生成n个连续复合数字的列表(一种有趣的方法)

给定数字n，生成n个复合数字的列表。
例子：

Input : 5
Output : 122, 123, 124, 125

Input : 10
Output : 3628802, 3628803, 3628804, 3628805, 3628806, 
         3628807, 3628808, 3628809, 3628810

这里的想法是使用$n!$ 。自从$n! = 1\times2\times3....(n-1)\times n$ ，然后是数字$1, 2, 3..... (n-1), n$ ，所有分裂$n!$ 。所以$n!+2$可被2整除$n!+3$可被3整除….. $n!+n$可被n整除。按照上述模式，它们是连续的复合材料。
我们找到(n + 1)!，然后打印数字(n + 1)！ + 2，(n + 1)！ + 3，…。 (n + 1)！ +(n + 1)。
下面是上述方法的实现：

C++

// CPP program to print n consecutive composite
// numbers.
#include 
using namespace std;
 
// function to find factorial of given
// number
unsigned long long int factorial(unsigned int n)
{   
    unsigned long long int res = 1;
    for (int i=2; i<=n; i++)
        res *= i;
    return res;
}
 
// Prints n consecutive numbers.
void printNComposite(int n)
{
    unsigned long long int fact = factorial(n+1);
    for (int i = 2; i <= n+1; ++i)
        cout << fact + i << " ";
}
 
// Driver program to test above function
int main()
{
    int n = 4;
    printNComposite(n);
    return 0;
}

Java

// Java program to print n consecutive composite
// numbers
 
class GFG {
 
// function to find factorial of given
// number
    static long factorial(int n) {
        long res = 1;
        for (int i = 2; i <= n; i++) {
            res *= i;
        }
        return res;
    }
 
// Prints n consecutive numbers.
    static void printNComposite(int n) {
        long fact = factorial(n + 1);
        for (int i = 2; i <= n + 1; ++i) {
            System.out.print(fact + i + " ");
        }
    }
 
// Driver program to test above function
    public static void main(String[] args) {
        int n = 4;
        printNComposite(n);
 
    }
}

Python3

# Python3 program to print n consecutive
# composite numbers.
 
# function to find factorial
# of given number
def factorial( n):
 
    res = 1;
    for i in range(2, n + 1):
        res *= i;
    return res;
 
# Prints n consecutive numbers.
def printNComposite(n):
    fact = factorial(n + 1);
    for i in range(2, n + 2):
        print(fact + i, end = " ");
 
# Driver Code
n = 4;
printNComposite(n);
     
# This code is contributed by mits

C#

// C# program to print n consecutive composite
// numbers
using System;
                     
public class Program{
  
// function to find factorial of given
// number
    static long factorial(int n) {
        long res = 1;
        for (int i = 2; i <= n; i++) {
            res *= i;
        }
        return res;
    }
  
// Prints n consecutive numbers.
    static void printNComposite(int n) {
        long fact = factorial(n + 1);
        for (int i = 2; i <= n + 1; ++i) {
            Console.Write(fact + i + " ");
        }
    }
  
// Driver program to test above function
    public static void Main() {
        int n = 4;
        printNComposite(n);
  
    }
}
 
// This code is contributed by Rajput-Ji

Javascript

 
// JavaScript program to print n consecutive composite
// numbers
 
// function to find factorial of given
// number
    function factorial(n) {
        let res = 1;
        for (let i = 2; i <= n; i++) {
            res *= i;
        }
        return res;
    }
   
// Prints n consecutive numbers.
    function printNComposite(n) {
        let fact = factorial(n + 1);
        for (let i = 2; i <= n + 1; ++i) {
            document.write(fact + i + " ");
        }
    }
 
// Driver code
     
        let n = 4;
        printNComposite(n);
     
    // This code is contributed by code_hunt.

输出：

122 123 124 125

上面的解决方案很快就会导致溢出(对于较小的n值)。我们可以使用技术来查找大量的阶乘以避免溢出。