阶乘及其邻域的 LCM
给定一个数，我们需要找到数及其邻域的阶乘的 LCM。如果数是 N，我们需要求(N-1)的 LCM！，N！和(N+1)！。这里 N 总是大于等于太 1 T5】例:
Input : N = 5
Output : 720
Explanation
Here the given number is 5, its neighbors
are 4 and 6\. The factorial of these three
numbers are 24, 120, and 720.so the LCM
of 24, 120, 720 is 720.
Input : N = 3
Output : 24
Explanation
Here the given number is 3, its Neighbors
are 2 and 4.the factorial of these three
numbers are 2, 6, and 24\. So the LCM of
2, 6 and 24 is 24.
方法 1(简单)。我们首先计算数和的阶乘及其相邻数的阶乘，然后 求这些阶乘数的 LCM。 方法 2(高效) 我们可以看到(N-1)的 LCM！，N！和(N+1)！永远是(N-1)！ N！ (N+1)！ 这个可以写成(N-1)！ N(N-1)！ (N+1)N(N-1)！ 所以 LCM 变成(N-1)！ N * (N+1) 也就是(N+1)！ 例 N = 5 我们需要找到 4 的 LCM！, 5!和 6！ LCM 4 的！, 5!和 6！ = 4！ 5! 6! = 4！ 54! 654! = 65*4！ = 720 所以我们可以说三个连续数的阶乘的 LCM 永远是最大数的阶乘，这种情况下(N+1)！。
C++
// CPP program to calculate the LCM of N!
// and its neighbor (N-1)! and (N+1)!
#include <bits/stdc++.h>
using namespace std;
// function to calculate the factorial
unsigned int factorial(unsigned int n)
{
if (n == 0)
return 1;
return n * factorial(n - 1);
}
int LCMOfNeighbourFact(int n)
{
// returning the factorial of the
// largest number in the given three
// consecutive numbers
return factorial(n + 1);
}
// Driver code
int main()
{
int N = 5;
cout << LCMOfNeighbourFact(N) << "\n";
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to calculate the LCM of N!
// and its neighbor (N-1)! and (N+1)!
import java.io.*;
class GFG {
// function to calculate the factorial
static int factorial(int n)
{
if (n == 0)
return 1;
return n * factorial(n - 1);
}
static int LCMOfNeighbourFact(int n)
{
// returning the factorial of the
// largest number in the given three
// consecutive numbers
return factorial(n + 1);
}
// Driver code
public static void main(String args[])
{
int N = 5;
System.out.println(LCMOfNeighbourFact(N));
}
}
/*This code is contributed by Nikita Tiwari.*/
Python 3
# Python3 program to calculate the LCM of N!
# and its neighbor (N-1)! and (N+1)!
# Function to calculate the factorial
def factorial(n):
if (n == 0):
return 1
return n * factorial(n - 1)
def LCMOfNeighbourFact(n):
# returning the factorial of the
# largest number in the given three
# consecutive numbers
return factorial(n + 1)
# Driver code
N = 5
print(LCMOfNeighbourFact(N))
# This code is contributed by Anant Agarwal.
C
// Program to calculate the LCM
// of N! and its neighbor (N-1)!
// and (N+1)!
using System;
class GFG
{
// function to calculate the factorial
static int factorial(int n) {
if (n == 0)
return 1;
return n * factorial(n - 1);
}
static int LCMOfNeighbourFact(int n) {
// returning the factorial of the
// largest number in the given three
// consecutive numbers
return factorial(n + 1);
}
// Driver code
public static void Main()
{
int N = 5;
Console.WriteLine(LCMOfNeighbourFact(N));
}
}
// This code is contributed by Anant Agarwal.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to calculate
// the LCM of N! and its neighbor
// (N-1)! and (N+1)!
// function to calculate
// the factorial
function factorial($n)
{
if ($n == 0)
return 1;
return $n * factorial($n - 1);
}
function LCMOfNeighbourFact($n)
{
// returning the factorial
// of the largest number in
// the given three
// consecutive numbers
return factorial($n + 1);
}
// Driver code
$N = 5;
echo(LCMOfNeighbourFact($N));
// This code is contributed by Ajit.
?>
java 描述语言
<script>
// javascript program to calculate the LCM of N!
// and its neighbor (N-1)! and (N+1)!
// function to calculate the factorial
function factorial(n)
{
if (n == 0)
return 1;
return n * factorial(n - 1);
}
function LCMOfNeighbourFact(n)
{
// returning the factorial of the
// largest number in the given three
// consecutive numbers
return factorial(n + 1);
}
// Driver code
var N = 5;
document.write(LCMOfNeighbourFact(N));
// This code is contributed by aashish1995
</script>
Output:
720
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