通过lcm(X，N)/ X获得的不同整数的数量

给定一个数字$N$ ，找到由lcm(X，N)/ X获得的不同整数的数量，其中X可以是任何正数。

例子：

Input: N = 2  
Output: 2
if X is 1, then lcm(1, 2)/1 is 2/1=2. 
if X is 2, then lcm(2, 2)/2 is 2/2=1. 
For any X greater than 2 we cannot 
obtain a distinct integer.
  
Input: N = 3
Output: 2 

已知lcm(x，y)= x * y / gcd(x，y) 。

所以，

lcm(X, N) = X*N/gcd(X, N)
or, lcm(X, N)/X = N/gcd(X, N)

因此，只有$N$可以是可能的不同整数。因此，计算包括1和N本身在内的N的不同因子的数量，这是必需的答案。

下面是上述方法的实现：

C++

// C++ program to find distinct integers
// ontained by lcm(x, num)/x
#include 
#include 
  
using namespace std;
  
// Function to count the number of distinct
// integers ontained by lcm(x, num)/x
int numberOfDistinct(int n)
{
    int ans = 0;
  
    // iterate to count the number of factors
    for (int i = 1; i <= sqrt(n); i++) {
        if (n % i == 0) {
            ans++;
            if ((n / i) != i)
                ans++;
        }
    }
  
    return ans;
}
  
// Driver Code
int main()
{
    int n = 3;
  
    cout << numberOfDistinct(n);
  
    return 0;
}

Java

// Java  program to find distinct integers 
// ontained by lcm(x, num)/x 
  
import java.io.*;
  
class GFG {
      
// Function to count the number of distinct 
// integers ontained by lcm(x, num)/x 
static int numberOfDistinct(int n) 
{ 
    int ans = 0; 
  
    // iterate to count the number of factors 
    for (int i = 1; i <= Math.sqrt(n); i++) { 
        if (n % i == 0) { 
            ans++; 
            if ((n / i) != i) 
                ans++; 
        } 
    } 
  
    return ans; 
} 
  
// Driver Code 
    public static void main (String[] args) {
        int n = 3; 
  
        System.out.println (numberOfDistinct(n)); 
  
  
    }
}

Python 3

# Python 3 program to find distinct integers
# ontained by lcm(x, num)/x
  
import math
  
# Function to count the number of distinct
# integers ontained by lcm(x, num)/x
def numberOfDistinct(n):
    ans = 0
  
    # iterate to count the number of factors
    for i in range( 1, int(math.sqrt(n))+1):
        if (n % i == 0) :
            ans += 1
            if ((n // i) != i):
                ans += 1
    return ans
  
# Driver Code
if __name__ == "__main__":
    n = 3
  
    print(numberOfDistinct(n))
  
# This code is contributed by
# ChitraNayal

C#

// C# program to find distinct integers 
// ontained by lcm(x, num)/x 
using System;
  
class GFG
{
      
// Function to count the number 
// of distinct integers ontained
// by lcm(x, num)/x 
static int numberOfDistinct(int n) 
{ 
    int ans = 0; 
  
    // iterate to count the number
    // of factors 
    for (int i = 1; i <= Math.Sqrt(n); i++) 
    { 
        if (n % i == 0)
        { 
            ans++; 
            if ((n / i) != i) 
                ans++; 
        } 
    } 
  
    return ans; 
} 
  
// Driver Code
static public void Main ()
{
    int n = 3; 
    Console.WriteLine(numberOfDistinct(n)); 
}
}
  
// This code is contributed by ajit

输出：

2

时间复杂度：O(sqrt(n))