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?
The first line contains a single integer n (1 ≤ n ≤ 105) — the size of the permutations. The second line contains permutation a as ndistinct numbers from 1 to n, inclusive. The numbers are separated with single spaces. The third line contains permutation b in the same format.
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.
2
1 2
2 1
1
0
4
2 1 3 4
3 4 2 1
2
1
0
1
题意：给出一个1——n的排列a，再给出一个1——n的排列b。若a[i]==b[j]，则dis=|i-j|
每次将b序列左移1位，移n-1次，序列第一个补到最后面，问初始序列以及每次左移后的序列最小的dis
若b[j]在a[i]的左边，随着每次左移，两点间dis+1
若b[j]在a[i]的右边，随着每次左移，两点间dis-1
可以统计到这一次左移，一共加了多少1，减了多少1，出队入队的时候再考虑这些1，
即延迟标记
每次的答案，从>=0的里面找最小的，<0的里面找最大的，两者再取最小
multiset 模拟每次把第一个挪到最后一个的过程
解释最后一行：n-a[x]是第一个拿到最后一个的实际距离，再加i+1是延迟标记
#include<cstdio>
#include<set>
#include<algorithm>
using namespace std;
multiset<int>s;
multiset<int>::iterator it;
int a[100001],ans,b[100001];
int main()
{
int n,x;
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
scanf("%d",&x);
a[x]=i;
}
for(int i=1;i<=n;i++)
{
scanf("%d",&b[i]);
s.insert(i-a[b[i]]);
}
for(int i=0;i<n;i++)
{
it=s.lower_bound(i);
ans=1e5+10;
if(it!=s.end()) ans=min(ans,*it-i);
if(it!=s.begin()) ans=min(ans,i-(*--it));
printf("%d\n",ans);
x=b[i+1];
s.erase(s.find(i+1-a[x]));
s.insert(i+1-a[x]+n);
}
}
