A. Taymyr is calling you

time limit per test:1 second
memory limit per test:256 megabytes
input:standard input
output:standard output

Comrade Dujikov is busy choosing artists for Timofey's birthday and is recieving calls from Taymyr from Ilia-alpinist.

Ilia-alpinist calls every n minutes, i.e. in minutes n, 2n, 3n and so on. Artists come to the comrade every m minutes, i.e. in minutes m, 2m, 3m and so on. The day is z minutes long, i.e. the day consists of minutes 1, 2, ..., z. How many artists should be killed so that there are no artists in the room when Ilia calls? Consider that a call and a talk with an artist take exactly one minute.

Input

The only string contains three integers — n, m and z (1 ≤ n, m, z ≤ 104).

Output

Print single integer — the minimum number of artists that should be killed so that there are no artists in the room when Ilia calls.

Examples
Input
1 1 10
Output
10
Input
1 2 5
Output
2
Input
2 3 9
Output
1
Note

Taymyr is a place in the north of Russia.

In the first test the artists come each minute, as well as the calls, so we need to kill all of them.

In the second test we need to kill artists which come on the second and the fourth minutes.

In the third test — only the artist which comes on the sixth minute.

题目链接:http://codeforces.com/contest/764/problem/A

分析:此题竟然是求n与m的最小公倍数,我TM是智障了!试了三次,开始以为就是t/(m×n),智障宝宝!

不说太多,都是泪啊!

下面给出AC代码:

 #include <bits/stdc++.h>
using namespace std;
int gcd(int a,int b)
{
return b==?a:gcd(b,a%b);
}
int main()
{
int n,m,t;
while(cin>>n>>m>>t)
{
int x=gcd(n,m);
int y=n*m/x;
int z=t/y;
cout<<z<<endl;
}
return ;
}

B. Timofey and cubes

time limit per test:1 second
memory limit per test:256 megabytes
input:standard input
output:standard output

Young Timofey has a birthday today! He got kit of n cubes as a birthday present from his parents. Every cube has a number ai, which is written on it. Timofey put all the cubes in a row and went to unpack other presents.

In this time, Timofey's elder brother, Dima reordered the cubes using the following rule. Suppose the cubes are numbered from 1 to n in their order. Dima performs several steps, on step i he reverses the segment of cubes from i-th to (n - i + 1)-th. He does this while i ≤ n - i + 1.

After performing the operations Dima went away, being very proud of himself. When Timofey returned to his cubes, he understood that their order was changed. Help Timofey as fast as you can and save the holiday — restore the initial order of the cubes using information of their current location.

Input

The first line contains single integer n (1 ≤ n ≤ 2·105) — the number of cubes.

The second line contains n integers a1, a2, ..., an ( - 109 ≤ ai ≤ 109), where ai is the number written on the i-th cube after Dima has changed their order.

Output

Print n integers, separated by spaces — the numbers written on the cubes in their initial order.

It can be shown that the answer is unique.

Examples
Input
7 
4 3 7 6 9 1 2
Output
2 3 9 6 7 1 4
Input
8 
6 1 4 2 5 6 9 2
Output
2 1 6 2 5 4 9 6
Note

Consider the first sample.

  1. At the begining row was [2, 3, 9, 6, 7, 1, 4].
  2. After first operation row was [4, 1, 7, 6, 9, 3, 2].
  3. After second operation row was [4, 3, 9, 6, 7, 1, 2].
  4. After third operation row was [4, 3, 7, 6, 9, 1, 2].
  5. At fourth operation we reverse just middle element, so nothing has changed. The final row is [4, 3, 7, 6, 9, 1, 2]. So the answer for this case is row [2, 3, 9, 6, 7, 1, 4].

题目链接:http://codeforces.com/contest/764/problem/B

分析:智障宝宝继续犯傻,看了半天,以为规律就是奇偶变换,结果样例都没过,再回头看了下题,傻了眼,原来就是每两项第i项和第n-i+1项交换!

智障宝宝第二摔!

下面给出AC代码:

 #include <bits/stdc++.h>
using namespace std;
int a[];
int main()
{
int n;
while(cin>>n)
{
for(int i=;i<=n;i++)
cin>>a[i];
for(int i=;i<=(n+)/;i+=)
{
swap(a[i],a[n-i+]);
}
for(int i=;i<=n-;i++)
cout<<a[i]<<" ";
cout<<a[n]<<endl;
}
return ;
}

Codeforces Round #395 (Div. 2)(A.思维,B,水)的更多相关文章

  1. Codeforces Round #367 (Div. 2) A. Beru-taxi (水题)

    Beru-taxi 题目链接: http://codeforces.com/contest/706/problem/A Description Vasiliy lives at point (a, b ...

  2. Codeforces Round #603 (Div. 2) A. Sweet Problem(水.......没做出来)+C题

    Codeforces Round #603 (Div. 2) A. Sweet Problem A. Sweet Problem time limit per test 1 second memory ...

  3. Codeforces Round #334 (Div. 2) A. Uncowed Forces 水题

    A. Uncowed Forces Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/604/pro ...

  4. Codeforces Round #285 (Div. 2) A, B , C 水, map ,拓扑

    A. Contest time limit per test 1 second memory limit per test 256 megabytes input standard input out ...

  5. 【CF1256】Codeforces Round #598 (Div. 3) 【思维+贪心+DP】

    https://codeforces.com/contest/1256 A:Payment Without Change[思维] 题意:给你a个价值n的物品和b个价值1的物品,问是否存在取物方案使得价 ...

  6. Codeforces Round #143 (Div. 2) (ABCD 思维场)

    题目连链接:http://codeforces.com/contest/231 A. Team time limit per test:2 seconds memory limit per test: ...

  7. Codeforces Round #416 (Div. 2)(A,思维题,暴力,B,思维题,暴力)

    A. Vladik and Courtesy time limit per test:2 seconds memory limit per test:256 megabytes input:stand ...

  8. Codeforces Round #533 (Div. 2) C.思维dp D. 多源BFS

    题目链接:https://codeforces.com/contest/1105 C. Ayoub and Lost Array 题目大意:一个长度为n的数组,数组的元素都在[L,R]之间,并且数组全 ...

  9. Codeforces Round #539 (Div. 2) D 思维

    https://codeforces.com/contest/1113/problem/D 题意 将一个回文串切成一段一段,重新拼接,组成一个新的回文串,问最少切几刀 题解 首先无论奇偶串,最多只会切 ...

随机推荐

  1. js登录滑动验证,不滑动无法登陆

    js的判断这里是根据滑块的位置进行判断,应该是用一个flag判断 <%@ page language="java" contentType="text/html; ...

  2. sqlserver 存储过程 递归查询分组+hierarchyid重建会员关系

    CREATE PROCEDURE [dbo].[GetGroupInfo] @s_code NVARCHAR() = --会员卡号 AS BEGIN declare @p int; --查询唯一性结果 ...

  3. WindowsServer2012 搭建域错误“本地Administraor账户不需要密码”

    标签:MSSQL/SQLServer/域控制器提升的先决条件验证失败/密码不符合要求 概述 在安装WindowsServer2012域控出现administrator账户密码不符合要求的错误,但是实际 ...

  4. 自己动手写把”锁”之---JMM和volatile

    一.JAVA内存模型 关于Java内存模型的文章,网上真的数不胜数.在这里我就不打算说的很详细.很严谨了.只力求大家能更好的理解和运用,为后边的技术点做铺垫.   内存模型并不是Java独有的概念,而 ...

  5. 在亚马逊linux环境上装mysql+添加启动项

    安装mysql sudo yum install mysql sudo yum install mysql-server sudo yum install mysql-devel 添加到系统启动项su ...

  6. 解析JSON的两种方法eval()和JSON.parse()

    解析JSON 一种方法是使用eval函数. var dataObj = eval("("+json+")"); 必须把文本包围在括号中,这样才能避免语法错误,迫 ...

  7. 日语编程语言"抚子" - 第三版特色初探

    原址: https://zhuanlan.zhihu.com/p/30800689 原文: 日语编程语言"抚子" - 第三版特色初探 它山之石可以攻玉. 学习其他的母语编程语言, ...

  8. C# 将DataTable一行放入另一个DataTable中

    http://blog.csdn.net/huyu107/article/details/53509171 概述 从一个DataTable中取一行放到另一个DataTable里报错: 该行已经属于另一 ...

  9. 一些JavaScript技巧

    1.获取浏览器的高度和宽度(不包括工具栏和滚动条): var w=window.innerWidth //现代浏览器 || document.documentElement.clientWidth / ...

  10. 关于 Python 入门的一些问题?

    一.用 python 能够做什么?解决什么问题? A1:理论上来说,计算机能做什么,python 语言就能让它做什么,也即 python能做什么. 数值计算.机器学习.爬虫.云相关开发.自动化测试.运 ...