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

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

Mike is the president of country What-The-Fatherland. There areNBears living in this country besides Mike. All of them are standing in a line and they are numbered from 1NFrom left to right.I-Th bear is exactlyAIFeet 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 eachXSuch that 1 limit ≤ limitXLimit ≤ limitNThe maximum strength among all groups of sizeX.

Input

The first line of input contains integerN(1 digit ≤ DigitNLimit ≤ limit 2 × hour 105), the number of bears.

The second line containsNIntegers separated by space,A1, bytes,A2, middle..., middle ,...,AN(1 digit ≤ DigitAIMemory ≤ memory 109), heights of bears.

Output

PrintNIntegers in one line. For eachXFrom 1N, Print the maximum strength among all groups of sizeX.

Sample test (s) input

101 2 3 4 5 4 3 2 1 6

Output

6 4 4 3 3 2 2 1 1

You can use a monotonous stack to maintain a stack and record minmum (minimum value of the interval) and count (total length of the interval ).

 

# Include
 
  
# Include
  
   
# Include
   
    
# Include
    
     
# Define deusing namespace std; # define maxn 200060int 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]
      
       
/* The ans [I] calculated here is the minimum value of the continuous length range with I, but this minimum value is the maximum value of all consecutive lengths with I, if ans [I + 1] is larger than ans [I], ans [I] can be updated to ans [I + 1]. if the maximum value of the minimum value of an I + continuous number interval is B, remove a number and the maximum value of the continuous number interval with the length of I is 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;}