Little Dima and Equation CodeForces

Little Dima and Equation CodeForces - 460B

题目描述
Little Dima misbehaved during a math lesson a lot and the nasty teacher Mr. Pickles gave him the following problem as a punishment.

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 10^9.

Example

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

大致题意
给你一个公式然后让你求所有满足条件的x的个数，并从小到大输出x的值。

思路 ： s（x）表示的是x的十进制表示中所有的数字和，又因为x<10^9，所以x最多有9位，当每一位上都是9时s（x）有最大值81，所以s（x）的取值范围为1到81.暴力即可。

注意 x的取值范围是x>0&&x<10^9,超过这个范围的x不记录。

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<queue>#include<vector>#include<cmath>#include<string>#define ll long long  #define maxn 100005using namespace std;ll p(int x,int y){    ll n=1;    for(int i=1;i<=y;i++){        n*=x;    }    return n;}int f[100];int main(){    int a,b,c,k=0;    scanf("%d %d %d",&a,&b,&c);    for(int i=1;i<=81;i++)    {        long long int sum,num=0,sum1;         sum1=sum=p(i,a)*b+c;        while(sum){            num+=sum%10;            sum/=10;        }        if(num==i&&sum1>0&&sum1<1e9)        f[k++]=sum1;    }    printf("%d\n",k);    for(int i=0;i<k;i++)    printf("%d ",f[i]);    return 0;}