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)

//ABC106-B
#include <stdio.h>

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