使用等价关系中的给定操作，使用范围 [2, N] 中的整数可能的集合计数

给定一个整数N ，从2到N范围内重复选择两个不同的整数，如果发现它们的 GCD 大于 1，则将它们尽可能长地插入同一个集合中。在等价关系中形成的集合。因此，如果整数a和b在同一个集合中，并且整数b和c在同一个集合中，那么整数a和c也被称为在同一个群中。任务是找到可以形成的此类集合的总数。

例子：

Input: N = 3 
Output: 2
Explanation: Sets formed are: {2}, {3}. They cannot be put in the same set because there GCD is 1.

Input: N = 9
Output : 3
Sets formed are : {2, 3, 4, 6, 8, 9}, {5}, {7}
As {2, 4, 6, 8} lies in same set and {3, 6, 9} also lies in same set. Hence, all these lie in one set together, because 6 is the common element in both the sets.

方法：解决问题的想法是基于以下观察，即所有小于或等于N/2的数都属于同一个集合，因为如果在其中乘以 2，它们将是偶数并且 GCD 大于1与2 。所以剩下的集合是由大于 N/2 的数组成并且是素数，因为如果它们不是素数，那么有一个小于或等于 N/2 的数是该数的除数。使用埃拉托色尼筛法可以找到从 2 到 N 的素数。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
bool prime[100001];
 
// Sieve of Eratosthenes to find
// primes less than or equal to N
void SieveOfEratosthenes(int n)
{
 
    memset(prime, true, sizeof(prime));
 
    for (int p = 2; p * p <= n; p++) {
 
        if (prime[p] == true) {
            for (int i = p * p; i <= n; i += p)
                prime[i] = false;
        }
    }
}
 
// Function to find number of Sets
void NumberofSets(int N)
{
    SieveOfEratosthenes(N);
 
    // Handle Base Case
    if (N == 2) {
        cout << 1 << endl;
    }
    else if (N == 3) {
        cout << 2 << endl;
    }
    else {
 
        // Set which contains less
        // than or equal to N/2
        int ans = 1;
 
        // Number greater than N/2 and
        // are prime increment it by 1
        for (int i = N / 2 + 1; i <= N; i++) {
 
            // If the number is prime
            // Increment answer by 1
            if (prime[i]) {
                ans += 1;
            }
        }
 
        cout << ans << endl;
    }
}
// Driver Code
int main()
{
    // Input
    int N = 9;
 
    // Function Call
    NumberofSets(N);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
     
static boolean prime[] = new boolean[100001];
 
// Sieve of Eratosthenes to find
// primes less than or equal to N
static void SieveOfEratosthenes(int n)
{
    Arrays.fill(prime, true);
 
    for(int p = 2; p * p <= n; p++)
    {
        if (prime[p] == true)
        {
            for(int i = p * p; i <= n; i += p)
                prime[i] = false;
        }
    }
}
 
// Function to find number of Sets
static void NumberofSets(int N)
{
    SieveOfEratosthenes(N);
 
    // Handle Base Case
    if (N == 2)
    {
        System.out.print(1);
    }
    else if (N == 3)
    {
        System.out.print(2);
    }
    else
    {
         
        // Set which contains less
        // than or equal to N/2
        int ans = 1;
 
        // Number greater than N/2 and
        // are prime increment it by 1
        for(int i = N / 2 + 1; i <= N; i++)
        {
             
            // If the number is prime
            // Increment answer by 1
            if (prime[i])
            {
                ans += 1;
            }
        }
        System.out.print(ans);
    }
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Input
    int N = 9;
 
    // Function Call
    NumberofSets(N);
}   
}
 
// This code is contributed by code_hunt

Python3

# Python3 program for the above approach
prime = [True] * 100001
 
# Sieve of Eratosthenes to find
# primes less than or equal to N
def SieveOfEratosthenes(n):
     
    global prime
     
    for p in range(2, n + 1):
        if p * p > n:
            break
 
        if (prime[p] == True):
            for i in range(p * p, n + 1, p):
                prime[i] = False
 
# Function to find number of Sets
def NumberofSets(N):
     
    SieveOfEratosthenes(N)
 
    # Handle Base Case
    if (N == 2):
        print(1)
    elif (N == 3):
        print(2)
    else:
         
        # Set which contains less
        # than or equal to N/2
        ans = 1
 
        # Number greater than N/2 and
        # are prime increment it by 1
        for i in range(N // 2, N + 1):
             
            # If the number is prime
            # Increment answer by 1
            if (prime[i]):
                ans += 1
 
        print(ans)
 
# Driver Code
if __name__ == '__main__':
     
    # Input
    N = 9
 
    # Function Call
    NumberofSets(N)
 
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
static bool []prime = new bool[100001];
 
// Sieve of Eratosthenes to find
// primes less than or equal to N
static void SieveOfEratosthenes(int n)
{
     
    for(int i=0;i<100001;i++)
        prime[i] = true;
 
    for (int p = 2; p * p <= n; p++) {
 
        if (prime[p] == true) {
            for (int i = p * p; i <= n; i += p)
                prime[i] = false;
        }
    }
}
 
// Function to find number of Sets
static void NumberofSets(int N)
{
    SieveOfEratosthenes(N);
 
    // Handle Base Case
    if (N == 2) {
        Console.Write(1);
    }
    else if (N == 3) {
        Console.Write(2);
    }
    else {
 
        // Set which contains less
        // than or equal to N/2
        int ans = 1;
 
        // Number greater than N/2 and
        // are prime increment it by 1
        for (int i = N / 2 + 1; i <= N; i++) {
 
            // If the number is prime
            // Increment answer by 1
            if (prime[i]) {
                ans += 1;
            }
        }
 
        Console.Write(ans);
    }
}
   
// Driver Code
public static void Main()
{
    // Input
    int N = 9;
 
    // Function Call
    NumberofSets(N);
}
}
 
// This code is contributed by SURENDRA_GANGWAR.

Javascript

// Javascript program for the above approach
 
let prime = new Array(100001);
 
// Sieve of Eratosthenes to find
// primes less than or equal to N
function SieveOfEratosthenes(n)
{
    for(let i=0;i<prime.length;i++)
    {
        prime[i]=true;
    }
  
    for(let p = 2; p * p <= n; p++)
    {
        if (prime[p] == true)
        {
            for(let i = p * p; i <= n; i += p)
                prime[i] = false;
        }
    }
}
 
// Function to find number of Sets
function NumberofSets(N)
{
    SieveOfEratosthenes(N);
  
    // Handle Base Case
    if (N == 2)
    {
        document.write(1);
    }
    else if (N == 3)
    {
        document.write(2);
    }
    else
    {
          
        // Set which contains less
        // than or equal to N/2
        let ans = 1;
  
        // Number greater than N/2 and
        // are prime increment it by 1
        for(let i = Math.floor(N / 2) + 1; i <= N; i++)
        {
              
            // If the number is prime
            // Increment answer by 1
            if (prime[i])
            {
                ans += 1;
            }
        }
        document.write(ans);
    }
}
 
// Driver Code
// Input
let N = 9;
 
// Function Call
NumberofSets(N);
 
// This code is contributed by unknown2108

输出：

3

时间复杂度： O(N)
辅助空间： O(K)