B. Little Dima and Equation

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Little Dima misbehaved during a math lesson a lot and the nasty teacher Mr. Pickles gave him the following problem as a punishment.

Find all integer solutions x (0 < x < 109) of the equation:

x = b·s(x)a + c, 

where a, b, c are some predetermined constant values and function s(x) determines the sum of all digits in the decimal representation of number x.

The teacher gives this problem to Dima for each lesson. He changes only the parameters of the equation: a, b, c. Dima got sick of getting bad marks and he asks you to help him solve this challenging problem.

Input

The first line contains three space-separated integers: a, b, c (1 ≤ a ≤ 5; 1 ≤ b ≤ 10000;  - 10000 ≤ c ≤ 10000).

Output

Print integer n — the number of the solutions that you've found. Next print n integers in the increasing order — the solutions of the given equation. Print only integer solutions that are larger than zero and strictly less than 109.

Sample test(s)

input

3 2 8

output

3
10 2008 13726 

input

1 2 -18

output

0

input

2 2 -1

output

4
1 31 337 967 

题目地址：http://codeforces.com/contest/460/problem/B

这题乍一看没思路，但是仔细分析下会发现，s(x)是一个从1到81的数，无论x是多少。所以可以枚举1到81，这样就转化成了一个一元一次方程，直接求解x就可以了。这时候还要判断x是否在1到10^9之间，并且它的各位数之和是s(x)。

这题脑残了两次。。。第一次是写成了<=10^9。。然后发现错误后，就又改了回来。。但是发现10^9是不可能的

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <ctype.h>
#include <algorithm>
#include <queue>
using namespace std;
#define LL __int64
LL aa[100];
int main()
{
    LL n, a, b, c, ans=0, i, y, zz, j;
    LL x, z;
    scanf("%I64d%I64d%I64d",&a,&b,&c);
    for(i=1;i<=81;i++)
    {
        zz=1;
        for(j=1;j<=a;j++)
            zz*=i;
        x=zz*b+c;
        y=0;
        z=x;
        while(z)
        {
            y+=z%10;
            z/=10;
        }
        if(y==i&&x>=1&&x<1e9)
        {
            aa[ans++]=x;
        }
    }
    printf("%I64d
",ans);
    for(i=0;i<ans;i++)
    {
        printf("%I64d ",aa[i]);
    }
    return 0;
}