B. Mike and Feet

 

B. Mike and Feet

time limit per test

1 second

memory limit per test

256 megabytes

 

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.

input

10
1 2 3 4 5 4 3 2 1 6

output

6 4 4 3 3 2 2 1 1 1 

 Solution

题意:给你一个长度为N的序列, 对于（1，n）长度，让你找到线段的最小值的最大值是多少

题解:我们处理出L,R两个数组,表示一个数字为线段最小值时的最右边界和最左边界,然后DP一下就好了.

#include<bits/stdc++.h>
using namespace std;
const int maxn = 200000;
int n,a[maxn + 99],f[maxn + 99],l[maxn + 99],r[maxn + 99];
int main(){
    scanf("%d",&n);
    for (int i = 1 ; i <= n ; ++i) scanf("%d",&a[i]);
    a[0] = -1; a[n + 1] = -1;
    for (int i = 1 ; i <= n ; ++i){
        int j = i - 1;
        while (a[i] <= a[j]) j = l[j];
        l[i] = j;
    }
    for (int i = n ; i >= 1 ; --i){
        int j = i + 1;
        while (a[i] <= a[j]) j = r[j];
        r[i] = j;
    }
    for (int i = 1 ; i <= n ; ++i)    f[r[i] - l[i] - 1] = max(f[r[i] - l[i] - 1],a[i]);
    for (int i = n - 1 ; i >= 1 ; --i) f[i] = max(f[i],f[i+1]);
      for (int i = 1 ; i <= n ; ++i) cout<<f[i]<<" ";
    return 0;
}