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.
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.
Print "YES" (without quotes) if the rearrangement can be done in the existing network, and "NO" otherwise.
3
1 0 2
2 0 1
YES
2
1 0
0 1
YES
4
1 2 3 0
0 3 2 1
NO
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.
回过头来想想这题其实挺水的，然而我却没做出来
题意：就是n个数编号，并且首尾相连，其中0能和的交换，判断能不能通过交换达到下面的目的序列
分析：其实不管怎么怎么于0交换，他们的顺序是不变的，就是从第一组数0后面的数的顺序跟第二组是一样的
1 #include <cstdio> 2 #include <algorithm> 3 #include <iostream> 4 using namespace std; 5 typedef long long LL; 6 const int Max = 200000; 7 int a[Max + 5], b[Max + 5]; 8 int main() 9 { 10 int n, s, t; 11 scanf("%d", &n); 12 for(int i = 0; i < n; i++) 13 { 14 scanf("%d", &a[i]); 15 if(a[i] == 0) 16 s = i; 17 } 18 s = (s + 1) % n; 19 for(int i = 0; i < n; i++) 20 { 21 scanf("%d", &b[i]); 22 if(b[i] == a[s]) 23 t = i; 24 } 25 int m = n - 1; 26 int flag = false; 27 while(m--) 28 { 29 if(a[s] == 0) 30 s = (s + 1) % n; 31 if(b[t] == 0) 32 t = (t + 1) % n; 33 if(a[s] == b[t]) 34 { 35 s = (s + 1) % n; 36 t = (t + 1) % n; 37 continue; 38 } 39 else 40 { 41 flag = true; 42 break; 43 } 44 } 45 if(flag) 46 printf("NO "); 47 else 48 printf("YES "); 49 return 0; 50 }