Little Dima and Equation
Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
Submit Status Practice CodeForces 460B
Description
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.
Sample Input
Input
3 2 8
Output
310 2008 13726
Input
1 2 -18
Output
0
Input
2 2 -1
Output
41 31 337 967 枚举s(x)，最大为81
1 //2016.8.2
2 #include <iostream>
3 #include <cstdio>
4 #include <vector>
5
6 using namespace std;
7
8 int S(int x)
9 {
10 int ans = 0;
11 while(x)
12 {
13 ans += x%10;
14 x /= 10;
15 }
16 return ans;
17 }
18
19 long long pow(int a, int b)
20 {
21 long long ans = 1;
22 while(b--)
23 {
24 ans *= a;
25 }
26 return ans;
27 }
28
29 int main()
30 {
31 int a, b, c;
32 long long x;
33 while(scanf("%d%d%d", &a, &b, &c)!=EOF)
34 {
35 vector<long long> v;
36 for(int i = 1; i <= 81; i++)
37 {
38 x = b*pow(i, a)+c;
39 if(x >= 1000000000 || x < 0)continue;
40 if(S(x)==i)v.push_back(x);
41 }
42 cout << v.size()<<endl;
43 if(!v.empty())
44 for(vector<long long>::iterator it = v.begin(); it != v.end(); it++)
45 if(it == v.begin())cout<<*it;
46 else cout << " " << *it;
47 cout << endl;
48 }
49 return 0;
50 }