Codeforces Round #262 (Div. 2)

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 

题目意思是：给出a,b,c,求满足等式 x=b*s(x)^a+c 的所有x（s(x)表示x的十进制各数位之和）。没有就输出0。
一开始一点思路都没有，后来仔细想了想，s(x)最大只有81，所以可以从s(x)开始枚举就可以了。
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
int ss[1000];
int main()
{
    int a,b,c;
    int k=0;
    scanf("%d%d%d",&a,&b,&c);
    for(int i=1;i<=81;i++)
    {
        int bbb=b;
        for(int j=1;j<=a;j++)
            bbb*=i;
        int ans=c+bbb;  //cout<<ans<<" ";
        int tans=ans;   //cout<<tans<<" ";
        if(ans<=0) continue;
        if(ans>=1000000000) break;
        int sum=0;
        while(ans){
            sum+=ans%10;
            ans/=10;
        }
        if(sum==i){// cout<<sum<<endl;
            ss[k++]=tans;//printf("%d ",tans);
        }
    }
    if(ss[0]==0) printf("0\n");
    else{
        printf("%d\n",k);
        for(int i=0;i<k;i++)
        {
           if(i==k-1) printf("%d\n",ss[i]);
           else       printf("%d ",ss[i]);
        }
    }
    return 0;
}