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>
#include<stdlib.h>
#include<string.h>
#include<math.h>

#define INT_MAX 2147483647

int main(void){
    int N;
    scanf("%d", &N);
    int j;
    int count = 0;
    for(j=1; j<=N; j++) {
        int H[201] = {0};
        int x = j;
        int i = 3;
        while(i <= x && i != 1) {
            if(i % 2 != 0) {
                if(x % i == 0) {
                    H[i]++;
                    x = x / i;
                }
                else{
                    i++;
                }
            }
            else{
                i++;
            }
        }

        int sum = 1;
        for(i=1; i <= j; i++){ 
            if(H[i] != 0) {
                sum *= H[i] + 1;
            }
        }
        if(j % 2 != 0 ) {
            if(sum == 8) {
                count++;
            }
        }

    }
    printf("%d\n", count);

    return 0;
}