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. linux系统实现多个进程监听同一个端口

    通过 fork 创建子进程的方式可以实现父子进程监听相同的端口. 方法:在绑定端口号(bind函数)之后,监听端口号之前(listen函数),用fork()函数生成子进程,这样子进程就可以克隆父进程, ...

  2. Linux终极shell-zsh的完美配置方案!——oh-my-zsh

    Zsh 介绍 Zsh 兼容 Bash,据传说 99% 的 Bash 操作 和 Zsh 是相同的 Zsh 官网:http://www.zsh.org/ 先看下你的 Linux支持哪些 shell:cat ...

  3. maven 中配置多个mirror的问题

    公司搭建的maven私服做镜像,有使用aliyun的镜像,还有其他地方的, 默认情况下配置多个mirror的情况下,只有第一个生效.那么我们可以将最后一个作为默认值,前面配置的使用环境变量动态切换. ...

  4. SSH连接服务器时,长时间不操作就会断开的解决方案

    最近在配置服务器相关内容时候,不同的事情导致长时间不操作,页面就断开了连接,不能操作,只能关闭窗口,最后通过以下命令解决. SSH连接linux时,长时间不操作就断开的解决方案: 1.修改/etc/s ...

  5. 循环(数组循环、获取json数据循环)、each()循环详解

    return; // 退出循环(不满足,退出此次循环.下次满足条件,依然会走此循环)return false; //退出函数(退出所有) 一. 数组循环: html: <div class=&q ...

  6. 05 多继承、object类

    多继承 Python中一个类可以继承多个父类,并且获得全部父类的属性和方法. class A: def demo(self): print("demo") class B: def ...

  7. 机器学习xgboost参数解释笔记

    首先xgboost有两种接口,xgboost自带API和Scikit-Learn的API,具体用法有细微的差别但不大. 在运行 XGBoost 之前, 我们必须设置三种类型的参数: (常规参数)gen ...

  8. Android--圆角背景style

    <?xml version="1.0" encoding="utf-8"?> <shape xmlns:android="http: ...

  9. 【exgcd】卡片

    卡片 题目描述 你有一叠标号为1到n的卡片.你有一种操作,可以重排列这些卡片,操作如下:1.将卡片分为前半部分和后半部分.2.依次从后半部分,前半部分中各取一张卡片,放到新的序列中.例如,对卡片序列( ...

  10. Itemchanged事件

    Itemchanged事件:当数据窗口控件中某个域被修改并且该域失去输入焦点该事件返回的意义为: 0--(缺省返回值),接收新修改的值: 1--不接收新修改的值且不允许改变输入焦点: 2--不接收新修 ...