B. sland Puzzle
题目连接：
http://www.codeforces.com/contest/635/problem/B
Description
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 = 0 indicates 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.
Sample Input
3
1 0 2
2 0 1
Sample Output
YES
Hint
题意
给你一个起始串，然后这个起始串的0可以和他的周围的数字交换位置（是环状的）
问你经过若干次交换之后
能否变成下面那个目标串
题解：
显然交换不改变相对位置
所以只要相对位置相同就好了
不相同就是NO
代码
#include<bits/stdc++.h>
using namespace std;
const int maxn = 200005;
int a[maxn],b[maxn],tot1,tot2;
int main()
{
int n;
scanf("%d",&n);
for(int i=0;i<n;i++)
{
int x;scanf("%d",&x);
if(x==0)continue;
a[tot1++]=x;
}
for(int i=0;i<n;i++)
{
int x;scanf("%d",&x);
if(x==0)continue;
b[tot2++]=x;
}
int s=0;
for(int i=0;i<tot2;i++)
{
if(b[i]==a[0])
{
s=i;
break;
}
}
for(int i=0;i<n-1;i++)
if(a[i]!=b[(i+s)%(n-1)])
return puts("NO");
return puts("YES");
}