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: Wrong Answer)

#include <stdio.h>
#include <string.h>
#include <stdbool.h>
#include <stdint.h>
#include <stdlib.h>
#include <limits.h>
#include <math.h>
int acs(const void *a, const void *b){return *(int*)a - *(int*)b;} /* 1,2,3,4.. */
int des(const void *a, const void *b){return *(int*)b - *(int*)a;} /* 8,7,6,5.. */
int cmp_char(const void *a, const void *b){return *(char*)a - *(char*)b;} /* a,b,c,d.. */
int cmp_str(const void *a, const void *b){return strcmp(*(const char **)a, *(const char **)b);} /* aaa,aab.. */
#define min(a,b) (a < b ? a: b)
#define max(a,b) (a > b ? a: b)
#define rep(i, l, r) for (int i = l; i < r; i++)
#define MAX 100001
#define MOD 1000000007

typedef long long int lli;

int main(void) {
  // 105, 165, 195
  int n;
  scanf("%d", &n);
  int ans = 0;
  if ( n >= 105 ) ans++;
  if ( n >= 165 ) ans++;
  if ( n >= 195 ) ans++;
  printf("%d\n", ans);
  return 0;
}