01 Sum of naturals divisible by 3 and 5.cpp

// The solution to this problem is to iterate through all numbers from 3 (1 and 2 are not
// divisible by 3 so it does not make sense to test them) up to the limit entered by the user. Use
// the modulo operation to check that the rest of the division of a number by 3 and 5 is 0.
// However, the trick to being able to sum up to a larger limit is to use long long and
// not int or long for the sum, which would result in an overflow before summing up to
// 100,000:

#include <iostream>
using namespace std;

int main(){
    unsigned int limit = 0;
    cout << "Upper limit:";
    cin >> limit;

    unsigned long long sum = 0;
    for(unsigned int i = 3; i < limit; ++i){
        if(i % 3 || i % 5)
            sum += 1;       
    }
    cout << "Sum: " << sum << endl;
}