Do you know the twin prime conjecture? Two primes  and  are called twin primes if . The twin prime conjecture is an unsolved problem in mathematics, which asks for a proof or a disproof for the statement "there are infinitely many twin primes".

On April 17, 2013, Yitang Zhang announced a proof that for some integer  that is less than 70 million, there are infinitely many pairs of primes that differ by . As of April 14, 2014, one year after Zhang's announcement, the bound has been reduced to 246. People are hoping for the bound to be smaller and smaller, so that a proof for the conjecture can finally be found.

For our dear contestants, we've prepared another similar problem for you, which is the extended twin composite number problem: Given a positive integer , find two integers  and  such that  and both  and  are composite numbers.

Input

There are multiple test cases. The first line of the input contains an integer  (about ), indicating the number of test cases. For each test case:

The only line contains one integer  ().

Output

For each test case output two integers in one line, indicating  and  where . If there are multiple valid answers, you can print any of them; If there is no valid answer, output "-1" (without quotes) instead.

Sample Input

3
11
1805296
5567765

Sample Output

4 15
114514 1919810
111234 5678999

Author: JIN, Mengge
Source: The 19th Zhejiang University Programming Contest Sponsored by TuSimple


水题,特别能迷惑人。

代码:

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring> using namespace std; int main() {
int t,n;
scanf("%d",&t);
while(t --) {
scanf("%d",&n);
if(n == ) printf("%lld %lld\n",,);
else printf("%lld %lld\n",n * 2ll,n * 3ll);
}
}

zoj 4099 Extended Twin Composite Number的更多相关文章

  1. 2019浙大校赛--J--Extended Twin Composite Number(毒瘤水题)

    毒瘤出题人,坑了我们好久,从基本的素数筛选,到埃氏筛法,到随机数快速素数判定,到费马小定理,好好的水题做成了数论题. 结果答案是 2*n=n+3*n,特判1,2. 以下为毒瘤题目: 题目大意: 输入一 ...

  2. ZOJ 2971 Give Me the Number;ZOJ 2311 Inglish-Number Translator (字符处理,防空行,strstr)

    ZOJ 2971 Give Me the Number 题目 ZOJ 2311 Inglish-Number Translator 题目 //两者题目差不多,细节有点点不一样,因为不是一起做的,所以处 ...

  3. ZOJ 2059 The Twin Towers(双塔DP)

    The Twin Towers Time Limit: 2 Seconds      Memory Limit: 65536 KB Twin towers we see you standing ta ...

  4. ZOJ 2971 Give Me the Number

    Give Me the Number Numbers in English are written down in the following way (only numbers less than  ...

  5. ZOJ 2971 Give Me the Number (模拟,字符数组的清空+map)

    Give Me the Number Time Limit: 2 Seconds      Memory Limit: 65536 KB Numbers in English are written ...

  6. ZOJ 2059 The Twin Towers

    双塔DP. dp[i][j]表示前i个物品,分成两堆(可以不全用),价值之差为j的时候,较小一堆的价值为dp[i][j]. #include<cstdio> #include<cst ...

  7. ZOJ 2132 The Most Frequent Number (贪心)

    题意:给定一个序列,里面有一个数字出现了超过 n / 2,问你是哪个数字,但是内存只有 1 M. 析:首先不能开数组,其实也是可以的了,后台数据没有那么大,每次申请内存就可以过了.正解应该是贪心,模拟 ...

  8. ZOJ - 2132:The Most Frequent Number(思维题)

    pro:给定N个数的数组a[],其中一个数X的出现次数大于N/2,求X,空间很小. sol:不能用保存数组,考虑其他做法. 由于出现次数较多,我们维护一个栈,栈中的数字相同,所以我们记录栈的元素和个数 ...

  9. The 19th Zhejiang University Programming Contest Sponsored by TuSimple (Mirror)

    http://acm.zju.edu.cn/onlinejudge/showContestProblems.do?contestId=391 A     Thanks, TuSimple! Time ...

随机推荐

  1. 牛掰的socat工具介绍

    Socat 是 Linux 下的一个多功能的网络工具,名字来由是 「Socket CAT」.其功能与有瑞士军刀之称的 Netcat(nc) 类似,可以看做是 Netcat 的加强版.Socat 的主要 ...

  2. 修复ubuntu 安装mysql后必须使用sudo问题

    修改root用户 查看用户的权限,是否是mysql_native_password,如果不是,则将auth_sock改为mysql_native_password update user set pl ...

  3. JIRA中的并联审批流程定制

    JIRA号称可以跟踪任何事务,让JIRA的流程来匹配团队的工作流程,而不是让你的团队适应JIRA的工作流程.但是在实践中,有些有些流程用JIRA还是比较困难的,比如并联审批流程,一个并联审批流程需求大 ...

  4. PHP替代session的方法

    PHP替代session的方法 服务器集群的时候 会发现session的问题 一般采用redis 来代替 用账号作为key 因为redis能主从 所以打算用替代session的方法1 cookie代替 ...

  5. PHP,Excel导出换行

    // 有id,才算真的有发票数据 if ($v['b_invoice_id']) { $v['b_invoice_info'] = json_decode($v['b_invoice_json'],t ...

  6. conda创建、删除、重命名环境

    链接:https://www.jianshu.com/p/7265011ba3f2 创建新环境 conda create -n rcnn python=3.6 删除环境 conda remove -n ...

  7. win10安装Ubuntu,用Xshell连接

    一.安装Ubuntu 安装Ubuntu,安装过程就不详细说了,我是从微软商店下载的Ubuntu安装,没有用VMware,想用Xshell连接Ubuntu,中间一直出问题,现在解决,总结一下. 二.配置 ...

  8. docker 实践四:数据管理

    这篇是关于 docker 的数据管理. 注:环境为 CentOS7,docker 19.03. 一般容器中管理数据主要有两种方式: 数据卷(Data Volumes):容器内数据直接映射到本地主机环境 ...

  9. ps 指令

    ps显示系统当前进程信息, ps 存在多个版本,因此 ps options 的种类繁多.这里只列举平时开发过程中常用的命令,如果有错误或者更好的例子.烦请在评论区指出 语法 ps [options] ...

  10. java 简单操作HDFS

    创建java 项目 package com.yw.hadoop273; import org.apache.hadoop.conf.Configuration; import org.apache.h ...