Problem Description

The Little Elephant has two permutations a and b of length n, consisting of numbers from 1 to n, inclusive. Let's denote the i-th (1?≤?i?≤?n) element of the permutation a as ai, the j-th (1?≤?j?≤?n) element of the permutation b — as bj.

The distance between permutations a and b is the minimum absolute value of the difference between the positions of the occurrences of some number in a and in b. More formally, it's such minimum |i?-?j|, that ai?=?bj.

A cyclic shift number i (1?≤?i?≤?n) of permutation b consisting from n elements is a permutation bibi?+?1... bnb1b2... bi?-?1. Overall a permutation has n cyclic shifts.

The Little Elephant wonders, for all cyclic shifts of permutation b, what is the distance between the cyclic shift and permutation a?

Input

The first line contains a single integer n (1 ≤ n ≤ 105) — the size of the permutations. The second line contains permutation a as n distinct numbers from 1 to n, inclusive. The numbers are separated with single spaces. The third line contains permutation b in the same format.

Output

In n lines print n integers — the answers for cyclic shifts. Print the answers to the shifts in the order of the shifts' numeration in permutation b, that is, first for the 1-st cyclic shift, then for the 2-nd, and so on.

Examples

Input

2
1 2
2 1

Output

1
0

Input

4
2 1 3 4
3 4 2 1

Output

2
1
0
1

題意：給出兩個長度為 n 的序列 a 、b，定義距離為：若 a[i]==b[i]，則 dis=|i-j|，現在要進行 n-1 次操作，每次將 b 序列進行循環左移，要求輸出初始序列以及每次左移后的序列的最小的 dis

思路：

若 b[j] 在 a[i] 的左邊，那么每次移動，dis+1，反之，若 b[j] 在 a[i] 的右邊，每次移動，dis-1

因此使用 multiset 模擬每次挪動的過程，統計每一次左移一共加了多少、減了多少，然后在每次的答案中找大于 0 中最小的，小于 0 中最大的，兩者再取最小

Source Program

#include<iostream> #include<cstdio> #include<cstdlib> #include<string> #include<cstring> #include<cmath> #include<ctime> #include<algorithm> #include<utility> #include<stack> #include<queue> #include<vector> #include<set> #include<map> #include<bitset> #define EPS 1e-9 #define PI acos(-1.0) #define INF 0x3f3f3f3f #define LL long long const int MOD = 1E9+7; const int N = 1000000+5; const int dx[] = {-1,1,0,0,-1,-1,1,1}; const int dy[] = {0,0,-1,1,-1,1,-1,1}; using namespace std;multiset<int> st; int bucket[N],b[N]; int main(){int n,x;scanf("%d",&n);for(int i=1;i<=n;i++){int x;scanf("%d",&x);bucket[x]=i;}for(int i=1;i<=n;i++){scanf("%d",&b[i]);int pos=i-bucket[b[i]];st.insert(pos);}multiset<int>::iterator it;for(int i=0;i<n;i++){it=st.lower_bound(i);//尋找位置int res=INF;if(it!=st.end()){int temp=(*it)-i;res=min(res,temp);}if(it!=st.begin()){it--;int temp=i-(*it);res=min(res,temp);}printf("%d\n",res);int pos=b[i+1];int temp=i+1-bucket[pos];it=st.find(temp);st.erase(it);//刪除第一個temp+=n;st.insert(temp);//插入到最后}return 0; }

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