最小的数字可被前n个数字整除

给定一个数字n，找到可被1到n均等除的最小数字。
例子：

Input : n = 4
Output : 12
Explanation : 12 is the smallest numbers divisible
              by all numbers from 1 to 4

Input : n = 10
Output : 2520

Input :  n = 20
Output : 232792560

如果仔细观察， ans必须是数字1到n的LCM 。
要查找从1到n的数字的LCM –

1. 初始化ans = 1。
2. 迭代从i = 1到i = n的所有数字。
   在第i次迭代中， ans = LCM(1，2，……..，i) 。这很容易做到，因为LCM(1,2，…。，i)= LCM(ans，i) 。
   因此，在第i次迭代中，我们只需要做–

ans = LCM(ans, i) 
         = ans * i / gcd(ans, i) [Using the below property,
                                 a*b = gcd(a,b) * lcm(a,b)]

注意：在C++代码中，答案很快超过了整数限制，甚至超过long long限制。
下面是逻辑的实现。

C++

// C++ program to find smallest number evenly divisible by
// all numbers 1 to n
#include
using namespace std;
 
// Function returns the lcm of first n numbers
long long lcm(long long n)
{
    long long ans = 1;   
    for (long long i = 1; i <= n; i++)
        ans = (ans * i)/(__gcd(ans, i));
    return ans;
}
 
// Driver program to test the above function
int main()
{
    long long n = 20;
    cout << lcm(n);
    return 0;
}

Java

// Java program to find the smallest number evenly divisible by
// all numbers 1 to n
 
 class GFG{
 
static long gcd(long a, long b)
{
   if(a%b != 0)
      return gcd(b,a%b);
   else
      return b;
}
 
// Function returns the lcm of first n numbers
static long lcm(long n)
{
    long ans = 1;   
    for (long i = 1; i <= n; i++)
        ans = (ans * i)/(gcd(ans, i));
    return ans;
}
  
// Driver program to test the above function
public static void main(String []args)
{
    long n = 20;
    System.out.println(lcm(n));
 
}
}

C#

// C#  program to find smallest number
// evenly divisible by
// all numbers 1 to n
using System;
 
public class GFG{
    static long gcd(long a, long b)
{
if(a%b != 0)
    return gcd(b,a%b);
else
    return b;
}
 
// Function returns the lcm of first n numbers
static long lcm(long n)
{
    long ans = 1;    
    for (long i = 1; i <= n; i++)
        ans = (ans * i)/(gcd(ans, i));
    return ans;
}
 
// Driver program to test the above function
    static public void Main (){
        long n = 20;
        Console.WriteLine(lcm(n));
    }
//This code is contributed by akt_mit   
}

Python3

# Python program to find the smallest number evenly 
# divisible by all number 1 to n
import math
   
# Returns the lcm of first n numbers
def lcm(n):
    ans = 1
    for i in range(1, n + 1):
        ans = int((ans * i)/math.gcd(ans, i))        
    return ans
   
# main
n = 20
print (lcm(n))

Javascript

 
// Javascript program to find the smallest number evenly divisible by
// all numbers 1 to n
 
function gcd(a, b)
{
   if(a%b != 0)
      return gcd(b,a%b);
   else
      return b;
}
   
// Function returns the lcm of first n numbers
function lcm(n)
{
    let ans = 1;   
    for (let i = 1; i <= n; i++)
        ans = (ans * i)/(gcd(ans, i));
    return ans;
}
       
// function call
     
    let n = 20;
    document.write(lcm(n));
     

输出 ：

232792560