Problem:
Score: 200 points

Problem Statement

The number 105 is quite special - it is odd but still it has eight divisors.
Now, your task is this: how many odd numbers with exactly eight positive divisors are there between 1 and N (inclusive)?

Constraints

N is an integer between 1 and 200 (inclusive).

Input

Input is given from Standard Input in the following format:

N

Output

Print the count.

Sample Input 1

105

Sample Output 1

1

Among the numbers between 1 and 105, the only number that is odd and has exactly eight divisors is 105.

Sample Input 2

7

Sample Output 2

0

1 has one divisor. 3, 5 and 7 are all prime and have two divisors. Thus, there is no number that satisfies the condition.

Solution in C (GCC 9.2.1) (judge verdict: Accepted)

#include <stdio.h>
#include <math.h>
#define rep(i,l,r) for(int i=l; i<(int)(r); i++)

int main(){
    int N;
    scanf("%d", &N);
    int ans = 0;
    for(int i=105; i<=N; i+=2){
        int tmp = 0;
        for(int j=1; j<=i; j++){
            if(i%j == 0) tmp++;
        }
        if(tmp == 8) ans++;
    }

    printf("%d", ans);
    return 0;
}