Codeforces Round #262 (Div. 2) B

B - Little Dima and Equation

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

题意:给出一个公式x = b·s(x)a + c,其中s(x)为x的各个位的数字之和。给出a,b,c,求x在1~10^9内的所有解。

题解:枚举s(x),算出来x后看看x和s(x)符不符合,x在不在范围内。因为x<=10^9,所以s(x)为0~81,一下就枚举完了。

注意a<=5,81^5已经超int了,要用long long……血的教训

代码:

 //#pragma comment(linker, "/STACK:102400000,102400000")
#include<cstdio>
#include<cmath>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<map>
#include<set>
#include<stack>
#include<queue>
using namespace std;
#define ll long long
#define usll unsigned ll
#define mz(array) memset(array, 0, sizeof(array))
#define minf(array) memset(array, 0x3f, sizeof(array))
#define REP(i,n) for(i=0;i<(n);i++)
#define FOR(i,x,n) for(i=(x);i<=(n);i++)
#define RD(x) scanf("%d",&x)
#define RD2(x,y) scanf("%d%d",&x,&y)
#define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define WN(x) prllf("%d\n",x);
#define RE freopen("D.in","r",stdin)
#define WE freopen("1biao.out","w",stdout)
#define mp make_pair
#define pb push_back
vector<int>v;
int a,b,c;
int main(){
int i,j;
ll k;
scanf("%d%d%d",&a,&b,&c);
v.clear();
for(i=;i<=;i++){
k=;
REP(j,a)k*=i;
ll x=k*b+c;
if(x<= || x>=)continue;
int y=x,sum=;
while(y){
sum+=y%;
y/=;
}
if(sum==i)v.pb(x);
}
int maxi=v.size();
printf("%d\n",maxi);
if(maxi>)printf("%d",v[]);
for(i=;i<maxi;i++)
printf(" %d",v[i]);
return ;
}

CF460B Little Dima and Equation (水题?的更多相关文章

  1. fzu 1909 An Equation(水题)

    题目链接:fzu 1909 An Equation 典型的签到题. #include <stdio.h> #include <string.h> bool judge(int ...

  2. B - 楼下水题(扩展欧几里德)

    B - 楼下水题 Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u Submit St ...

  3. HDOJ 2317. Nasty Hacks 模拟水题

    Nasty Hacks Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Tota ...

  4. ACM :漫漫上学路 -DP -水题

    CSU 1772 漫漫上学路 Time Limit: 1000MS   Memory Limit: 131072KB   64bit IO Format: %lld & %llu Submit ...

  5. ytu 1050:写一个函数,使给定的一个二维数组(3×3)转置,即行列互换(水题)

    1050: 写一个函数,使给定的一个二维数组(3×3)转置,即行列互换 Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 154  Solved: 112[ ...

  6. CodeForces460B. Little Dima and Equation

    B. Little Dima and Equation time limit per test 1 second memory limit per test 256 megabytes input s ...

  7. [poj2247] Humble Numbers (DP水题)

    DP 水题 Description A number whose only prime factors are 2,3,5 or 7 is called a humble number. The se ...

  8. gdutcode 1195: 相信我这是水题 GDUT中有个风云人物pigofzhou,是冰点奇迹队的主代码手,

    1195: 相信我这是水题 Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 821  Solved: 219 Description GDUT中有个风云人 ...

  9. BZOJ 1303 CQOI2009 中位数图 水题

    1303: [CQOI2009]中位数图 Time Limit: 1 Sec  Memory Limit: 162 MBSubmit: 2340  Solved: 1464[Submit][Statu ...

随机推荐

  1. Leetcode #2 Add two number

    Q: You are given two linked lists representing two non-negative numbers. The digits are stored in re ...

  2. Leetcode 130. Surrounded Regions

    Given a 2D board containing 'X' and 'O' (the letter O), capture all regions surrounded by 'X'. A reg ...

  3. struts2 CVE-2012-0838 S2-007 Remote Code Execution && Hotfix

    catalog . Description . Effected Scope . Exploit Analysis . Principle Of Vulnerability . Patch Fix 1 ...

  4. Eclipse 代码自动补全配置方法

    Eclipse默认只有在出现 “.”时才进行代码提示,更改方法为: 即把“.”改为“.abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ”,表示输 ...

  5. Markdown编辑器简单总结

    字体大小: #1 ##2 添加链接: 文字链接 [blog](http://yalantis.com/blog/how_we_created_tab_bar_animation_for_ios/?ut ...

  6. XmlReader读取XML

    StringBuilder output = new StringBuilder(); String xmlString = @"<bookstore> <book gen ...

  7. [USACO2009 NOV GOLD]奶牛的图片

    校内题,不给传送门了. 以前做完NOIp2013的火柴排队那道题后,当时很担心NOIp会出那种题,因为贪心的规则能不能看出来真的要看运气.但是这类题做多了后发现其实那道题的规则其实是很多题都已经用到了 ...

  8. 3 构建Mysql+heartbeat+DRBD+LVS集群应用系统系列之heartbeat的搭建

    preface 在上节的说了mysql的搭建,这节我们在上节的基础上,继续搭建heartbeat. 安装和配置heartbeat 采用yum安装,dbmaster81和dbbackup136上都安装, ...

  9. HtmlAgilityPack使用

    http://stackoverflow.com/questions/5876825/htmlagilitypack-and-timeouts-on-load http://stackoverflow ...

  10. [Java]double初始化问题

    如下: 1. 直接初始化 double[][] embossFilter = {{-1/9, 0, 1/9}, {-1/9, 1/9, 1/9}, {-1/9, 0, 1/9}}; 2. 赋值初始化 ...