Codeforces 460B Little Dima and Equation【思维+暴力】

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 functions(x) determines the sum of all digits in the decimal representation of numberx.

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 printn integers in the increasing order — the solutions of the given equation. Print only integer solutions that are larger than zero and strictly less than10⁹.

Examples

Input

3 2 8

Output

310 2008 13726 

Input

1 2 -18

Output

0

Input

2 2 -1

Output

41 31 337 967 

题目大意：

让你找到所有合法的X（1e9>X>0）；使得X=b*s（x）^a+c.这里s（x）表示数字X所有位子上的数字的累加和。

思路：

我们直接枚举X肯定是不行的，我们可以逆序思维，考虑枚举S（x）的值来计算一个值tmp=b*s（x）^a+c，显然，如果s（tmp）==S（x），那么tmp==X.

考虑到X最大也就是1e9.那么我们枚举的范围就十分的有限了。

Ac代码：

#include<stdio.h>#include<string.h>#include<math.h>using namespace std;#define ll __int64ll a,b,c;ll ans[4545];ll poww(ll a,ll b){    ll ans=1;    for(int i=0;i<b;i++)    {        ans*=a;    }    return ans;}int main(){    while(~scanf("%I64d%I64d%I64d",&a,&b,&c))    {        int cont=0;        for(int i=1;i<=100;i++)        {            ll tmp=b*(ll)poww(i,a)+c;            ll tmpp=tmp;            ll sum=0;            while(tmp)            {                sum+=tmp%10;                tmp/=10;            }            if(sum==i)            {                if(tmpp>0&&tmpp<1000000000)                ans[cont++]=tmpp;            }        }        printf("%d\n",cont);        for(int i=0;i<cont;i++)        {            printf("%I64d ",ans[i]);        }        printf("\n");    }}