Geeksforgeeks Solution For " 3 Divisors "

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Problem :- Given a value N, you need to find how many numbers less than or equal to N have numbers of divisors exactly equal to 3.

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Solution :- 

#include<iostream>
#include<math.h>
using namespace std;
int isprime(int n)
{
    for(int i=2; i<n; i++)
        if(n%i==0)
            return 0;
    return 1;
}
int check(int n)

{
    int i,c;
    for(i=2,c=0; i<=sqrt(n); i++)
    {
        if(isprime(i))
            c++;
    }
    return c;
}
int main()
 {
int t,n,i;
    cin>>t;
    while(t--){
        cin>>n;
        cout<<check(n)<<'\n';
    }
return 0;
}

Output:-