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 main(void){

	int N,i,j,x=0;
	int ans[200]={0};
	
	scanf("%d",&N);
	
	for(i=1;i<=N;i++){
		for(j=1;j<=i;j++){
			if(i%2!=0 && i%j==0)
				ans[i-1]++;
	}	}
	for(i=0;i<200;i++){
		if(ans[i]==8)
			x++;
	}
	printf("%d",x);
	
	return 0;
}