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>

void run() {
	int N, i, j, cnt1 = 0, cnt2 = 0;
	scanf("%d", &N);
	for(j = 0; j < N; ++j){
		for(i = 1; i <= N-j; ++i) {
			if((N-j)%2 == 0) continue;
			if(((N-j) % i) == 0) cnt1++;
			if(cnt1 == 8) cnt2++;
			if(cnt1 == 9) cnt2--;
		}
		cnt1 = 0;
	}
	printf("%d", cnt2);
}
int main() {
	run();
	return 0;
}