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>
#define rep(i,n)for(int i=0;i<n;++i)
#include<string>
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }

int Digit(int n) {
	int sum = 0;
	while (n > 0) {
		sum += n % 10;
		n /= 10;
	}
	return sum;
}

int main() {
	int n, a, b;
	cin >> n >> a >> b;
	int ans = 0;
	for (int i = 1;i <= n;++i) {
		if (a <= Digit(i) && Digit(i) <= b)ans+=i;
	}
	cout << ans << endl;
}