劍指offer上的一題，但是感覺這位兄弟的解法更好

#include<iostream>
using namespace std;
#define N 2000

int cnt(int n)
{
    if (n < 1)
    {
        return 0;
    }
    int count = 0;
    int base = 1;
    int round = n;
    while (round > 0)
    {
        int weight = round % 10;
        round /= 10;
        count += round * base
 
;
        if (weight == 1)
        {
            count += (n % base) + 1;
        }
        else if (weight > 1)
        {
            count += base;
        }
        base *= 10;
    }   
    return count;
}


int main()
{
    int n;
    cin >> n;
    cout << cnt(n) << endl;
    return
 
 0;
}