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 Gcount=0;

void kei(int n)
{
    int i;
    int count = 0;
    for(i=1;i<=n;i++){
        if(n % i == 0) count++;
    }
    if(count==8) Gcount++;
}

int main(void)
{
    int i,N;
    scanf("%d", &N);

    for(i=1;i<=N;i++){
        if(i%2==0) continue;
        kei(i);
    }
    printf("%d\n", Gcount);
}