Home >Web Front-end >HTML Tutorial >Codeforces Round #262 (Div. 2) 460B. Little Dima and Equation(枚举)_html/css_WEB-ITnose

Codeforces Round #262 (Div. 2) 460B. Little Dima and Equation(枚举)_html/css_WEB-ITnose

WBOY
WBOYOriginal
2016-06-24 11:59:31926browse

题目链接:http://codeforces.com/problemset/problem/460/B


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?=?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

310 2008 13726 

input

1 2 -18

output

input

2 2 -1

output

41 31 337 967 


代码如下:

#include #include #include #include #include #include using namespace std;typedef __int64 LL;LL ss[10017];LL F(LL num){    LL s = 0;    while(num)    {        s += num%10;        num/=10;    }    return s;}LL Fac(LL num, LL n){    LL s = 1;    for(LL i = 1; i <= n; i++)    {        s*=num;    }    return s;}int main(){    LL a, b, c;    while(~scanf("%I64d%I64d%I64d",&a,&b,&c))    {        memset(ss,0,sizeof(ss));        LL l = 0, i;        for(i = 1; i <= 81; i++)//枚举        {            LL x = b*Fac(i,a)+c;            if(x >= 1000000000 || x <= 0)                continue;            LL ans = (int)F(x);            if(ans == i)                ss[l++] = x;        }        printf("%I64d\n",l);        for(i = 0; i < l; i++)        {            if(i == 0)                printf("%I64d",ss[i]);            else                printf(" %I64d",ss[i]);        }        printf("\n");    }    return 0;}  

Statement:
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn