Little Dima and Equation（枚举）

原题链接
B. Little Dima and Equation
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard 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.

Examples
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

这个题要是枚举x那么会发现这几乎是一件不可能的事情，但是反过来枚举s(x)则要简单很多，因为s(x)最多也就81，但是它却对应唯一的x

#include <cstdio>#include <algorithm>#include <vector>using namespace std;vector<int> s;int integerpow(long long  dishu,int zhishu){        long long t = dishu;        for(int i=1;i<zhishu;i++)                dishu *= t;        return dishu;}int main(){        int a,b,c,res=0;        scanf("%d%d%d",&a,&b,&c);        for(int sx=1;sx<82;sx++){                long long x = b * integerpow(sx,a) + c;                if (x >= 1000000000) break;//这一点非常重要，否则会过不了，因为可能当x超过后，其1~9位上加上去刚好会有答案，但是9位之后的数字去没有算，这是不符合题意的                long long t = x;                int m[10] = {0};                int sum=0;                while(x){                        sum += x%10;                        x/=10;                }                if(sx == sum){                        res++;                        s.push_back(t);                }        }        sort(s.begin(),s.end());        printf("%d\n",res);        for(int i = 0;i<res;i++){                printf("%d",s[i]);                if(i != res - 1) printf(" ");                else printf("\n");        }        return 0;}