Codeforces548d:mike and Feet (monotonic stack)

Mike is the president of country What-the-fatherland. There isNBears living in this country besides Mike. All of them is standing in a line and they is numbered from1ToNFrom 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 "the Bear" that group.

Mike is a curious to know for each x such this 1?≤? x? ≤? n The maximum strength among all groups of size X.

Input

The first line of input contains integer n (1?≤? N? ≤?2?x?105), the number of bears.

The second line containsNIntegers separated by space, a1,? a 2,?...,? a N (1?≤? a i? ≤?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

101 2 3 4 5 4 3 2 1 6

Output

Test instructions

Given the number of n, the n number can be divided into intervals when the interval length is I (1~n), each of which will have a minimum value, and the maximum value of these intervals in the same length

Ideas:

Use a monotonic stack to keep the number in the stack monotonically increasing

#include <iostream> #include <stdio.h> #include <string.h> #include <stack> #include <queue > #include <map> #include <set> #include <vector> #include <math.h> #include <bitset># Include <algorithm> #include <climits>using namespace std; #define LS 2*i#define RS 2*i+1#define Up (i,x,y) for (i=x;i<=y;i++) #define DOWN (i,x,y) for (i=x;i>=y;i--) #define MEM (a,x) memset (A,x,sizeof (a)) #define W (a) while (a) #define GCD (A, B) __gcd (A, b) #define LL long long#define N 200005#define MOD 1000000007#define INF 0x3f3f3f3f#define EXP 1e- 8#define lowbit (x) (x&-x)/*num stack the element at the top of the width, and the number of times separated from the current into the stack, because a continuous interval is required, so this also must be recorded */struct node{int num,width    ;    Node () {};    Node (int _num,int _width): num (_num), Width (_width) {}};stack<node> s;int a[n],ans[n];int Main () {int n,i,j;    scanf ("%d", &n);    for (i = 0; i<n; i++) scanf ("%d", &a[i]);    a[n++] = 0;    MEM (ans,0); for (i = 0; i<n; i++) {int len = 0;Length of continuous interval node k; while (!            S.empty ()) {k = S.top ();                if (K.num<a[i]) break; The new stack element is smaller than the top of the stack, so for this continuous interval, this is more than the new stack of elements is useless, you can go out of the stack int ls=k.width+len;//out the stack and get its length if (K.num>ans[ls]            The maximum value of//ans recording ls interval {ans[ls]=k.num;            } len+=k.width;        S.pop ();    } S.push (Node (a[i],len+1));    } for (i = n-1; i>=1; i--)//Because the above only updates a subset of the points, the Ans[i]=max (Ans[i],ans[i+1]) is now updated for those points that are not updated;    printf ("%d", ans[1]);    for (i = 2; i<n; i++) printf ("%d", ans[i]);    printf ("\ n"); return 0;}

Codeforces548d:mike and Feet (monotonic stack)