Problem:
Dudeney Number

　

A Dudeney number is a positive integer for which the sum of its decimal digits is equal to the cube root of the number. For example, $512$ is a Dudeney number because it is the cube of $8$, which is the sum of its decimal digits ($5 + 1 + 2$).

In this problem, we think of a type similar to Dudeney numbers and try to enumerate them.

Given a non-negative integer $a$, an integer $n$ greater than or equal to 2 and an upper limit $m$, make a program to enumerate all $x$’s such that the sum of its decimal digits $y$ satisfies the relation $x = (y + a)^n$, and $x \leq m$.

Input

The input is given in the following format.

$a$ $n$ $m$

The input line provides three integers: $a$ ($0 \leq a \leq 50$), $n$ ($2 \leq n \leq 10$) and the upper limit $m$ ($1000 \leq m \leq 10^8$).

Output

Output the number of integers that meet the above criteria.

Sample Input 1

16 2 1000

Sample Output 1

2

Two: $400 = (4 + 0 + 0 + 16)^2$ and $841 = (8 + 4 + 1 + 16)^2$

Sample Input 2

0 3 5000

Sample Output 2

3

Three: $1=1^3$, $512 = (5 + 1 + 2)^3$ and $4913 = (4 + 9 + 1 + 3)^3$.

Sample Input 3

2 3 100000

Sample Output 3

0

There is no such number $x$ in the range below $100,000$ such that its sum of decimal digits $y$ satisfies the relation $(y+2)^3 = x$.

Solution in C++ (judge verdict: Time Limit Exceeded)

#include<iostream>
#include<cmath>
using namespace std;

int main(){

	int a, n, m;
	int ans = 0;

	cin >> a >> n >> m;

	int kai = powl(a,n)+1;

	for( kai; kai <= m; kai++) {

		int  sum = 0;

		while(kai>0){
			sum = kai % 10;
			kai /10;
		}

		if(kai - powl(a,n) == sum * (sum + 2*a)) ans++;
	}

	cout << ans << endl;

	return 0;
}