【Codeforces Round 262 (Div 2)B】【暴力枚举】Little Dima and Equation b乘数位之和的a次方+c=数字

Little Dima and Equation

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard 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

310 2008 13726 

input

1 2 -18

output

0

input

2 2 -1

output

41 31 337 967 

#include<stdio.h>#include<iostream>#include<string.h>#include<string>#include<ctype.h>#include<math.h>#include<set>#include<map>#include<vector>#include<queue>#include<bitset>#include<algorithm>#include<time.h>using namespace std;void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }#define MS(x,y) memset(x,y,sizeof(x))#define MC(x,y) memcpy(x,y,sizeof(x))#define MP(x,y) make_pair(x,y)#define ls o<<1#define rs o<<1|1typedef long long LL;typedef unsigned long long UL;typedef unsigned int UI;template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b>a)a = b; }template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b<a)a = b; }const int N = 0, M = 0, Z = 1e9 + 7, ms63 = 0x3f3f3f3f;int a, b, c;LL mul(int x, int p){LL y = 1;while (p--)y = y*x;return y;}int d[100];void solve(){int ans = 0;for (int i = 1; i <= 81; ++i){LL x = b*mul(i, a) + c;if (x <= 0 || x >= 1e9)continue;int y = x;int sum = 0;while (x){sum += x % 10;x /= 10;}if (sum == i)d[++ans] = y;}sort(d + 1, d + ans + 1);printf("%d\n", ans);for (int i = 1; i <= ans; ++i)printf("%d ", d[i]);puts("");}int main(){while (~scanf("%d%d%d", &a,&b,&c)){solve();}return 0;}/*【trick&&吐槽】枚举范围比较小的参量是很好的思想，比如位数。*/