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;
 
int sumBit(int n){
  int sum = 0;
  while(n > 0){
    sum += n % 10;
    n /= 10;
  }
  return sum;
}
 
int main() {
  int n,a,b;
  long long total = 0;
  
  cin >> n >> a >> b;
  
  int tmpSum = 0;
  for(int i = 1; i <= n; i++){
    tmpSum = sumBit(i);
    if(a <= tmpSum && tmpSum <= b){
      total += i;
    }
  }
  
  cout << total;
  return 0;
}