Codeforces Round #262 (Div. 2) B

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本文探讨了一种数学问题，即在给定的参数范围内寻找满足特定等式的整数解。通过详细解释解题过程和关键步骤，作者提供了一种有效的算法来解决此类问题，并通过实例演示了如何应用这种方法。

题目：

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 

题意分析：

题意很清楚，就是计算x和x各位和s（x）的一个关系等式是否成立，成立就输出x。这题思路形成得还是比较快，枚举s（x）。s（x）的范围是1到81（我最开始SB的以为10的9次方是9位数，开始写的72

）。然后就对着式子敲出来就行了，注意long long和x的范围小于1e9，这两个地方是这道题的cha点，各种血腥的cha啊。我提交的过的代码没有判断小于1e9，然后我就天真的把lock了。呵呵呵呵呵，然后被cha了，呵呵呵呵呵呵。还好考5 2 100 cha到一个人挽回点损失~

代码：

#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");
        }
}
}