计算具有 GCD 且 N 等于数字本身的数字

给定一个正整数N ，任务是找到正整数的数量，其 GCD 与给定的整数N是数字本身。

例子：

Input: N = 5
Output: 2
Explanation:
Following are the numbers whose GCD with N is the number itself:

1. Number 1: GCD(1, 5) = 1.
2. Number 1: GCD(5, 5) = 5.

Therefore, the total count is 2.

Input: N = 10
Output: 4

方法：给定的问题可以基于以下观察来解决：任何数字（比如K ）的 GCD 与N的必要条件是K当且仅当K是N 的一个因子。因此，想法是找到N的因子数。请按照以下步骤解决问题：

• 初始化一个变量，比如count为0 ，以计算N的因子数。
• 迭代范围[1, sqrt(N)]并执行以下步骤：
  • 如果当前数字i除以给定的整数N ，则将count增加1 。
  • 如果i和N / i的值不同，则将count增加1 。
• 完成以上步骤后，打印count的值作为结果。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to count numbers whose
// GCD with N is the number itself
int countNumbers(int N)
{
    // Stores the count of factors of N
    int count = 0;
 
    // Iterate over the range [1, sqrt(N)]
    for (int i = 1; i * i <= N; i++) {
 
        // If i is divisible by i
        if (N % i == 0) {
 
            // Increment count
            count++;
 
            // Avoid counting the
            // same factor twice
            if (N / i != i) {
                count++;
            }
        }
    }
 
    // Return the resultant count
    return count;
}
 
// Driver Code
int main()
{
    int N = 10;
    cout << countNumbers(N);
 
    return 0;
}

Java

// java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
public class GFG {
 
    // Function to count numbers whose
    // GCD with N is the number itself
    static int countNumbers(int N)
    {
        // Stores the count of factors of N
        int count = 0;
 
        // Iterate over the range [1, sqrt(N)]
        for (int i = 1; i * i <= N; i++) {
 
            // If i is divisible by i
            if (N % i == 0) {
 
                // Increment count
                count++;
 
                // Avoid counting the
                // same factor twice
                if (N / i != i) {
                    count++;
                }
            }
        }
 
        // Return the resultant count
        return count;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int N = 10;
        System.out.println(countNumbers(N));
    }
}
 
// This code is contributed by Kingash.

Python3

# Python3 program for the above approach
 
# Function to count numbers whose
# GCD with N is the number itself
def countNumbers(N):
     
    # Stores the count of factors of N
    count = 0
     
    # Iterate over the range [1, sqrt(N)]
    for i in range(1, N + 1):
        if i * i > N:
            break
 
        # If i is divisible by i
        if (N % i == 0):
 
            # Increment count
            count += 1
             
            # Avoid counting the
            # same factor twice
            if (N // i != i):
                count += 1
 
    # Return the resultant count
    return count
 
# Driver Code
if __name__ == '__main__':
     
    N = 10
    print(countNumbers(N))
 
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
public class GFG {
 
    // Function to count numbers whose
    // GCD with N is the number itself
    static int countNumbers(int N)
    {
        // Stores the count of factors of N
        int count = 0;
 
        // Iterate over the range [1, sqrt(N)]
        for (int i = 1; i * i <= N; i++) {
 
            // If i is divisible by i
            if (N % i == 0) {
 
                // Increment count
                count++;
 
                // Avoid counting the
                // same factor twice
                if (N / i != i) {
                    count++;
                }
            }
        }
 
        // Return the resultant count
        return count;
    }
 
    // Driver Code
    public static void Main(string[] args)
    {
        int N = 10;
        Console.WriteLine(countNumbers(N));
    }
}
 
// This code is contributed by ukasp.

Javascript

 
// Javascript program for the above approach
 
 
// Function to count numbers whose
// GCD with N is the number itself
function countNumbers(N)
{
    // Stores the count of factors of N
    var count = 0;
 
    // Iterate over the range [1, sqrt(N)]
    for (i = 1; i * i <= N; i++) {
 
        // If i is divisible by i
        if (N % i == 0) {
 
            // Increment count
            count++;
 
            // Avoid counting the
            // same factor twice
            if (parseInt(N / i) != i) {
                count++;
            }
        }
    }
 
    // Return the resultant count
    return count;
}
 
// Driver Code
var N = 10;
document.write(countNumbers(N));
 
// This code is contributed by Amit Katiyar
 

输出：

4

时间复杂度： O(N ^1/2 )
辅助空间： O(1)