Codeforces Round #262 (Div. 2) B

題目：

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 

題意分析：

題意很清楚，就是計算x和x各位和s（x）的一個關係等式是否成立，成立就輸出x。這題思路形成得還是比較快，枚舉s（x）。s（x）的範圍是1到81（我最開始SB的以爲10的9次方是9位數，開始寫的72）。然後就對着式子敲出來就行了，注意long long和x的範圍小於1e9，這兩個地方是這道題的cha點，各種血腥的cha啊。我提交的過的代碼沒有判斷小於1e9，然後我就天真的把lock了。呵呵呵呵呵，然後被cha了，呵呵呵呵呵呵。還好考5 2 100 cha到一個人挽回點損失~

代碼：

#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <iostream>

using namespace std;

int d[100000];
int main()
{
    int a,b,c,sum,cou;
    long long temp,flag;
    while(cin>>a>>b>>c)
    {
        memset(d,0,sizeof(d));
        cou=0;
        for(int i=1;i<=81;i++)
        {
            sum=0;
            temp=1;
            for(int j=0;j<a;j++)
            {
                temp*=i;
            }
            temp=b*temp;
            temp=c+temp;
            //printf("%d ",temp);
            flag=temp;
            while(temp)
            {
                sum+=temp%10;
                temp/=10;
            }
            if(sum==i&&flag>0&&flag<(1e9))
                d[cou++]=flag;
        }
        if(cou==0)
            printf("0\n");
        else
        {
            printf("%d\n%d",cou,d[0]);
            for(int i=1;i<cou;i++)
            {
                printf(" %d",d[i]);
            }
            printf("\n");
        }
}
}