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()
{
	int i,j,n,temp,count[201][14],yaku[201],ans=0;
	scanf("%d",&n);
	for(i=3;i<=n;i++){
		yaku[i] = 1;
		temp = i;
		if(i%2==1){
			for(j=3;j<14;j++){
				count[i][j] = 0;
				while(temp%j==0){
					count[i][j]++;
					temp = temp/j;
				}
				yaku[i] = yaku[i]*(count[i][j]+1);
			}
			if(yaku[i]==8)ans++;
		}
	}
	printf("%d",ans);
	return 0;
}