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>
int prime[7]={2,3,5,7,11,13,17};
int Index[7]={1,1,1,1,1,1,1};
int A = 0;
int Ans = 0;
int count = 0;
int number = 105;
int Nnumber = 0;
int i = 0;
int prints = 0;
int main(void){
  scanf("%d",&A);
    if(A>=105){
      prints=1;
      while(number==A){
        if(number%2){
      Ans = 1;
      Nnumber = number;
      while(Nnumber==1){
        if(!Nnumber%prime[count]){
          Nnumber %=prime[count];
          Index[count] = Index[count]+1;
        }else{
          count++;
        }
     }
    for(i=6;i >= 0; i--){
       Ans*=Index[i];
     }
     if(Ans==8)prints++;
   }
     number++;
   }
 }
 printf("%d",prints);
}