给定数字n，我们必须找出小于或等于n的正好有3个除数的数字

Question

i have used sieve of eratosthenes method to calculate such numbers that have exactly 3 divisors and are less than or equal to n where n is the given number.

the problem with this code is that it give answer 37386 for n=999999,but the correct answer is 168

public static void main (String[] args) throws java.lang.Exception
{
    Scanner scn =new Scanner(System .in);
    print("enter the prime no");
    int n= scn.nextint();
    int arr[]= new int[n+1];
    for(int i=2;i<arr.length;i++)
    {
        arr[i]=1;
    }

    for(int i=2;i*i<=n;i++)
    {
        if(arr[i]==1)
        {

            for(int p=2;p*i<=n;p++)
            {
                arr[p*i]=0;
            }
        }
    }

    int count=0;
    for(int i=2;i<=n;i++)
    {
        if(arr[i]==1&&i*i<n)
        {
            count++;
        }
    }

    System.out.print(count);
}

code for calculating numbers that have exactly 3 divisors

Answer 1

The Sieve of Eratosthenes is used to calculate prime numbers. Your code correctly calculates that there are 37386 prime numbers less than 999999. It should be possible to modify the code to find numbers with exactly three divisors rather than one, but I will leave that as an exercise to the asker.