Compute sum of digits in all numbers from 1 to n

Problem

   Given a number x, find sum of digits in all numbers from 1 to n.

Solution
     take 389 for example;
     count[1,299) + [300,389)'s 3 + [1,89)
Code

#include <stdio.h>
#include <iostream>
#include <cmath>
using namespace std;


int f(int n) {
    if (n < 10) {
        return n * (n + 1) / 2;
    }
    
    int d = log10(n);
    
    int a[15];
    a[1] = 45;
    for (int i = 2;i <= d;i ++) {
        a[i] = a[i - 1] * 10 + 45 * pow(10,i - 1);
    }
    int p = ceil(pow(10,d));
    
    int msd = n / p;
    
    int p1 = msd * a[d] + (msd * (msd - 1)) / 2 * p;
    //int p1 = f(msd * p - 1);
    int p2 = msd * (n % p + 1) + f(n % p);
    return p1 + p2;
}

int main() {
 //code
 cout << f(389) << endl;
 return 0;
}