找出在范围内不可被任何数整除的数字，使用C++

Input : num = 14
Output : 3
Explanation: There are three numbers, 1, 11, and 13, which are not divisible.

Input : num = 21
Output : 5
Explanation: There are five numbers 1, 11, 13, 17, and 19, which are not divisible.

= NUM – X + Y – Z + A.

X = num divisible by 2, 3, 5, 7 ( [num / 2] + [num / 3] + [num / 5] + [num / 7] )

Y = num divisible by (2,3), (2, 5), (2, 7), (3, 5), (3, 5), (3, 7) and (5, 7) = ( [num / (2 * 3)] + [num / (2 * 5)] + [num / (2 * 7)] + [num / (3 * 5)] + num / (3 * 7)] + [num / (5 * 7)] ).

Z = num divisible by (2, 3, 5), (2, 3, 7), (2, 5, 7) and (3, 5, 7) = ( [num / (2 * 3 * 5)] + [num / (2 * 3 * 7)] + [num / (2 * 5 * 7)] + [num / (3 * 5 * 7)] )

A = num divisible by (2, 3, 5, 7) = ( [num / (2 * 3 * 5 * 7)] )

#include <bits/stdc++.h>
using namespace std;

int main() {
   int n = 21, result;
   // applying formula from inclusion - exclusion principle
   // to find the count of numbers not divisible by any number from 2 to 10.
   result = n - n / 2 - n / 3 - n / 5 - n / 7
      + n / 6 + n / 10 + n / 14 + n / 15 + n / 21 + n / 35
      - n / 30 - n / 42 - n / 70 - n / 105 + n / 210;
   cout << "The count of numbers, not div by [2, 10] is: " << result;

   return 0;
}

The count of numbers, not div by [2, 10] is: 5