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 <stdlib.h>

int main(){
	int n;
	if(scanf("%d", &n)==EOF){return 0;}
	int count = 0;
	int count_sub;
	
	if(n<105){printf("0\n"); return 0;}
	for(int i = 105; i <= n; i++){
		count_sub = 0;
		if(i % 2 == 0){continue;}
		for(int j = 3; j < i; j++){
			if(i % j == 0){count_sub++;}
		}
		//その他j=1, i で割り切れる
		if(count_sub == 6){count++;}
	}

	printf("%d\n", count);
	return 0;

}