Sum of all the multiples of 3 or 5 below 1000 gives a wrong answer in C

Question

Project Euler problem:

If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9

Answer 1

Your solution with for-loop will be O(n).

I found a more generic solution O(1).

Here we can use another multiplies, even non-primes.

#include 

long gcd(long a, long b) {  
    return b == 0 ? a : gcd(b, a % b);  
}

long lcm(long a, long b) {  
    return a / gcd(a, b) * b;  
}  

long sumMultiple(long mult, long to) {
    to = to / mult;
    to *= to + 1;
    return (to >> 1) * mult;
}

long calc(long a, long b, long n) {
    n--;
    return sumMultiple(a, n) + sumMultiple(b, n) - sumMultiple(lcm(a,b), n);
}

int main() {
    int n = 1000;
    printf("Sum of multiplies of 3 and 5 is %ld\n", calc(3,5,n));
    return 0;
}