cf B. Little Dima and Equation
http://codeforces.com/contest/460/problem/B
import java.util.*;
import java.math.*;
public class Main { public static void main(String []args)
{
Scanner cin=new Scanner(System.in);
int a,b,c;
BigInteger a1=new BigInteger("");
BigInteger a10=new BigInteger("");
BigInteger a3=new BigInteger("");
BigInteger [][]cc=new BigInteger[][];
BigInteger []f1=new BigInteger[];
for(int i=; i<=; i++)
{
for(int j=; j<=; j++)
{
BigInteger sum=new BigInteger("");
BigInteger h=new BigInteger(((Integer)i).toString());
int k=;
while(k<=j)
{
sum=sum.multiply(h);
k++;
}
cc[i] [j] =sum;
}
}
a=cin.nextInt(); b=cin.nextInt();c=cin.nextInt();
BigInteger h1=new BigInteger(((Integer)b).toString());
BigInteger h2=new BigInteger(((Integer)c).toString());
int cnt=;
for(int i=; i<=; i++)
{
BigInteger a2=new BigInteger("");
BigInteger hh=new BigInteger(((Integer)i).toString());
BigInteger ans,ans1;
ans=cc[i] [a] .multiply(h1).add(h2);
ans1=ans;
if(ans.compareTo(a1)<=||ans.compareTo(a10)>) continue;
while(!ans.equals(a1))
{
a2=a2.add(ans.mod(a3));
ans=ans.divide(a3);
}
if(a2.equals(hh))
{
f1[cnt] =ans1;
cnt++;
}
}
System.out.println(cnt);
for(int i=; i<cnt; i++)
{
if(i==) System.out.print(f1[i] );
else System.out.print(" "+f1[i] );
}
System.out.print("\n");
}
}
cf B. Little Dima and Equation的更多相关文章
- CodeForces460B. Little Dima and Equation
B. Little Dima and Equation time limit per test 1 second memory limit per test 256 megabytes input s ...
- CF460B Little Dima and Equation (水题?
Codeforces Round #262 (Div. 2) B B - Little Dima and Equation B. Little Dima and Equation time limit ...
- Codeforces Little Dima and Equation 数学题解
B. Little Dima and Equation time limit per test 1 second memory limit per test 256 megabytes input s ...
- B. Little Dima and Equation
time limit per test 1 second memory limit per test 256 megabytes input standard input output standar ...
- cf 366D D. Dima and Trap Graph (计算所有线段共同覆盖的某段区间)
http://codeforces.com/problemset/problem/366/D 题意:给出n个点,m条边,a,b,ll,rr分别代表点a,点b相连,点a和点b的区间范围(ll,rr),然 ...
- CF#214 C. Dima and Salad 01背包变形
C. Dima and Salad 题意 有n种水果,第i个水果有一个美味度ai和能量值bi,现在要选择部分水果做沙拉,假如此时选择了m个水果,要保证\(\frac{\sum_{i=1}^ma_i}{ ...
- cf 366C C. Dima and Salad(01背包)
http://codeforces.com/contest/366/problem/C 题意:给出n个水果的两种属性a属性和b属性,然后挑选苹果,选择的苹果必须要满足这样一个条件:,现在给出n,k,要 ...
- codeforces 460B Little Dima and Equation 解题报告
题目链接:http://codeforces.com/problemset/problem/460/B 题目意思:给出a, b, c三个数,要你找出所有在 1 ≤ x ≤ 1e9 范围内满足 x = ...
- codeforces #262 DIV2 B题 Little Dima and Equation
题目地址:http://codeforces.com/contest/460/problem/B 这题乍一看没思路.可是细致分析下会发现,s(x)是一个从1到81的数,不管x是多少.所以能够枚举1到8 ...
随机推荐
- android资料
http://bbs.51cto.com/thread-903936-1.html http://zhidao.baidu.com/question/195697097.html?sort=4& ...
- hdu Number Sequence
这道题是寻找规律.别的方法一般都是超时. #include <cstdio> #include <cstring> #include <algorithm> usi ...
- 转:Android模拟器连接电脑网络
原文地址:http://www.it165.net/pro/html/201212/4444.html 第一步: 在命令行(就是开始——运行——输入cmd)模式下输入adb shell命令一般会报两种 ...
- 工作中用到的linux命令
都是工作中用到的,解决问题至上,不求甚解,怕再忘了,所以记录一下,勿喷. .log |,,,,|,| 先说一下这条命令: cat:打印文件内容 grep:查找,用到的有\s匹配空白字符 sed:刚用到 ...
- 【剑指offer】二叉树深度
转载请注明出处:http://blog.csdn.net/ns_code/article/details/27249675 题目描写叙述: 输入一棵二叉树,求该树的深度.从根结点到叶结点依次经过的结点 ...
- 退役笔记一#MySQL = lambda sql : sql + ' Source Code 4 Explain Plan '
Mysql 查询运行过程 大致分为4个阶段吧: 语法分析(sql_parse.cc<词法分析, 语法分析, 语义检查 >) >>sql_resolver.cc # JOIN.p ...
- extern C的作用详解
extern "C"的主要作用就是为了能够正确实现C++代码调用其他C语言代码.加上extern "C"后,会指示编译器这部分代码按C语言的进行编译,而不是C+ ...
- GDI+(Graphics Device Interface)例子
使用SolidBrush 单色画笔 Bitmap bitmap = new Bitmap(800, 600); Graphics graphics = Graphics.From ...
- c# 取得扩展名
string KZM=files[0].FileName.Substring(files[0].FileName.LastIndexOf(".") + 1);
- JS 图片预览功能
<script type="text/javascript"> function DisplayImage(fileTag) { document. ...