Codeforces Round #305 (Div. 1)-B. Mike and Feet(单调栈)

原题链接

B. 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.

Examples

input

10
1 2 3 4 5 4 3 2 1 6

output

6 4 4 3 3 2 2 1 1 1 

对与输入的每个数num[i]用单调栈可以找到最大的l[i](l[i] < i, num[l[i]] < num[i])和最小的r[i](r[i] > i, num[r[i]] < num[i]),

那么num[i]为最小值的区间的最大长度为r[i] - l[i] - 1;

用ans[k]表示长度为K的区间的最小值的最大值.

因为如果有一个长度为6的窗口，里面的最小值的最大值是10，那么长度为5的窗口的最小值的最大值肯定不会小于10. 
所以我们在计算完各个长度的答案后还要从后往前更新一遍值。

#include 
#define INF 1e9
#define maxn 200005
using namespace std;
typedef long long ll;

int num[200005], l[maxn], r[maxn], ans[maxn];
stack s;
int main(){
	
//	freopen("in.txt", "r", stdin);
	int n;
	
	scanf("%d", &n);
	for(int i = 1; i <= n; i++)
	 scanf("%d", num+i);
	for(int i = 1; i <= n; i++){
		while(!s.empty() && num[s.top()] >= num[i])
		 s.pop();
		if(s.empty())
		 l[i] = 0;
		else
		 l[i] = s.top();
		s.push(i);
	}
	while(!s.empty())
	 s.pop();
	for(int i = n; i >= 1; i--){
		while(!s.empty() && num[s.top()] >= num[i])
		 s.pop();
		if(s.empty())
		 r[i] = n + 1;
		else
		 r[i] = s.top();
		s.push(i);
	}
	for(int i = 1; i <= n; i++){
		int len = r[i] - l[i] - 1;
		ans[len] = max(ans[len], num[i]);
	}
	for(int i = n-1; i >= 1; i--){
		ans[i] = max(ans[i], ans[i+1]);
	}
	printf("%d", ans[1]);
	for(int i = 2; i <= n; i++)
	 printf(" %d", ans[i]);
	puts("");
	
	return 0;
}