Codeforces#262_1002

Codeforces#262_1002

B. 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.

Sample test(s)

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

 

虽然是简单的枚举，其实也没那么简单！

这位同学说的好：

不算难的题目，就是暴力枚举，不过枚举也没有那么容易的，而是需要很好的逻辑思维能力，才能在这么段时间内想出问题答案的。

思考：
1 如何找到规律？
2 没有找到规律，暴力搜索？
3 如何暴力搜索？遍历？以那个值作为遍历？
4 以x作为遍历？范围太大，肯定超时。
5 以s(x)作为遍历，s(x)表示x的数位值相加，一个数字的数位值相加范围肯定是很少的，故此可以选定这个值遍历。
6 第5步是关键思考转折点，有点逆向思维的味道。暴力枚举也是可以很巧妙。没那么容易掌握好。

 1 #include <cstring>
 2 
 3 #include <iostream>
 4 
 5 #include <algorithm>
 6 
 7 #include <cstdio>
 8 
 9 #include <cmath>
10 
11 #include <map>
12 
13 #include <cstdlib>
14 
15 #define M(a,b) memset(a,b,sizeof(a))
16 
17 using namespace std;
18 
19 
20 
21 int res[2000006];
22 
23 
24 
25 int main()
26 
27 {
28 
29   long long a,b,c;
30 
31   scanf("%I64d%I64d%I64d",&a,&b,&c);
32 
33   int count = 0;
34 
35   for(int i = 1;i<=81;i++)
36 
37   {
38 
39       long long aa = 1;
40 
41       for(int t = 0;t<a;t++)
42 
43       {
44 
45           aa*=i;
46 
47       }
48 
49       if(b*aa+c>1000000000||b*aa+c<0) continue;
50 
51       int tem = (int)(b*aa+c);
52 
53       int tem1 = tem;
54 
55       //cout<<i<<' '<<tem<<' ';
56 
57       if(tem<1000000000&&tem>0)
58 
59       {
60 
61           int ans = 0;
62 
63           while(tem)
64 
65           {
66 
67              ans += tem%10;
68 
69              tem/=10;
70 
71           }
72 
73           //cout<<ans<<endl;
74 
75           if(ans==i)
76 
77             res[count++] = tem1;
78 
79       }
80 
81   }
82 
83   printf("%d\n",count);
84 
85   for(int i = 0;i<count;i++)
86 
87   {
88 
89       printf("%d ",res[i]);
90 
91   }
92 
93   if(count>0)
94 
95     cout<<endl;
96 
97   return 0;
98 
99 }

 

转载于:https://www.cnblogs.com/haohaooo/p/4035326.html