ACM思维题训练集合 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 这个题预处理一下初始的dis，然后用mutiset模拟即可
代码语言：javascript复制#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("%dn", ans);
po = ms.find(i - a[b[i]]);
ms.erase(po);
ms.insert(i - a[b[i]] n);
}
}
