Little Dima and Equation

最新推荐文章于 2022-07-11 23:57:10 发布

转载最新推荐文章于 2022-07-11 23:57:10 发布 · 120 阅读

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原文链接：http://www.cnblogs.com/xzxj/p/7236011.html

本文介绍了一个数学惩罚问题：寻找特定方程的所有整数解。通过枚举数字和的方法，使用C++实现了解决方案并详细展示了代码实现过程。

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.

Example

Input

3 2 8

Output

3
10 2008 13726

Input

1 2 -18

Output

Input

2 2 -1

Output

4
1 31 337 967 

因为x的范围是0到10的9次方，所以 0<s(x)<=81,
然后就可以枚举s（x）求出x的值，然和求出x的每位上数字之和，
和s（x）比较，看是否满足。

#include<stdio.h>
#include<algorithm>
#include<iostream>

using namespace std;

int main()
{
    long long a,b,c;
    long long s=1;
    long long t[50000],l=0;
    long long x;
    scanf("%I64d %I64d %I64d",&a,&b,&c);
    for(long long i=1;i<=81;i++)
    {
        s=1;
        for(long long j=1;j<=a;j++)
            s=s*i;
        x=b*s;
        x+=c;
        long long t1=0;
        long long t2=x;
        if(x>0&&x<1000000000)
        {
            while(x)
            {
                t1+=x%10;
                x/=10;
            }
            if(t1==i)
                t[l++]=t2;
        }

    }
    sort(t,t+l);
    printf("%I64d",l);
    if(l!=0)
        printf("\n");
    for(long long i=0;i<l;i++)
    {
        printf("%I64d",t[i]);
        if(i!=l-1)
            printf(" ");
    }
    return 0;
}