B. Case of Fake Numbers

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/556/problem/B

Description

Andrewid the Android is a galaxy-famous detective. He is now investigating a case of frauds who make fake copies of the famous Stolp's gears, puzzles that are as famous as the Rubik's cube once was.

Its most important components are a button and a line of n similar gears. Each gear has n teeth containing all numbers from 0 to n - 1 in the counter-clockwise order. When you push a button, the first gear rotates clockwise, then the second gear rotates counter-clockwise, the the third gear rotates clockwise an so on.

Besides, each gear has exactly one active tooth. When a gear turns, a new active tooth is the one following after the current active tooth according to the direction of the rotation. For example, if n = 5, and the active tooth is the one containing number 0, then clockwise rotation makes the tooth with number 1 active, or the counter-clockwise rotating makes the tooth number 4 active.

Andrewid remembers that the real puzzle has the following property: you can push the button multiple times in such a way that in the end the numbers on the active teeth of the gears from first to last form sequence 0, 1, 2, ..., n - 1. Write a program that determines whether the given puzzle is real or fake.

Input

The first line contains integer n (1 ≤ n ≤ 1000) — the number of gears.

The second line contains n digits a1, a2, ..., an (0 ≤ ai ≤ n - 1) — the sequence of active teeth: the active tooth of the i-th gear contains number ai.

Output

In a single line print "Yes" (without the quotes), if the given Stolp's gears puzzle is real, and "No" (without the quotes) otherwise.

Sample Input

3
1 0 0

Sample Output

Yes

HINT

题意

给n个齿轮,每按一下,这些齿轮都会转

奇数位和偶数位的转动方向不一样

最后问能不能变成0,1,2,3,4...n-1

题解:

假设第一位方向是+1,那么就实质上第一个数就确定了转多少下了

于是判一下就好了

代码

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef unsigned long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 2000001
#define mod 1000000007
#define eps 1e-9
int Num;
char CH[];
const int inf=0x3f3f3f3f;
inline ll read()
{
int x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
} //************************************************************************************** int main()
{
int n,d;
cin >> n >> d;
for(int i = ; i < n; ++i)
{
int a;
cin >> a;
if(i!=(a+d*(i&?:-)+n)%n)
{
cout << "No" << endl;
return ;
}
}
cout << "Yes" << endl;
}

Codeforces Round #310 (Div. 2) B. Case of Fake Numbers 水题的更多相关文章

  1. 构造 Codeforces Round #310 (Div. 2) B. Case of Fake Numbers

    题目传送门 /* 题意:n个数字转盘,刚开始每个转盘指向一个数字(0~n-1,逆时针排序),然后每一次转动,奇数的+1,偶数的-1,问多少次使第i个数字转盘指向i-1 构造:先求出使第1个指向0要多少 ...

  2. 贪心/思维题 Codeforces Round #310 (Div. 2) C. Case of Matryoshkas

    题目传送门 /* 题意:套娃娃,可以套一个单独的娃娃,或者把最后面的娃娃取出,最后使得0-1-2-...-(n-1),问最少要几步 贪心/思维题:娃娃的状态:取出+套上(2),套上(1), 已套上(0 ...

  3. 找规律/贪心 Codeforces Round #310 (Div. 2) A. Case of the Zeros and Ones

    题目传送门 /* 找规律/贪心:ans = n - 01匹配的总数,水 */ #include <cstdio> #include <iostream> #include &l ...

  4. Codeforces Round #368 (Div. 2) A. Brain's Photos (水题)

    Brain's Photos 题目链接: http://codeforces.com/contest/707/problem/A Description Small, but very brave, ...

  5. Codeforces Round #373 (Div. 2) C. Efim and Strange Grade 水题

    C. Efim and Strange Grade 题目连接: http://codeforces.com/contest/719/problem/C Description Efim just re ...

  6. Codeforces Round #185 (Div. 2) A. Whose sentence is it? 水题

    A. Whose sentence is it? Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/ ...

  7. Codeforces Round #373 (Div. 2) A. Vitya in the Countryside 水题

    A. Vitya in the Countryside 题目连接: http://codeforces.com/contest/719/problem/A Description Every summ ...

  8. Codeforces Round #371 (Div. 2) A. Meeting of Old Friends 水题

    A. Meeting of Old Friends 题目连接: http://codeforces.com/contest/714/problem/A Description Today an out ...

  9. Codeforces Round #355 (Div. 2) B. Vanya and Food Processor 水题

    B. Vanya and Food Processor 题目连接: http://www.codeforces.com/contest/677/problem/B Description Vanya ...

随机推荐

  1. Groovy读取properties及txt

    昨晚帮老同事解决了一个SoapUI的代码问题,好长时间没用SoapUI,好多东西都忘了,今天先总结下Groovy读取properties 首先吐槽下SoapUI的apidocs,我几乎从中看不出什么东 ...

  2. [转]Linux之type命令

    转自:http://codingstandards.iteye.com/blog/831504 用途说明 type命令用来显示指定命令的类型.一个命令的类型可以是如下之一 alias 别名 keywo ...

  3. DIV背景半透明文字不半透明的样式

    DIV背景半透明,DIV中的字不半透明 代码如下:<body bgcolor="#336699"> <div style="filter:alpha(o ...

  4. Nodejs_day01

    helloworld.txt的内容: 我是nodejs 阻塞式调用 var fs = require('fs'); var data = fs.readFileSync('helloworld.txt ...

  5. oracle 查看表的相关信息

    1.查看当前用户的表 SELECT * FROM user_tables; 2.查看指定用户的表 SELECT * FROM all_tables WHERE owner = 'SYS';

  6. F#相关图书推荐

    C#与F#编程实践 作      者 [捷] Tomas Petricek,[英] Jon Skeet 著:贾洪峰 译 出 版 社 清华大学出版社 出版时间 2011-10-01 版      次 1 ...

  7. 轻松学习Linux之入门篇

    1.Linux概述: 2.Linux优点 3.linux历史待上传 4.linux部分发行版 5.linux政府扶持 本文出自 "李晨光原创技术博客" 博客,谢绝转载!

  8. 安装完 MySQL 后必须调整的 10 项配置(转)

    英文原文:10 MySQL settings to tune after installation 译文原文:安装完 MySQL 后必须调整的 10 项配置 当我们被人雇来监测MySQL性能时,人们希 ...

  9. uber优步提高成单率,轻松拿奖励!

    滴快车单单2.5倍,注册地址:http://www.udache.com/ 如何注册Uber司机(全国版最新最详细注册流程)/月入2万/不用抢单:http://www.cnblogs.com/mfry ...

  10. [翻译]创建ASP.NET WebApi RESTful 服务(7)

    实现资源分页 本章我们将介绍几种不同的结果集分页方式,实现手工分页,然后将Response通过两个不同的方式进行格式化(通过Response的Envelop元数据或header). 大家都知道一次查询 ...