在给定条件下生成数组的元素

给定一个整数N，对于2到N范围内的每个整数i ，分配一个正整数$a_i$满足以下条件：

• 对于任何一对索引(i，j) ，如果i和j是互质的，则$a_i \neq a_j$ 。
• 全部的最大值$a_i$应该最小化(即最大值应该尽可能小)。

如果一对整数的gcd(i，j)= 1，则称它们为互质。
例子：

Input: N = 4 
Output: 1 2 1 
Input: N = 10 
Output: 1 2 1 3 2 4 1 2 3 
 

方法：在从2到n的索引处分配值时，我们必须牢记两件事。首先，对于任何一对(i，j) ，如果i和j是互质的，则我们不能为这对索引分配相同的值。第二，我们必须最小化分配给每个索引的最大值从2到n 。
如果为每个素数分配了不同的数字，这是可以实现的，因为所有素数都是互质数。然后，在与其所有主要除数相同的所有复合位置上分配值。此解决方案适用于任何对(i，j) ， i被赋予相同数量的除数， j也是如此，因此，如果它们是互质的，则不能被赋予相同的数量。
可以通过在构建Eratosthenes筛时进行一些小的更改来实现此方法。
当我们第一次遇到素数时，请为其分配唯一的最小值及其所有倍数。这样，所有素数索引都应具有唯一值，并且所有综合索引的值均应与其任何素数因子相同。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to generate the required array
void specialSieve(int n)
{
 
    // Initialize cnt variable for assigning
    // unique value to prime and its multiples
    int cnt = 0;
    int prime[n + 1];
 
    for (int i = 0; i <= n; i++)
        prime[i] = 0;
 
    for (int i = 2; i <= n; i++) {
 
        // When we get a prime for the first time
        // then assign a unique smallest value to
        // it and all of its multiples
        if (!prime[i]) {
            cnt++;
            for (int j = i; j <= n; j += i)
                prime[j] = cnt;
        }
    }
 
    // Print the generated array
    for (int i = 2; i <= n; i++)
        cout << prime[i] << " ";
}
 
// Driver code
int main()
{
    int n = 6;
 
    specialSieve(n);
 
    return 0;
}

Java

// Java implementation of the approach
import java.io.*;
 
class GFG
{
 
// Function to generate the required array
static void specialSieve(int n)
{
 
    // Initialize cnt variable for assigning
    // unique value to prime and its multiples
    int cnt = 0;
    int prime[] = new int[n+1];
 
    for (int i = 0; i <= n; i++)
        prime[i] = 0;
 
    for (int i = 2; i <= n; i++)
    {
 
        // When we get a prime for the first time
        // then assign a unique smallest value to
        // it and all of its multiples
        if (!(prime[i]>0))
        {
            cnt++;
            for (int j = i; j <= n; j += i)
                prime[j] = cnt;
        }
    }
 
    // Print the generated array
    for (int i = 2; i <= n; i++)
        System.out.print(prime[i] + " ");
}
 
// Driver code
public static void main (String[] args)
{
    int n = 6;
 
specialSieve(n);
 
}
}
 
// This code is contrubuted by anuj_67..

Python3

# Python3 implementation of the approach
 
# Function to generate the required array
def specialSieve(n) :
 
    # Initialize cnt variable for assigning
    # unique value to prime and its multiples
    cnt = 0;
    prime = [0]*(n + 1);
 
    for i in range(2, n + 1) :
 
        # When we get a prime for the first time
        # then assign a unique smallest value to
        # it and all of its multiples
        if (not prime[i]) :
            cnt += 1;
            for j in range(i, n + 1, i) :
                prime[j] = cnt;
 
    # Print the generated array
    for i in range(2, n + 1) :
        print(prime[i],end = " ");
 
 
# Driver code
if __name__ == "__main__" :
 
    n = 6;
    specialSieve(n);
     
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
 
class GFG
{
 
// Function to generate the required array
static void specialSieve(int n)
{
 
    // Initialize cnt variable for assigning
    // unique value to prime and its multiples
    int cnt = 0;
    int []prime = new int[n+1];
 
    for (int i = 0; i <= n; i++)
        prime[i] = 0;
 
    for (int i = 2; i <= n; i++)
    {
 
        // When we get a prime for the first time
        // then assign a unique smallest value to
        // it and all of its multiples
        if (!(prime[i] > 0))
        {
            cnt++;
            for (int j = i; j <= n; j += i)
                prime[j] = cnt;
        }
    }
 
    // Print the generated array
    for (int i = 2; i <= n; i++)
        Console.Write(prime[i] + " ");
}
 
// Driver code
public static void Main ()
{
    int n = 6;
 
    specialSieve(n);
 
}
}
 
// This code is contrubuted by anuj_67..

Javascript

 
// Javascript implementation of the approach
 
// Function to generate the required array
function specialSieve(n)
{
 
    // Initialize cnt variable for assigning
    // unique value to prime and its multiples
    let cnt = 0;
    let prime = new Array(n + 1);
 
    for (let i = 0; i <= n; i++)
        prime[i] = 0;
 
    for (let i = 2; i <= n; i++) {
 
        // When we get a prime for the first time
        // then assign a unique smallest value to
        // it and all of its multiples
        if (!prime[i]) {
            cnt++;
            for (let j = i; j <= n; j += i)
                prime[j] = cnt;
        }
    }
 
    // Print the generated array
    for (let i = 2; i <= n; i++)
        document.write(prime[i] + " ");
}
 
// Driver code
    let n = 6;
 
    specialSieve(n);
 

输出：

1 2 1 3 2

时间复杂度： O(N Log N)

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