Mike and Feet（CF 547B）

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 strengthof 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.

Sample test(s)

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n个数。问连续区间长度为1,2,3,4,....n 所相应的区间长度最小值中的最大值是多少。

解题：单调栈。

#include<stdio.h>  
const int N = 200005;  
struct NODE  
{  
    int h,w;  
}S[N];  
int h[N],ans[N];  
int main()  
{  
    int n;  
    scanf("%d",&n);  
    for(int i=0; i<n; i++)  
        scanf("%d",&h[i]),ans[i]=0;  
    h[n++]=0;ans[n]=0;  
    int sum,top=0;  
    for(int i=0; i<n; i++){  
        sum=0;  
        while(top>0 && S[top].h>=h[i]){  
            sum+=S[top].w;  
            if(ans[sum]<S[top].h)  
                ans[sum]=S[top].h;  
            --top;  
        }  
        S[++top].h=h[i]; S[top].w=sum+1;  
    }  
    n--;  
      
    /* 
     长度为i 的连续数中ans[i]是这i个数的最小数，但却是全部长度为i 的连续数中 
     最小中的最大数。长度i能够依据长度i+1更新大小。原因是ans[i+1]比在此区间内 
     的数都要小于等于，所以去掉边上的一个数答案不影响。
对于假设要求区间内的 
        最大值   ，仅仅需对以下的循环倒过来且比較符取反就可以。同理。
 
     */  
    for(int i=n-1; i>=1; i--)  
        if(ans[i]<ans[i+1])  
        ans[i]=ans[i+1];  
  
    for(int i=1; i<n; i++)  
        printf("%d ",ans[i]);  
    printf("%d\n",ans[n]);  
}