# Codeforces Little Dima and Equation 数学题解

2022年11月23日
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B. Little Dima and Equationtime 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 abc 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: abc.
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

`310 2008 13726 `

http://codeforces.com/contest/460/problem/B

1 怎样找到规律？

2 没有找到规律，暴力搜索？

3 怎样暴力搜索？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非常少的。故此能够选定这个值遍历。

6 第5步是关键思考转折点，有点逆向思维的味道。暴力枚举也是能够非常巧妙。没那么easy掌握好。

`#include <stdio.h>#include <vector>#include <string.h>#include <algorithm>#include <iostream>#include <string>#include <limits.h>#include <stack>#include <queue>#include <set>#include <map>#include <math.h>using namespace std;const int MAX_N = 1000;const int MAX_VAL = 1000000000;int digit[MAX_N];int sumDigits(int num){int sum = 0;while (num){sum += num %10;num /= 10;}return sum;}int main(){int a, b, c, N;long long num;while (scanf("%d %d %d", &a, &b, &c) != EOF){N = 0;for (int i = 1; i < MAX_N; i++)//enumerate i, which is the sum of digits{num = (long long) pow(double(i), double(a));num = num * b + c;if (num >= MAX_VAL) break;if (sumDigits((int)num) == i) digit[N++] = (int)num;}sort(digit, digit+N);printf("%d\n", N);for (int i = 0; i < N; i++){printf("%d ", digit[i]);}if (N) putchar('\n');}return 0;}`

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