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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
 
int digit_sum(int num) {
  int sum = 0;
  while(num > 0) {
    sum += num % 10;
    num /= 10;
  }
  return sum;
}
 
int main(void) {
  int i, N, A, B;
  int sum = 0;
  scanf("%d %d %d", &N, &A, &B);
  for(i = 1; i <= N; ++i) {
    int k = digit_sum(i);
    if (A <= i && i <= B) sum += i;
  }
  printf("%d\n", sum);
  return 0;
}