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

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.

Examples
Input
yes
no
yes
Output
2
80
5
composite
Input
no
yes
no
no
no
Output
58
59
78
78
2
prime
Note

The hidden number in the first query is 30. In a table below you can see a better form of the provided example of the communication process.

The hidden number is divisible by both 2 and 5. Thus, it must be composite. Note that it isn't necessary to know the exact value of the hidden number. In this test, the hidden number is 30.

59 is a divisor of the hidden number. In the interval [2, 100] there is only one number with this divisor. The hidden number must be 59, which is prime. Note that the answer is known even after the second query and you could print it then and terminate. Though, it isn't forbidden to ask unnecessary queries (unless you exceed the limit of 20 queries).

题目链接:http://codeforces.com/contest/680/problem/C


题意:人机交互。codeforces评测系统随机给出一个区间在[2,100]的数x。通过一些操作来判断x是质数还是合数。操作为:你输出一个数m,OJ进行判断,如果x%m==0,OJ输入yes,否则OJ输出no。操作不能操作20次。

思路:这种题目理解题意才是难点(▄█▀█●给跪了)。题意就是最多通过20个数可以判断[2,100]区间内任意一个数是素数还是合数。只要x分别除以2,3,5,7,11,13,17,19,23,29,31,37,41,43,47(2*47=94)来进行判断。x%sign[i]==0的话,cou++。如果cou>1,那么就是合数。光这几个数还是不够的,还需要一些特殊的数4,9,25。

代码:

#include<bits/stdc++.h>
using namespace std;
string str;
int sign[]= {,,,,,,,,,,,,,,,,,,};
int main()
{
int i,ans=;
for(i=; i<; i++)
{
printf("%d\n",sign[i]);
fflush(stdout);
cin>>str;
if(str=="yes") ans++;
}
if(ans>=) printf("composite\n"),fflush(stdout);
else printf("prime\n"),fflush(stdout);
return ;
}

680C. Bear and Prime 100 数学的更多相关文章

  1. codeforces 356 div2 C.Bear and Prime 100 数学

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

  2. codeforces 680C C. Bear and Prime 100(数论)

    题目链接: C. Bear and Prime 100 time limit per test 1 second memory limit per test 256 megabytes input s ...

  3. 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 ...

  4. Codeforces A - Bear and Prime 100(交互题)

    A - Bear and Prime 100 思路:任何一个合数都可以写成2个以上质数的乘积.在2-100中,除了4,9,25,49外都可以写成两个以上不同质数的乘积. 所以打一个质数加这四个数的表: ...

  5. Codeforces Round #356 (Div. 2) C. Bear and Prime 100 水题

    C. Bear and Prime 100 题目连接: http://www.codeforces.com/contest/680/problem/C Description This is an i ...

  6. 【CodeForces】679 A. Bear and Prime 100

    [题目]A. Bear and Prime 100 [题意]有一数字x,每次可询问一个数字y是否x的因子,最后输出数字x是否素数,要求询问次数<=20. [题解]容易发现[2,100]范围内的非 ...

  7. 暑假练习赛 006 B Bear and Prime 100

    Bear and Prime 100Crawling in process... Crawling failed Time Limit:1000MS     Memory Limit:262144KB ...

  8. CF #356 div1 A. Bear and Prime 100

    题目链接:http://codeforces.com/contest/679/problem/A CF有史以来第一次出现交互式的题目,大致意思为选择2到100中某一个数字作为隐藏数,你可以询问最多20 ...

  9. CF385C Bear and Prime Numbers 数学

    题意翻译 给你一串数列a.对于一个质数p,定义函数f(p)=a数列中能被p整除的数的个数.给出m组询问l,r,询问[l,r]区间内所有素数p的f(p)之和. 题目描述 Recently, the be ...

随机推荐

  1. 「SHOI2015」自动刷题机

    /* 有理有据的二分答案 因为在过程中最多减到零 所以n越小显然就能刷更多的题 无解时就是无论如何也无法得到k , 这个特判一下即可 */ #include<cstdio> #includ ...

  2. [Mysql]查看版本号的五种方式

    [Mysql]查看版本号的五种方式   目录(?)[+]   查看版本信息 #1 使用命令行模式进入mysql会看到最开始的提示符 Your MySQL connection id is 3Serve ...

  3. linux系统下修改文件夹目录权限-chmod

    Linux.Fedora.Ubuntu修改文件.文件夹权限的方法差不多.很多人开始接触Linux时都很头痛Linux的文件权限问题.这里告诉大家如何修改Linux文件-文件夹权限.以主文件夹下的一个名 ...

  4. 《GPU高性能编程CUDA实战》第九章 原子性

    ▶ 本章介绍了原子操作,给出了基于原子操作的直方图计算的例子. ● 章节代码 #include <stdio.h> #include "cuda_runtime.h" ...

  5. 加密算法之AES算法(转)

    转载http://www.mamicode.com/info-detail-514466.html 0 AES简介 美国国家标准技术研究所在2001年发布了高级加密标准(AES).AES是一个对称分组 ...

  6. 文件的基本操作(python)

    文件打开: 1. f = open('yesterday,‘r+’,encoding = ‘utf-8’) 读取的方式加载为Utf-8 r    打开文件并写, 只适用于文字类 r+   打开文件并读 ...

  7. vue-cli 上手

    1.cnpm install --global vue-cli 安装脚手架 2.vue init webpack baoge 创建 3.选择配置项 Project name (baoge): ---- ...

  8. ADO接口简介

    源地址:http://blog.csdn.net/xiaobai1593/article/details/7449151 参考: 1. 百度文库:http://wenku.baidu.com/view ...

  9. servlet第三篇

    Servlet是一个供其他Java程序(Servlet引擎)调用的Java类,它不能独立运行,它的运行完全由Servlet引擎来控制和调度. 针对客户端的多次Servlet请求,通常情况下,服务器只会 ...

  10. VB.net 与 C# 的对应逻辑运算符

    And:对两个Boolean表达式执行逻辑和.AndAlso:与AndAlso类似,关键差异是AndAlso显示短路行为,如果AndAlso中的第一个表达式为False,则不计算第二个表达式.Or:对 ...