Codeforces Round #305 (Div. 2) D 维护单调栈

Mike is the president of country What-The-Fatherland. There are n bears living in this country besides Mike. All of them are standing in a line and they are numbered from 1 to n from left to right. i-th bear is exactly ai feet high.

$\text{[math]}$

A group of bears is a non-empty contiguous segment of the line. The size of a group is the number of bears in that group. The strength of a group is the minimum height of the bear in that group.

Mike is a curious to know for each x such that 1 ≤ x ≤ n the maximum strength among all groups of size x.

Input

The first line of input contains integer n (1 ≤ n ≤ 2 × 105), the number of bears.

The second line contains n integers separated by space, a1, a2, ..., an (1 ≤ ai ≤ 109), heights of bears.

Output

Print n integers in one line. For each x from 1 to n, print the maximum strength among all groups of size x.

Examples

input

10
1 2 3 4 5 4 3 2 1 6

output

6 4 4 3 3 2 2 1 1 1

题意：给你一个长度为n的数列求长度为x x取值(1,n) 的区段最小值的最大值

题解：求以a[i]为最小值的区段的左界右界;

dp[r[i]-l[i]+1]=max(dp[r[i]-l[i]+1],a[i]) 倒序取max得到每个长度的答案;

 1 #pragma comment(linker, "/STACK:102400000,102400000")
 2 #include <cstdio>
 3 #include <iostream>
 4 #include <cstdlib>
 5 #include <cstring>
 6 #include <algorithm>
 7 #include <cmath>
 8 #include <cctype>
 9 #include <map>
10 #include <set>
11 #include <queue>
12 #include <bitset>
13 #include <string>
14 #include <complex>
15 #define ll long long
16 #define mod 1000000007
17 using namespace std;
18 int n;
19 int a[200005];
20 int l[200005];
21 int r[200005];
22 int ans[200005];
23 int dp[200005];
24 int main()
25 {
26         scanf("%d",&n);
27         for(int i=1;i<=n;i++)
28         scanf("%d",&a[i]);
29         a[0]=-1;
30         a[n+1]=-1;
31         l[1]=1;
32         for(int i=2; i<=n; i++) //关键********
33         {
34             int temp=i-1;
35             while(a[temp]>=a[i])//维护一个递增的序列
36                 temp=l[temp]-1;
37             l[i]=temp+1;
38         }
39         r[n]=n;
40         for (int i=n-1; i>=1; i--)
41         {
42             int temp=i+1;
43             while(a[temp]>=a[i])
44                 temp=r[temp]+1;
45             r[i]=temp-1;
46         }
47         for(int i=1;i<=n;i++)
48             dp[r[i]-l[i]+1]=max(dp[r[i]-l[i]+1],a[i]);
49         int res=0;
50         for(int i=n;i>=1;i--){
51             res=max(res,dp[i]);
52             ans[i]=res;
53         }
54         for(int i=1;i<=n;i++)
55             printf("%d ",ans[i]);
56         return 0;
57 }