C. Bear and Prime 100

题目连接:

http://www.codeforces.com/contest/680/problem/C

Description

This is an interactive problem. In the output section below you will see the information about flushing the output.

Bear Limak thinks of some hidden number — an integer from interval [2, 100]. Your task is to say if the hidden number is prime or composite.

Integer x > 1 is called prime if it has exactly two distinct divisors, 1 and x. If integer x > 1 is not prime, it's called composite.

You can ask up to 20 queries about divisors of the hidden number. In each query you should print an integer from interval [2, 100]. The system will answer "yes" if your integer is a divisor of the hidden number. Otherwise, the answer will be "no".

For example, if the hidden number is 14 then the system will answer "yes" only if you print 2, 7 or 14.

When you are done asking queries, print "prime" or "composite" and terminate your program.

You will get the Wrong Answer verdict if you ask more than 20 queries, or if you print an integer not from the range [2, 100]. Also, you will get the Wrong Answer verdict if the printed answer isn't correct.

You will get the Idleness Limit Exceeded verdict if you don't print anything (but you should) or if you forget about flushing the output (more info below).

Input

After each query you should read one string from the input. It will be "yes" if the printed integer is a divisor of the hidden number, and "no" otherwise.

Output

Up to 20 times you can ask a query — print an integer from interval [2, 100] in one line. You have to both print the end-of-line character and flush the output. After flushing you should read a response from the input.

In any moment you can print the answer "prime" or "composite" (without the quotes). After that, flush the output and terminate your program.

To flush you can use (just after printing an integer and end-of-line):

fflush(stdout) in C++;

System.out.flush() in Java;

stdout.flush() in Python;

flush(output) in Pascal;

See the documentation for other languages.

Hacking. To hack someone, as the input you should print the hidden number — one integer from the interval [2, 100]. Of course, his/her solution won't be able to read the hidden number from the input.

Sample Input

yes

no

yes

Sample Output

2

80

5

composite

Hint

题意

现在在[2,100]里面有一个隐藏的数字,你现在可以问最多20个问题,每个问题可以提出一个数,如果这个数是隐藏的数字的因子的话

那么就会返回yes

否则就会返回no

让你判断这个数是合数,还是素数

题解:

把小于50的素数全部问了一遍,且把4 9 25 49这四个小于100的素数的平方问一遍就好了

如果超过1次回答为yes的话,那么就是合数。

道理很显然,因为一个合数肯定是一个素数乘以另外一个素数,所以至少有2嘛,第二个数就是小于等于50的了

代码

#include<bits/stdc++.h>
using namespace std;
vector<int> ans; void TAT()
{
for(int i=2;i<=50;i++)
{
int flag = 0;
for(int j=2;j<i;j++)
if(i%j==0)flag = 1;
if(flag==0)ans.push_back(i);
}
ans.push_back(4);
ans.push_back(9);
ans.push_back(25);
ans.push_back(49);
int tmp = 0;
for(int i=0;i<ans.size();i++)
{
cout<<ans[i]<<endl;
string s;
cin>>s;
if(s=="yes")tmp++;
}
if(tmp<2)cout<<"prime"<<endl;
else cout<<"composite"<<endl;
}
int main()
{
TAT();
return 0;
}

Codeforces Round #356 (Div. 2) C. Bear and Prime 100 水题的更多相关文章

  1. Codeforces Round #356 (Div. 2) C. Bear and Prime 100(转)

    C. Bear and Prime 100 time limit per test 1 second memory limit per test 256 megabytes input standar ...

  2. Codeforces Round #356 (Div. 2) B. Bear and Finding Criminal 水题

    B. Bear and Finding Criminals 题目连接: http://www.codeforces.com/contest/680/problem/B Description Ther ...

  3. Codeforces Round #356 (Div. 2) A. Bear and Five Cards 水题

    A. Bear and Five Cards 题目连接: http://www.codeforces.com/contest/680/problem/A Description A little be ...

  4. Codeforces Round #368 (Div. 2) A. Brain's Photos (水题)

    Brain's Photos 题目链接: http://codeforces.com/contest/707/problem/A Description Small, but very brave, ...

  5. Codeforces Round #356 (Div. 2)B. Bear and Finding Criminals(水题)

    B. Bear and Finding Criminals time limit per test 2 seconds memory limit per test 256 megabytes inpu ...

  6. Codeforces Round #373 (Div. 2) C. Efim and Strange Grade 水题

    C. Efim and Strange Grade 题目连接: http://codeforces.com/contest/719/problem/C Description Efim just re ...

  7. Codeforces Round #185 (Div. 2) A. Whose sentence is it? 水题

    A. Whose sentence is it? Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/ ...

  8. Codeforces Round #373 (Div. 2) A. Vitya in the Countryside 水题

    A. Vitya in the Countryside 题目连接: http://codeforces.com/contest/719/problem/A Description Every summ ...

  9. Codeforces Round #371 (Div. 2) A. Meeting of Old Friends 水题

    A. Meeting of Old Friends 题目连接: http://codeforces.com/contest/714/problem/A Description Today an out ...

随机推荐

  1. 源码安装postgresql数据库

    一般情况下,postgresql由非root用户启动. 1.创建postgres用户 groupadd postgres useradd -g postgres postgres 下面的操作都在pos ...

  2. Fiddler大师之路系列(一)

    江湖传言,Fiddler是捕获客户端与服务器之间的所有HTTP(S) 请求的利器,但是在具体使用过程中,发现使用Fiddler进行抓包时有一部分请求总是没到,多方苦寻之下发现客户端使用WinINET这 ...

  3. mysql测试工具 -> mysqlslap

    一.简介 mysqlslap是mysql自带的基准测试工具 优点:查询数据,语法简单,灵活容易使用.该工具可以模拟多个客户端同时并发的向服务器发出查询更新,给出了性能测试数据而且提供了多种引擎的性能比 ...

  4. 阿里云slb+https 实践操作练习

    如果只是练习按照文档步骤逐步执行即可. 如果是业务需要,只供参考. 有道笔记链接->

  5. SQLAlchemy-对象关系教程ORM-一对多(外键),一对一,多对多

    一:一对多 表示一对多的关系时,在子表类中通过 foreign key (外键)引用父表类,然后,在父表类中通过 relationship() 方法来引用子表的类. 在一对多的关系中建立双向的关系,这 ...

  6. DOS命令大全(一)

    #1 一: net use \\ip\ipc$ " " /user:" " 建立IPC空链接 net use \\ip\ipc$ "密码" ...

  7. 最简单删除SQL Server中所有数据的方法(不用考虑表之间的约束条件,即主表与子表的关系)

    其实删除数据库中数据的方法并不复杂,为什么我还要多此一举呢,一是我这里介绍的是删除数据库的所有数据,因为数据之间可能形成相互约束关系,删除操作可能陷入死循环,二是这里使用了微软未正式公开的sp_MSF ...

  8. JAVA类课后练习

    1.Pg168--2 package com.hanqi; import java.util.Scanner; public class Rectangle { Rectangle() { //完成初 ...

  9. Oracle学习笔记——点滴汇总

    Oracle学习笔记——点滴汇总 http://www.botangdb.com/ Oracle GI = Grid Infrastructure = ASM + Cluster

  10. 009 HDFS的shell命令(里面有一个安全模式)

    一:dfs的命令 1.所有的领命 2.解释 -appendToFile <localsrc> ... <dst>   :追加文件 -checksum <src> . ...