Codeforces Round #262 (Div. 2)B. Little Dima and Equation

通过枚举x可能各位数字之和， 即0（0）， 81（999999999） 对于 x = b·s(x)a + c，
逆向求出x， 再判断x各位数字和是否等于所枚举的值

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
#include <list>
#include <deque>
#include <queue>
#include <cctype>
#include <map>
#include <set>
#include <bitset>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iomanip>
#include <cstdlib>
#include <ctime>
#include <cassert>
#include <limits>

using namespace std;

#define CLR(A) memset(A,0,sizeof(A))
#define pb(x) push_back(x)
#define mp(x,y) make_pair((x),(y))
#define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i))
#define rer(i,l,u) for(int (i)=(int)(l);(i)<=(int)(u);++(i))
#define reu(i,l,u) for(int (i)=(int)(l);(i)<(int)(u);++(i))
typedef long long int64;
int gcd(const int64 &a, const int64 &b) {return b == 0 ? a : gcd(b, a % b);}
int64 int64pow(int64 a, int64 b){if(b == 0) return 1;int64 t = int64pow(a, b / 2);if(b % 2) return t * t * a;return t * t;}
const int inf = 1e9;
const double eps = 1e-8;

int64 a, b, c;
vector<int64> v;

void init()
{

}
void solve()
{
v.clear();

for (int64 i = 0; i <= 81; ++i) {
int64 x = 1;
x = int64pow(i, a);
x *= b; x += c;
int64 sum = 0, tmp = x;
while (tmp) {
sum += tmp % 10;
tmp /= 10;
}
if (x > 0 && x < 1000000000 && sum == i) v.pb(x);
}
}
void print()
{
cout << v.size() << endl;
rep(i, (int)v.size()) cout << v[i] << " ";
cout << endl;
}
int main()
{
while (cin >> a >> b >> c) {
solve();
print();
}
return 0;
}