A. Island Puzzle
time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

A remote island chain contains n islands, labeled 1 through n. Bidirectional bridges connect the islands to form a simple cycle — a bridge connects islands 1 and 2, islands 2 and 3, and so on, and additionally a bridge connects islands n and 1. The center of each island contains an identical pedestal, and all but one of the islands has a fragile, uniquely colored statue currently held on the pedestal. The remaining island holds only an empty pedestal.

The islanders want to rearrange the statues in a new order. To do this, they repeat the following process: First, they choose an island directly adjacent to the island containing an empty pedestal. Then, they painstakingly carry the statue on this island across the adjoining bridge and place it on the empty pedestal.

Determine if it is possible for the islanders to arrange the statues in the desired order.

Input

The first line contains a single integer n (2 ≤ n ≤ 200 000) — the total number of islands.

The second line contains n space-separated integers ai (0 ≤ ai ≤ n - 1) — the statue currently placed on the i-th island. If ai = 0, then the island has no statue. It is guaranteed that the ai are distinct.

The third line contains n space-separated integers bi (0 ≤ bi ≤ n - 1) — the desired statues of the ith island. Once again, bi = 0indicates the island desires no statue. It is guaranteed that the bi are distinct.

Output

Print "YES" (without quotes) if the rearrangement can be done in the existing network, and "NO" otherwise.

Examples
input

Copy
3
1 0 2
2 0 1
output

Copy
YES
input

Copy
2
1 0
0 1
output

Copy
YES
input

Copy
4
1 2 3 0
0 3 2 1
output

Copy
NO
Note

In the first sample, the islanders can first move statue 1 from island 1 to island 2, then move statue 2 from island 3 to island 1, and finally move statue 1 from island 2 to island 3.

In the second sample, the islanders can simply move statue 1 from island 1 to island 2.

In the third sample, no sequence of movements results in the desired position.

【题意】

给定一个字符串你,其中为0的字符可以和相邻的交换位置(环形),问可以不可以变换成给定的字符串

【分析】

显然的交换并不会影响字符串的相对位置,所以记录一下相对位置就可以啦

【代码】

 

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N=2e5+5;
int n,cn1,cn2,a[N],b[N];
inline void Init(){
scanf("%d",&n);
for(int i=1,x;i<=n;i++){scanf("%d",&x);if(x) a[cn1++]=x;}
for(int i=1,x;i<=n;i++){scanf("%d",&x);if(x) b[cn2++]=x;}
}
inline void Solve(){
int pos=-1;
for(int i=0;i<n;i++){
if(b[i]==a[0]){
pos=i;break;
}
}
for(int i=0;i<n-1;i++){
if(a[i]!=b[(i+pos)%(n-1)]){
puts("NO");
return ;
}
}
puts("YES");
}
int main(){
Init();
Solve();
return 0;
}

CF 634A Island Puzzle的更多相关文章

  1. codeforce B Island Puzzle

    B. Island Puzzle time limit per test 2 seconds memory limit per test 256 megabytes input standard in ...

  2. codeforces A. Orchestra B. Island Puzzle

    A. Orchestra time limit per test 2 seconds memory limit per test 256 megabytes input standard input ...

  3. codeforces B. Island Puzzle

    B. Island Puzzle time limit per test 2 seconds memory limit per test 256 megabytes input standard in ...

  4. CF 936C Lock Puzzle——构造

    题目:http://codeforces.com/contest/936/problem/C 玩了一个小时,只能想出 5*n 的方法. 经过一番观察?考虑这样构造:已经使得 A 串的一个后缀 = B ...

  5. 贪心/构造/DP 杂题选做Ⅱ

    由于换了台电脑,而我的贪心 & 构造能力依然很拉跨,所以决定再开一个坑( 前传: 贪心/构造/DP 杂题选做 u1s1 我预感还有Ⅲ(欸,这不是我在多项式Ⅱ中说过的原话吗) 24. P5912 ...

  6. 8VC Venture Cup 2016 - Final Round (Div. 2 Edition)

    暴力 A - Orchestra import java.io.*; import java.util.*; public class Main { public static void main(S ...

  7. 冬训 day2

    模拟枚举... A - New Year and Buggy Bot(http://codeforces.com/problemset/problem/908/B) 暴力枚举即可,但是直接手动暴力会非 ...

  8. cf.301.D. Bad Luck Island(dp + probabilities)

    D. Bad Luck Island time limit per test 2 seconds memory limit per test 256 megabytes input standard ...

  9. CF#301 D:Bad Luck Island (概率dp)

    D:Bad Luck Island 一个岛上有r个石头,s个剪子,p个布,他们之间随机挑出两个相遇,如果不是相同物种,就会有一个消失,分别求出最后这座岛上只剩下一个物种的概率. 我们用dp[i][j] ...

随机推荐

  1. cookie实现用户登录验证

    cookie实现用户登录验证 1, INSTALLED_APPS中注册app03 2,在主程序中新建映射关系到app3的url中 from django.conf.urls import url,in ...

  2. leetcode笔记--水箱问题

    类型的引用:Solution *s=new Solution(); 1.Container With Most Water Given n non-negative integers a1, a2, ...

  3. CSS3-loading动画

    (二) 上次分享了四个CSS3的加载动画,今天继续(标题接上一次). 在线demo:http://liyunpei.xyz/loading.html   (持续更新) 请注意:代码中的关键帧动画有的用 ...

  4. android应用程序中获取view的位置

    我们重点在获取view的y坐标,你懂的... 依次介绍以下四个方法: 1.getLocationInWindow int[] position = new int[2]; textview.getLo ...

  5. OpenSceneGraphic 着色器中数组的应用【转】

    https://blog.csdn.net/zsq306650083/article/details/50533480 //osg的写法osg::ref_ptr<osg::StateSet> ...

  6. jquery append 和appendTo

    原文: https://www.cnblogs.com/stitchgogo/p/5721551.html ---------------------------------------------- ...

  7. shell编程学习笔记(十一):Shell中的while/until循环

    shell中也可以实现类似java的while循环 while循环是指满足条件时,进行循环 示例: #! /bin/sh index=10 while [ $index -gt 0 ] do inde ...

  8. 每天一个linux命令(14):head命令

    1.命令简介 head (head) 用来显示档案的开头至标准输出中.如果指定了多于一个文件,在每一段输出前会给出文件名作为文件头.如果不指定文件,或者文件为"-",则从标准输入读 ...

  9. 内核中的锁机制--RCU

    一. 引言 众所周知,为了保护共享数据,需要一些同步机制,如自旋锁(spinlock),读写锁(rwlock),它们使用起来非常简单,而且是一种很有效的同步机制,在UNIX系统和Linux系统中得到了 ...

  10. JAVA8 之 Stream sorted() 示例

    下面代码以自然序排序一个listlist.stream().sorted() 自然序逆序元素,使用Comparator 提供的reverseOrder() 方法list.stream().sorted ...