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.

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.

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

10
1 2 3 4 5 4 3 2 1 6

6 4 4 3 3 2 2 1 1 1

这题可以用单调栈做，维护一个栈，记录minmum(该区间的最小值）和count（区间的总长度）。

?

#include
  
   
#include
   
     #include
    
      #include
     
       #include
      
        using namespace std; #define maxn 200060 int ans[maxn]; struct node{ int count,minmum; }stack[maxn]; int main() { int n,m,i,j,top,count,b; while(scanf(%d,&n)!=EOF) { memset(ans,0,sizeof(ans)); top=0; for(i=1;i<=n;i++){ scanf(%d,&b); count=0; while(top>0 && stack[top].minmum>=b){ stack[top].count+=count; count=stack[top].count; if(ans[count]
       
        0){ stack[top].count+=count; count=stack[top].count; if(ans[count]
        
         
                /*这里算出来的ans[i]是连续长度为i的区间的最小值，但这个最小值是所有连续长度为i的区间长度的最大值，下面如果ans[i+1]比ans[i]大，那么ans[i]可以更新为ans[i+1],因为如果i+1个连续数区间的最小值的最大值是b,那么去掉一个数，一定可以做到长度为i的连续数区间的最大值是b。*/
 
         
		for(i=n;i>=2;i--){
			if(ans[i]>ans[i-1]){
				ans[i-1]=ans[i];
			}
		}
		for(i=1;i<=n-1;i++){
			printf(%d ,ans[i]);
		}
		printf(%d
,ans[i]);
	}
	return 0;
}
 
         
 
         
 
         
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