Mike and Feet CodeForces - 548D （单调栈）

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

思路：

我们可以用单调栈O（N）维护出每一个数在数字中左边第一个比它大的坐标和右边第一个比它大的坐标。

那么每一个数a[i] 在数组中为最小值的最大区间就可以求出，

然后我们可以计算每一个数a[i] 对答案的贡献，

那么肯定还有一些区间长度没有被更新到，

可以根据ans 数组的定义来获得，

如果 没有一个数a[i] 为最小值的区间长度为3，而又一个数a[j] 为最小值的区间长度为4

那么ans[3] 就可以等于 ans[4] ，因为 a[j] 长度为4的区间 去掉开头或者末尾任意一个位置，就可以得到长度为3的区间。

细节见代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) {ll ans = 1; while (b) {if (b % 2)ans = ans * a % MOD; a = a * a % MOD; b /= 2;} return ans;}
inline void getInt(int* p);
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
stack<int> st;
int l[maxn];
int r[maxn];
int a[maxn];
int n;
void solveinfo()
{
	a[0]=a[n+1]=-inf;
	while(st.size())
	{
		st.pop();
	}
	repd(i,0,n+1)
	{
		while(st.size()&&a[i]<=a[st.top()])
		{
			st.pop();
		}
		if(st.size())
		{
			l[i]=st.top();
		}else
		{
			l[i]=-1;
		}
		st.push(i);
	}
	while(st.size())
	{
		st.pop();
	}
	for(int i=n+1;i>=1;--i)
	{
		while(st.size()&&a[i]<=a[st.top()])
		{
			st.pop();
		}
		if(st.size())
		{
			r[i]=st.top();
		}else
		{
			r[i]=-1;
		}
		st.push(i);
	}
}
int ans[maxn];
int main()
{
	//freopen("D:\\code\\text\\input.txt","r",stdin);
	//freopen("D:\\code\\text\\output.txt","w",stdout);
	gbtb;
	cin>>n;
	repd(i,1,n)
	{
		cin>>a[i];
	}
	solveinfo();
	MS0(ans);
	repd(i,1,n)
	{
		int len=r[i]-l[i]-1;
		// cout<<i<<" "<<r[i]<<" "<<l[i]<<" "<<" "<<len<<endl;
		ans[len]=max(ans[len],a[i]);
	}
	for(int i=n-1;i>=1;--i)
	{
		ans[i]=max(ans[i],ans[i+1]);
	}
	repd(i,1,n)
	{
		printf("%d%c",ans[i],i==n?'\n':' ');
	}
	return 0;

}

inline void getInt(int* p) {
	char ch;
	do {
		ch = getchar();
	} while (ch == ' ' || ch == '\n');
	if (ch == '-') {
		*p = -(getchar() - '0');
		while ((ch = getchar()) >= '0' && ch <= '9') {
			*p = *p * 10 - ch + '0';
		}
	}
	else {
		*p = ch - '0';
		while ((ch = getchar()) >= '0' && ch <= '9') {
			*p = *p * 10 + ch - '0';
		}
	}
}