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[200005];
int l[200005];
int r[200005];
int ans[200005];
int dp[200005];
int main()
{
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
        a[0]=-1;
        a[n+1]=-1;
        l[1]=1;
        for(int i=2; i<=n; i++) //关键********
        {
            int temp=i-1;
            while(a[temp]>=a[i])//维护一个递增的序列
                temp=l[temp]-1;
            l[i]=temp+1;
        }
        r[n]=n;
        for (int i=n-1; i>=1; i--)
        {
            int temp=i+1;
            while(a[temp]>=a[i])
                temp=r[temp]+1;
            r[i]=temp-1;
        }
        for(int i=1;i<=n;i++)
            dp[r[i]-l[i]+1]=max(dp[r[i]-l[i]+1],a[i]);
        int res=0;
        for(int i=n;i>=1;i--){
            res=max(res,dp[i]);
            ans[i]=res;
        }
        for(int i=1;i<=n;i++)
            printf("%d ",ans[i]);
        return 0;
}