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

D. Mike and Feet

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

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;

 #pragma comment(linker, "/STACK:102400000,102400000")
 #include <cstdio>
 #include <iostream>
 #include <cstdlib>
 #include <cstring>
 #include <algorithm>
 #include <cmath>
 #include <cctype>
 #include <map>
 #include <set>
 #include <queue>
 #include <bitset>
 #include <string>
 #include <complex>
 #define ll long long
 #define mod 1000000007
 using namespace std;
 int n;
 int a[];
 int l[];
 int r[];
 int ans[];
 int dp[];
 int main()
 {
         scanf("%d",&n);
         for(int i=;i<=n;i++)
         scanf("%d",&a[i]);
         a[]=-;
         a[n+]=-;
         l[]=;
         for(int i=; i<=n; i++) //关键********
         {
             int temp=i-;
             while(a[temp]>=a[i])//维护一个递增的序列
                 temp=l[temp]-;
             l[i]=temp+;
         }
         r[n]=n;
         for (int i=n-; i>=; i--)
         {
             int temp=i+;
             while(a[temp]>=a[i])
                 temp=r[temp]+;
             r[i]=temp-;
         }
         for(int i=;i<=n;i++)
             dp[r[i]-l[i]+]=max(dp[r[i]-l[i]+],a[i]);
         int res=;
         for(int i=n;i>=;i--){
             res=max(res,dp[i]);
             ans[i]=res;
         }
         for(int i=;i<=n;i++)
             printf("%d ",ans[i]);
         return ;
 }