;
        }
      }
      x++;
    }
    return x - 1;
  }
 
  // Function to find the number
  static int findN(int D)
  {
     
    // Assuming 1+D as first required
    // divisor because it is
    // at distance D from 1
    int a = 1 + D;
 
    // Checking whether 1+D is prime or not
    // otherwise it will return prime number
    // just next to it
    a = next_prime(a);
 
    // Checking the same for next divisor
    int b = a + D;
    b = next_prime(b);
 
    int N = a * b;
    return N;
  }
 
  // Driver Code
  public static void Main () {
    int D = 2;
 
    int ans = findN(D);
    Console.WriteLine(ans);
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

        // JavaScript code for the above approach
        function next_prime(x)
        {
            // Edge case
            if (x == 1 || x == 2) {
                return 2;
            }
 
            // Checking if x is prime
            let is_prime = false;
 
            // Loop until next prime is found
            while (!is_prime) {
                is_prime = true;
                for (let i = 2; i <= Math.sqrt(x); i++) {
                    if (x % i == 0) {
                        is_prime = false;
                    }
                }
                x++;
            }
            return x - 1;
        }
 
        // Function to find the number
        function findN(D)
        {
         
            // Assuming 1+D as first required
            // divisor because it is
            // at distance D from 1
            let a = 1 + D;
 
            // Checking whether 1+D is prime or not
            // otherwise it will return prime number
            // just next to it
            a = next_prime(a);
 
            // Checking the same for next divisor
            let b = a + D;
            b = next_prime(b);
 
            let N = a * b;
            return N;
        }
 
        // Driver Code
        let D = 2;
 
        let ans = findN(D);
        document.write(ans);
 
       // This code is contributed by Potta Lokesh
    

输出

15

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