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 5.4.1) (judge verdict: Accepted)

#include<stdio.h>

int divisor(int a){
    int i,cnt=0;
    for(i=1;i<=a;i++){
        if(a%i==0)cnt++;
    }
    return cnt;
}

int main(){
    int n,i,j,div,cnt=0;
    scanf("%d",&n);
    for(i=1;i<=n;i+=2){
        div=divisor(i);
        if(div==8)cnt++;
    }
    printf("%d\n",cnt);
    return 0;

}