Problem:
Score : 200 points

Problem Statement

Find the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive).

Constraints

1 \leq N \leq 10^4

1 \leq A \leq B \leq 36

All input values are integers.

Input

Input is given from Standard Input in the following format:

N A B

Output

Print the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive).

Sample Input 1

20 2 5

Sample Output 1

84

Among the integers not greater than 20, the ones whose sums of digits are between 2 and 5, are: 2,3,4,5,11,12,13,14 and 20. We should print the sum of these, 84.

Sample Input 2

10 1 2

Sample Output 2

13

Sample Input 3

100 4 16

Sample Output 3

4554

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
 
using namespace std;
typedef long long ll;
typedef pair<ll, ll> p_ll;
 
template<class T>
void debug(T itr1, T itr2) { auto now = itr1; while(now<itr2) { cout << *now << " "; now++; } cout << endl; }
#define repr(i,from,to) for (int i=(int)from; i<(int)to; i++)
#define rep(i,N) repr(i,0,N)
#define per(i,N) for (int i=(int)N-1; i>=0; i--)
 
const ll MOD = pow(10,9)+7;
const ll LLINF = pow(2,61)-1;
const int INF = pow(2,30)-1;
  

int main() {
    int N, A, B;
    cin >> N >> A >> B;

    int res = 0;
    repr(i,1,N+1){
        int sum = 0;
        int n = i;
        while (n > 0) {
            sum += n % 10;
            n = n / 10;
        }
        if (A <= sum && sum <= B) res += i;
    }

    cout << res << endl;
    return 0;
}