Codeforces Round #384 (Div. 2) C. Vladik and fractions(构造题)
Description
Vladik and Chloe decided to determine who of them is better at math. Vladik claimed that for any positive integer n he can represent fraction
as a sum of three distinct positive fractions in form
.
Help Vladik with that, i.e for a given n find three distinct positive integers x, y and z such that
. Because Chloe can't check Vladik's answer if the numbers are large, he asks you to print numbers not exceeding 109.
If there is no such answer, print -1.
Input
The single line contains single integer n (1 ≤ n ≤ 104).
Output
If the answer exists, print 3 distinct numbers x, y and z (1 ≤ x, y, z ≤ 109, x ≠ y, x ≠ z, y ≠ z). Otherwise print -1.
If there are multiple answers, print any of them.
Sample Input
3 7
Sample Output
2 7 42 7 8 56
思路
题意:
给出一个数 n ,问是否存在三个不同的整数 x ,y ,z 满足
。
题解:
,因此,我们可以让1/z = 1 / n,剩下的为 1/x + 1/y = 1/n,通过恒等变换可以知道 1/x = ny / (y - n),可以发现,y = n + 1时候,x 必然为整数,因此x = n + n * n,y = n + 1,z = n。需要注意的是,当 n = 1时,x = 2,y = 2,不符合题意。
#include<bits/stdc++.h>
using namespace std;
int main()
{
int n;
scanf("%d",&n);
if (n == 1) printf("-1\n");
else printf("%d %d %d\n",n*n + n,n + 1,n);
return 0;
}
Codeforces Round #384 (Div. 2) C. Vladik and fractions(构造题)的更多相关文章
- Codeforces Round #384 (Div. 2) C. Vladik and fractions 构造题
C. Vladik and fractions 题目链接 http://codeforces.com/contest/743/problem/C 题面 Vladik and Chloe decided ...
- Codeforces Round #384 (Div. 2) A. Vladik and flights 水题
A. Vladik and flights 题目链接 http://codeforces.com/contest/743/problem/A 题面 Vladik is a competitive pr ...
- Codeforces Round #384 (Div. 2) E. Vladik and cards 状压dp
E. Vladik and cards 题目链接 http://codeforces.com/contest/743/problem/E 题面 Vladik was bored on his way ...
- Codeforces Round #384 (Div. 2) 734E Vladik and cards
E. Vladik and cards time limit per test 2 seconds memory limit per test 256 megabytes input standard ...
- Codeforces Round #275 (Div. 2) C - Diverse Permutation (构造)
题目链接:Codeforces Round #275 (Div. 2) C - Diverse Permutation 题意:一串排列1~n.求一个序列当中相邻两项差的绝对值的个数(指绝对值不同的个数 ...
- Codeforces Round #297 (Div. 2)A. Vitaliy and Pie 水题
Codeforces Round #297 (Div. 2)A. Vitaliy and Pie Time Limit: 2 Sec Memory Limit: 256 MBSubmit: xxx ...
- Codeforces Round #298 (Div. 2) A、B、C题
题目链接:Codeforces Round #298 (Div. 2) A. Exam An exam for n students will take place in a long and nar ...
- Codeforces Round #384 (Div. 2) //复习状压... 罚时爆炸 BOOM _DONE
不想欠题了..... 多打打CF才知道自己智商不足啊... A. Vladik and flights 给你一个01串 相同之间随便飞 没有费用 不同的飞需要费用为 abs i-j 真是题意杀啊, ...
- Codeforces Round #384 (Div. 2)A,B,C,D
A. Vladik and flights time limit per test 2 seconds memory limit per test 256 megabytes input standa ...
随机推荐
- AlloyTouch Button插件-不再愁click延迟和点击态
移动端不能使用click,因为click会有300ms.所有有了fastclick这样的解决方案.然后fastclick并没有解决点击态(用户点击的瞬间要有及时的外观变化反馈)的问题.hover会有不 ...
- jQuery的性能优化
原文链接:http://www.colotu.com/html/gcs/6.html 现在越来越多的人应用jQuery了,有些同学在享受爽快淋漓coding时就将性能问题忽略了, 比如我. jquer ...
- Android 手机卫士--导航界面3、4和功能列表界面跳转逻辑处理
刚刚花了一点时间,将导航界面3.4的布局和相应的跳转逻辑写了一下: Setup3Activity代码如下: /** * Created by wuyudong on 2016/10/10. */ pu ...
- Android开发案例 - 图库
本文不涉及UI方面的内容, 如果您是希望了解UI方面的访客, 请跳过此文. 本文将要详细介绍如何实现流畅加载本地图库. 像平时用得比较多应用, 如微信(见下图), 微博等应用, 都实现了图库功能, 其 ...
- [Erlang 0129] Erlang 杂记 VI
把之前阅读资料的时候记下的东西,整理了一下. Adding special-purpose processor support to the Erlang VM P23 简单介绍了Erlang C ...
- 在DBeaver中phoenix查询报错:org.apache.phoenix.exception.PhoenixIOException: The system cannot find the path specified
环境:Phoenix:4.4,win7系统 问题:Phoenix在查询hbase时,报"系统找不到指定路径". 解决: 请参见 https://distcp.quora.com/C ...
- Mysql常用函数,难点,注意
一.数学函数 ABS(x) 返回x的绝对值 BIN(x) 返回x的二进制(OCT返回八进制,HEX返回十六进制) CEILING(x) 返回大于x的最小整数值 EXP(x) 返回值e( ...
- Linux基础练习题(二)
Linux基础练习题(二) 1.复制/etc/skel目录为/home/tuer1,要求/home/tuser1及其内部文件的属组和其它用户均没有任何访问权限. [root@www ~]# cp -r ...
- Linux 关于Transparent Hugepages的介绍
透明大页介绍 Transparent Huge Pages的一些官方介绍资料: Transparent Huge Pages (THP) are enabled by default in RHEL ...
- 安全测试 - 端口嗅探工具Nmap
Nmap 在官网下载nmap端口检测工具https://nmap.org/,nmap是一个网络连接端扫描软件,用来扫描网上电脑开放的网络连接端. 使用: 通过cmd命令:nmap www.5i5j.c ...