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: Wrong Answer)

#include<bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define mod %
using namespace std;
using ll = long long;

int main(){
    int ans = 0;
    int n, a, b; cin >> n >> a >> b;
    
    for(int i = 1; i <= n; i++){
        int tmp = 0, j = i;
        /*
        for(int j = i; j > 0; j /= 10){
            tmp += j % 10;
        }*/

        tmp += j % 10;
        j /= 10;
        tmp += j % 10;
        j /= 10;
        tmp += j % 10;
        j /= 10;
        tmp += j % 10;
        j /= 10;
        
        if(tmp >= a && tmp <= b){
            ans += i;
        }
    }

    cout << ans << endl;
}