Codeforces Gym 100610 Problem A. Alien Communication Masterclass 构造
Problem A. Alien Communication Masterclass
Time Limit: 1 Sec
Memory Limit: 256 MB
题目连接
http://codeforces.com/gym/100610
Description
Andrea is a famous science fiction writer, who runs masterclasses for her beloved readers. The most popular one is the Alien Communication Masterclass (ACM), where she teaches how to behave if you encounter alien life forms or at least alien artifacts. One of the lectures concerns retrieving useful information based on aliens’ writings. Andrea teaches that based on alien mathematical formulas, one could derive the base of the numeral system used by the aliens, which in turn might give some knowledge about aliens’ organisms. (For example, we use numeral system with base 10, due to the fact that we have ten fingers on our upper extremities). Suppose for simplicity that aliens use the same digits as we do, and they understand and denote addition, subtraction, multiplication, parentheses and equality the same way as we do. For her lecture, Andrea wants an example of a mathematical equality that holds in numeral systems with bases a1, a2, · · · , an, but doesn’t hold in numeral systems with bases b1, b2, · · · , bm. Provide her with one such formula.
Input
The first line of the input file contains two integer numbers, n and m (1 ≤ n, m ≤ 8). The second line contains n numbers, a1, a2, · · · , an. The third line contains m numbers, b1, b2, · · · , bm. All ai and bi are distinct and lie between 2 and 10, inclusive.
Output
Output any syntactically correct mathematical equality that holds in numeral systems with bases a1, a2, · · · , an, but doesn’t hold in numeral systems with bases b1, b2, · · · , bm. The equality can contain only digits 0 through 9, addition (‘+’), subtraction and unary negation (‘-’), multiplication (‘*’), parentheses (‘(’ and ‘)’) and equality sign (‘=’). There must be exactly one equality sign in the output. Any whitespace characters in the output file will be ignored. The number of non-whitespace characters in the output file must not exceed 10 000.
Sample Input
1 2 2 3 9
Sample Output
(10 - 1) * (10 - 1) + 1 = 10
HINT
题意
要求你输出等式,满足N种进制,但是不满足任何的M种进制
题解:
直接输出(10-a1*1)*(10-a2*1)*……*(10-an*1)=0就好了
只有在n种进制中才能满足条件
代码:
//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <bitset>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 110000
#define mod 1001
#define eps 1e-9
#define pi 3.1415926
int Num;
//const int inf=0x7fffffff; //§ß§é§à§é¨f§³
const ll inf=;
inline ll read()
{
ll x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
//************************************************************************************* //(k-a1)(k-a2)(k-a3)*.....*(k-an) int a[];
int b[];
int main()
{
freopen("acm.in","r",stdin);
freopen("acm.out","w",stdout);
int n=read(),m=read();
for(int i=;i<n;i++)
cin>>a[i];
for(int i=;i<m;i++)
cin>>b[i];
for(int i=;i<n;i++)
{
cout<<"(10-";
for(int j=;j<=a[i];j++)
{
if(j!=a[i])
cout<<"1-";
else
cout<<"";
}
if(i!=n-)
cout<<")*";
else
cout<<")";
}
cout<<"=0";
}
Codeforces Gym 100610 Problem A. Alien Communication Masterclass 构造的更多相关文章
- Codeforces Gym 100610 Problem K. Kitchen Robot 状压DP
Problem K. Kitchen Robot Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/10061 ...
- Codeforces Gym 100610 Problem H. Horrible Truth 瞎搞
Problem H. Horrible Truth Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/1006 ...
- Codeforces Gym 100610 Problem E. Explicit Formula 水题
Problem E. Explicit Formula Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/10 ...
- 【Gym 100610A】Alien Communication Masterclass
题 Andrea is a famous science fiction writer, who runs masterclasses for her beloved readers. The mos ...
- Codeforces Gym 100342J Problem J. Triatrip 求三元环的数量 bitset
Problem J. Triatrip Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100342/at ...
- Codeforces Gym 100342C Problem C. Painting Cottages 转化题意
Problem C. Painting CottagesTime Limit: 2 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/10 ...
- Codeforces Gym 100342C Problem C. Painting Cottages 暴力
Problem C. Painting CottagesTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/1 ...
- Codeforces Gym 100500F Problem F. Door Lock 二分
Problem F. Door LockTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100500/at ...
- Codeforces Gym 100002 Problem F "Folding" 区间DP
Problem F "Folding" Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/ ...
随机推荐
- 【打表】HDOJ-2089-不要62
[题目链接:HDOJ-2089] 多组测试数据,所以可以先算出符合条件的所有数保存数组中,输入时则直接遍历数组. #include<iostream> #include<cstrin ...
- Delphi or函数的用法
function GetFlag(a: string): Integer;var I: Integer;begin Result := 0; for I := 0 to 3 - 1 do begin ...
- Java与WCF交互(一)补充:用WSImport生成WSDL的Java客户端代码
在<Java与WCF交互(一):Java客户端调用WCF服务>一 文中,我描述了用axis2的一个Eclipse控件生成WCF的Java客户端代理类,后来有朋友建议用Xfire.CXF,一 ...
- Eclipse “Invalid Project Description” when creating new project from existing source
1) File>Import>General>Existing Project into Workspace2) File>Import>Android>Exist ...
- XTUOJ 1246 Heartstone 贪心
题意:挺好懂得 分析:先计算出如果不能用(减2)操作,至少需要多少个(减3)操作,这个很好计算 然后就是尽量多的去减少(减3)操作,肯定先抹平 余2 和 余1 的,然后就可以了 #include &l ...
- SQL 语句记录
1.创建一个table @"create table rockTB(myId integer primary key autoincrement not null, time varchar ...
- qt 设置背景图片
博客出处:http://www.cppblog.com/qianqian/archive/2010/07/25/121238.htm 工作似乎走上正轨了,上周五的工作是做一个界面,用到QFrame和Q ...
- 【和我一起学python吧】Python安装、配置图文详解
Python安装.配置图文详解 目录: 一. Python简介 二. 安装python 1. 在windows下安装 2. 在Linux下安装 三. 在windows下配置python集成开发环境( ...
- Python 学习笔记(四)正则、闭合、生成器
(一)正则表达式 基本规则: ^ 匹配字符串开始位置. $ 匹配字符串结束位置. \b 匹配一个单词边界. \d 匹配一个数字. \D 匹配一个任意的非数字字符. x? 匹配可选的x字符.换句话说,就 ...
- 轻松突击ThreadLocal
本文出自 代码大湿 代码大湿 ThreadLocal是用来保存线程的本地变量,可以保证每个线程都有一个自己的变量(包括static变量). 本文所有代码请点击我 1 看个实际场景. 我们要设计一个序列 ...