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>
#include <string.h>

int souoc(int n) {
	int i;
	int sum = 0;

	for (i = 1; i <= n; i++) {
		if (n%i == 0) {
			sum++;
		}
	}

	return sum;
}

int main() {
	int n;
	int sum =0;
	int i;

	scanf("%d", &n);

	if (n >= 105) {
		for (i = 105; i <= n; i+=2) {
			if (souoc(i) == 8) {
				sum++;
			}
		}

		printf("%d", sum);
	}
	else {
		printf("0");
	}


	return 0;
}