ACM思維題訓練集合
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
這個題預處理一下初始的dis，然后用mutiset模擬即可

#include <bits/stdc++.h> using namespace std; const int maxn = 100005; int a[maxn],b[maxn]; multiset<int> ms; int main() {int n, x;scanf("%d", &n);for (int i = 0; i < n; ++i){scanf("%d", &x);a[x] = i;}for (int i = 0; i < n; ++i){scanf("%d", &b[i]);ms.insert(i - a[b[i]]);}for (int i = 0; i < n; ++i){auto po = ms.lower_bound(i);int ans = 0X3F3F3F3F;if (po != ms.end())ans = min(ans, *po - i);else if (po != ms.begin())ans = min(ans, i - *(--po));printf("%d\n", ans);po = ms.find(i - a[b[i]]);ms.erase(po);ms.insert(i - a[b[i]] + n);} }

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