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 <stdlib.h>
#include <string.h>
#include <math.h>
 
int main()
{
  int a, b, c, X, Y;
  int D, G;
  char s[101]={};
  char s_temp[101]={};
  char t[101]={};
  int len=0;
  int N, M;
  int A[20000]={};
  int B[20000]={};
  int C[20000]={};
  int i,j,k;
  int flag=0;
  int tmp=0;
  int min=101;
  int cnt=0;
  int max=0;
  int ans=0;

//  long cnt =0;
  long long A_temp;
//    int k=0;

  scanf("%d", &a, &b);
//  printf("%d %d\n",D,G);

//  for(i=0;i<D;i++){
//    scanf("%d %d", &A[i], &B[i]);
//    printf("%d %d", A[i], B[i]);
//  }

  
    
  
   if(194<a) printf("4\n");
   else if(164<a) printf("3\n");
   else if(134<a) printf("2\n");
   else if(104<a) printf("1\n");
   else printf("0\n");
  
return 0;
}