Codeforces Round #310 (Div. 2) B. Case of Fake Numbers 水题
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
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
- 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 ...
- 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 ...
- 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 ...
- 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, ...
- 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 ...
- 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/ ...
- 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 ...
- 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 ...
- 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
- 读取raw目录中的文件数据
try { InputStream is2 = getResources().openRawResource(R.raw.info); InputStreamReader isr2 = new Inp ...
- Delphi Val函数
在这里Val和iif都是你所用的数据库中的函数在delphi中Val是一个将字符串转换为数字的函数,Val(S; var V; var Code: Integer)第一个参数是要转换的字符串,第二个参 ...
- sqlserver数据可空插入报错
数据库和C#中均为可空类型. 这时候直接给字段赋值为null parameters[9].Value = null : 执行的时候报错了,一大堆,总之说它少了一个参数. 用sql ser ...
- 单源最短路径的Bellman-Ford 算法
1.算法标签 BFS 2.算法概念 Bellman-Ford算法有这么一个先验知识在里面,那就是最短路径至多在N步之内,其中N为节点数,否则说明图中有负权值的回路,这样的图是找不到最短路径的.因此Be ...
- Ubuntu14.04下安装QQ 国际版
在/etc/apt/source.list文件中添加: deb http://packages.linuxdeepin.com/deepin trusty main non-free universe ...
- PHP上传大文件和处理大数据
1. 上传大文件 /* 以1.5M/秒的速度写入文件,防止一次过写入文件过大导致服务器出错(chy/20150327) */ $is_large_file = false; if( strlen($x ...
- 详解Android定位
相信很多的朋友都有在APP中实现定位的需求,今天我就再次超炒冷饭,为大家献上国内开发者常用到的三种定位方式.它们分别为GPS,百度和高德,惯例先简单介绍下定位的背景知识. 什么是GPS定位.基站定位和 ...
- chrome浏览器插件window resizer调试webapp页面大小
chrome浏览器插件window resizer可以调整当前浏览器分辨率大小 可以自定义大小,以适合于andorid和iphone设备
- Flipping Parentheses(CSU1542 线段树)
http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1542 赛后发现这套题是2014东京区域赛的题目,看了排名才发现自己有多low = =! 题目大意 ...
- POJ2155Matrix(二维线段树)
链接http://poj.org/problem?id=2155 题目操作就是说,每次操作可以是编辑某个矩形区域,这个区域的0改为1,1改为0,每次查询只查询某一个点的值是0还是1. 方法:二维线段树 ...