查找给定长度的合成数字范围

给定一个整数n，我们需要找到一个正整数范围，以使该范围内的所有数字都是合成的，并且该范围的长度为n。如果答案不只一个，则可以打印任何范围。一个复合数字是一个正整数，它本身至少具有一个除1以外的除数(来源：wiki)

例子 ：

Input : 3
Output : [122, 124]
Explanation 122, 123, 124 are all composite numbers

解决方案有点棘手。由于有许多可能的答案，因此我们在这里讨论一个广义的解决方案。

Let the length of range be n and range starts 
from a then, a, a+1, a+2, ...., a+n-1 all should 
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive 
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be 
composite. So we end up finding p! + 2, p! + 3,
 .... p! + p-1 are all composite and continuous 
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2, 
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2 
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)! 
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)! 
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]

上面算法的例子

n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites

C++

// C++ program to find a range of
// composite numbers of given length
#include 
using namespace std;
 
// method to find factorial
// of given number
int factorial (int n)
{
    if (n == 0)
        return 1;
 
    return n * factorial(n-1);
}
 
// to print range of length n
// having all composite integers
int printRange(int n)
{
int a = factorial(n + 2) + 2;
int b = a + n - 1;
cout << "[" << a << ", " << b << "]";
return 0;
}
 
// Driver method
int main()
{
    int n = 3 ;
    printRange(n);
    return 0;
}
 
// This code is contributd by Anshika Goyal

Java

// Java program to find a range of composite
// numbers of given length
 
class Test
{
    // method to find factorial of given number
    static int factorial(int n)
    {
        if (n == 0)
          return 1;
          
        return n*factorial(n-1);
    }
     
    // to print range of length n
    //  having all composite integers
    static void printRange(int n)
    {
       int  a = factorial(n + 2) + 2;
       int  b = a + n - 1;
       System.out.println("[" + a + ", " + b + "]");
    }
     
    // Driver method
    public static void main(String args[]) throws Exception
    {
        int n = 3 ;
        printRange(n);
    }
}

Python

# Python program to find a range of composite
# numbers of given length
 
# function to calculate factorial
def factorial(n):
    a = 1
    for i in range(2, n + 1):
        a *= i
    return a
 
# to print range of length n
# having all composite integers
def printRange(n):
    a = factorial(n + 2) + 2
    b = a + n - 1
    print("["+str(a)+", "+str(b)+"]")
 
# driver code to test above functions
n = 3
printRange(n)

C#

// C# program to find a range of
// composite numbers of given
// length
using System;
 
class GFG {
     
    // method to find factorial
    // of given number
    static int factorial(int n)
    {
        if (n == 0)
        return 1;
         
        return n*factorial(n-1);
    }
     
    // to print range of length n
    // having all composite integers
    static void printRange(int n)
    {
    int a = factorial(n + 2) + 2;
    int b = a + n - 1;
    Console.WriteLine("[" + a +
                   ", " + b + "]");
    }
     
    // Driver method
    public static void Main()
    {
        int n = 3 ;
        printRange(n);
    }
}
 
// This code is contributed by anuj_67.

Javascript

     
// Javascript program to find a range of
// composite numbers of given length
 
// Method to find factorial
// of given number
function factorial(n)
{
    if (n == 0)
        return 1;
 
    return n * factorial(n - 1);
}
 
// To print range of length n
// having all composite integers
function printRange(n)
{
    let a = factorial(n + 2) + 2;
    let b = a + n - 1;
       
    document.write(`[${a}, ${b}]`);
       
    return 0;
}
 
// Driver Code
let n = 3;
 
printRange(n);
 
// This code is contributd by _saurabh_jaiswal.
 

输出 ：

[122, 124]

以上算法分析
时间复杂度：O(n)
辅助空间：O(1)