HackerEarth Number of divisors problem solution

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

using ll = long long;

long long f(int n, int k) {
    if (n == 0) return 0;
    long long res = (n/k);
    return f(n/k, k) + n * (ll)(n+1) / 2ll - (res * (res + 1) / 2ll) * k;
}                   

int main () {
    int T = 1;
    scanf("%d", &T);
    assert(T >= 1 && T <= 300000);
    while(T--) {

        int n, k;
        scanf("%d%d", &n, &k);
        assert(n <= 1e9);
        assert(k >= 2 && k <= 1e9);

        printf("%lldn", f(n, k));
        
    }

    return 0;
}