最多为 N 且至少有一个与 N 共有的素数的数的计数

给定一个整数N ，任务是计算范围[1, N]中至少有一个与N不是1的公质因数的整数个数。

例子：

Input: N = 5
Output: 1
Explanation: 
Since 5 is prime. Therefore, there is no other number which is at most N, that has at least one common factor with N except N itself. Therefore, the count is 1.

Input: N = 12
Output: 8

朴素方法：解决给定问题的最简单方法是遍历[1, N]范围内的所有数字，并且对于每个元素，如果每个数字的 GCD 与N大于1 ，则增加计数。否则，检查下一个数字。检查所有数字后，我们打印获得的总计数。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to count all the numbers
// in the range [1, N] having common
// factor with N other than 1
int countNumbers(int N)
{
    // Stores the count of numbers
    // having more than 1 factor with N
    int count = 0;
 
    // Iterate over the range [1, N]
    for (int i = 1; i <= N; i++) {
 
        // If gcd is not 1 then
        // increment the count
        if (__gcd(i, N) != 1)
            count++;
    }
 
    // Print the resultant count
    cout << count;
}
 
// Driver Code
int main()
{
    int N = 5;
    countNumbers(N);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to count all the numbers
// in the range [1, N] having common
// factor with N other than 1
static void countNumbers(int N)
{
     
    // Stores the count of numbers
    // having more than 1 factor with N
    int count = 0;
 
    // Iterate over the range [1, N]
    for(int i = 1; i <= N; i++)
    {
         
        // If gcd is not 1 then
        // increment the count
        if (__gcd(i, N) != 1)
            count++;
    }
 
    // Print the resultant count
    System.out.print(count);
}
 
static int __gcd(int a, int b)
{
    return b == 0 ? a : __gcd(b, a % b);
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 5;
    countNumbers(N);
}
}
 
// This code is contributed by shikhasingrajput

Python3

# Python 3 program for the above approach
import math
 
# Function to count all the numbers
# in the range [1, N] having common
# factor with N other than 1
def countNumbers(N):
 
    # Stores the count of numbers
    # having more than 1 factor with N
    count = 0
 
    # Iterate over the range [1, N]
    for i in range(1, N + 1):
 
        # If gcd is not 1 then
        # increment the count
        if (math.gcd(i, N) != 1):
            count += 1
 
    # Print the resultant count
    print(count)
 
 
# Driver Code
if __name__ == "__main__":
 
    N = 5
    countNumbers(N)
 
    # This code is contributed by ukasp.

C#

// C# program for the above approach
using System;
 
public class GFG{
 
// Function to count all the numbers
// in the range [1, N] having common
// factor with N other than 1
static void countNumbers(int N)
{
     
    // Stores the count of numbers
    // having more than 1 factor with N
    int count = 0;
 
    // Iterate over the range [1, N]
    for(int i = 1; i <= N; i++)
    {
         
        // If gcd is not 1 then
        // increment the count
        if (__gcd(i, N) != 1)
            count++;
    }
 
    // Print the resultant count
    Console.Write(count);
}
 
static int __gcd(int a, int b)
{
    return b == 0 ? a : __gcd(b, a % b);
}
 
// Driver Code
public static void Main(String[] args)
{
    int N = 5;
    countNumbers(N);
}
}
 
  
 
// This code is contributed by 29AjayKumar

Javascript

// javascript program for the above approach
 
// Function to count all the numbers
// in the range [1, N] having common
// factor with N other than 1
function countNumbers(N)
{
     
    // Stores the count of numbers
    // having more than 1 factor with N
    var count = 0;
 
    // Iterate over the range [1, N]
    for(var i = 1; i <= N; i++)
    {
         
        // If gcd is not 1 then
        // increment the count
        if (__gcd(i, N) != 1)
            count++;
    }
 
    // Print the resultant count
    document.write(count);
}
 
function __gcd(a , b)
{
    return b == 0 ? a : __gcd(b, a % b);
}
 
// Driver Code
var N = 5;
countNumbers(N);
 
// This code is contributed by 29AjayKumar

输出：

1

时间复杂度： O(N*log N)
辅助空间： O(1)

有效方法：上述方法也可以通过使用欧拉的 totient函数进行优化，该函数给出小于N且与N没有共同素因数的数字的计数。请按照以下步骤解决问题：

• 初始化一个变量，比如count ，它存储具有除1以外的共同质因数的数字。
• 定义一个函数phi()来计算欧拉函数的值，该函数表示小于N且与N没有公因数的整数的计数。
• 完成上述步骤后，打印(N – phi(N)的值作为结果计数。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to calculate the value of
// Euler's totient function
int phi(int N)
{
    // Initialize result with N
    int result = N;
 
    // Find all prime factors of N
    // and subtract their multiples
    for (int p = 2; p * p <= N; ++p) {
 
        // Check if p is a prime factor
        if (N % p == 0) {
 
            // If found to be true,
            // then update N and result
            while (N % p == 0)
                N /= p;
 
            result -= result / p;
        }
    }
 
    // If N has a prime factor greater
    // than sqrt(N), then there can be
    // at-most one such prime factor
    if (N > 1)
        result -= result / N;
 
    return result;
}
 
// Function to count all the numbers
// in the range [1, N] having common
// factor with N other than 1
int countNumbers(int N)
{
    // Stores the resultant count
    int count = N - phi(N);
 
    // Print the count
    cout << count;
}
 
// Driver Code
int main()
{
    int N = 5;
    countNumbers(N);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG
{
     
 // Function to calculate the value of
// Euler's totient function
static int phi(int N)
{
   
    // Initialize result with N
    int result = N;
 
    // Find all prime factors of N
    // and subtract their multiples
    for (int p = 2; p * p <= N; ++p)
    {
 
        // Check if p is a prime factor
        if (N % p == 0) {
 
            // If found to be true,
            // then update N and result
            while (N % p == 0)
                N /= p;
 
            result -= result / p;
        }
    }
 
    // If N has a prime factor greater
    // than sqrt(N), then there can be
    // at-most one such prime factor
    if (N > 1)
        result -= result / N;
 
    return result;
}
 
// Function to count all the numbers
// in the range [1, N] having common
// factor with N other than 1
static void countNumbers(int N)
{
   
    // Stores the resultant count
    int count = N - phi(N);
 
    // Print the count
   System.out.print(count);
}
 
  // Driver code
public static void main (String[] args)
{
     int N = 5;
    countNumbers(N);
}
}
 
// This code is contributed by offbeat.

Python3

# Python3 program for the above approach
 
# Function to calculate the value of
# Euler's totient function
def phi(N):
     
    # Initialize result with N
    result = N
     
    # Find all prime factors of N
    # and subtract their multiples
    for p in range(2, int(pow(N, 1 / 2)) + 1):
         
        # Check if p is a prime factor
        if (N % p == 0):
             
            # If found to be true,
            # then update N and result
            while (N % p == 0):
                N = N / p
 
            result -= result // p
         
    # If N has a prime factor greater
    # than sqrt(N), then there can be
    # at-most one such prime factor
    if (N > 1):
        result -= result // N
 
    return result
 
# Function to count all the numbers
# in the range [1, N] having common
# factor with N other than 1
def countNumbers(N):
     
    # Stores the resultant count
    count = N - phi(N)
 
    # Print the count
    print(count)
 
# Driver Code
N = 5
 
countNumbers(N)
 
# This code is contributed by SoumikMondal

C＃

// C# program for the above approach
using System;
 
public class GFG
{
     
 // Function to calculate the value of
// Euler's totient function
static int phi(int N)
{
   
    // Initialize result with N
    int result = N;
 
    // Find all prime factors of N
    // and subtract their multiples
    for (int p = 2; p * p <= N; ++p)
    {
 
        // Check if p is a prime factor
        if (N % p == 0) {
 
            // If found to be true,
            // then update N and result
            while (N % p == 0)
                N /= p;
 
            result -= result / p;
        }
    }
 
    // If N has a prime factor greater
    // than sqrt(N), then there can be
    // at-most one such prime factor
    if (N > 1)
        result -= result / N;
 
    return result;
}
 
// Function to count all the numbers
// in the range [1, N] having common
// factor with N other than 1
static void countNumbers(int N)
{
   
    // Stores the resultant count
    int count = N - phi(N);
 
    // Print the count
   Console.Write(count);
}
 
  // Driver code
public static void Main(String[] args)
{
     int N = 5;
    countNumbers(N);
}
}
 
// This code contributed by shikhasingrajput

Javascript

 
// Javascript implementation
// for the above approach
 
// Function to calculate the value of
// Euler's totient function
function phi(N)
{
    
    // Initialize result with N
    let result = N;
  
    // Find all prime factors of N
    // and subtract their multiples
    for (let p = 2; p * p <= N; ++p)
    {
  
        // Check if p is a prime factor
        if (N % p == 0) {
  
            // If found to be true,
            // then update N and result
            while (N % p == 0)
                N /= p;
  
            result -= result / p;
        }
    }
  
    // If N has a prime factor greater
    // than sqrt(N), then there can be
    // at-most one such prime factor
    if (N > 1)
        result -= result / N;
  
    return result;
}
  
// Function to count all the numbers
// in the range [1, N] having common
// factor with N other than 1
function countNumbers(N)
{
    
    // Stores the resultant count
    let count = N - phi(N);
  
    // Print the count
       document.write(count);
}
 
// Driver Code
     
    let N = 5;
    countNumbers(N);
 
// This code is contributed by susmitakundugoaldanga.

输出：

1

时间复杂度： O(N ^1/2 )
辅助空间： O(1)