CF #305(Div.2) D. Mike and Feet(数学推导）

D. 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 × 10^5), the number of bears.

The second line contains n integers separated by space, a1, a2, ..., an (1 ≤ ai ≤ 10^9), 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

10
1 2 3 4 5 4 3 2 1 6

output

6 4 4 3 3 2 2 1 1 1 

#include<stdio.h>
#include<string.h>
#include<algorithm>
const int M = 2e5 + 10 ;
int l[M] , r[M] , a[M] ;
int b[M] ;
int n ;
int main ()
{
   // freopen ("a.txt" , "r" , stdin ) ;
    scanf ("%d" , &n) ;
    for (int i = 1 ; i <= n ; i ++) scanf ("%d" , &a[i]) ;
    a[0] = 0 ; a[n + 1] = 0 ;
    memset (b , - 1 , sizeof(b)) ;
    for (int i = 1 ; i <= n ; i ++) {
        l[i] = i ;
        while (a[l[i] - 1] >= a[i]) l[i] = l[l[i] - 1] ;
    }
    for (int i = n ; i >= 1 ; i --) {
        r[i] = i ;
        while (a[r[i] + 1] >= a[i]) r[i] = r[r[i] + 1] ;
    }
    for (int i = 1 ; i <= n ; i ++) {
        int id = r[i] - l[i] + 1 ;
        b[id] = std::max (b[id] , a[i]);
    }
    for (int i = n ; i >= 2 ; i --) {
        if (b[i - 1] < b[i] ) b[i - 1] = b[i] ;
    }
    for (int i = 1 ; i <= n ; i ++) printf ("%d%c" , b[i] , i == n ? '
' : ' ') ;
    return 0 ;
}
/*10
1 2 3 4 5 4 3 2 1 6
*/

1. 最质补的想法，我们先考虑a[i]能影响的区间大小为w[i]，(另b[i] 为长度为i的区间的最小的最大值）

for i = 1 to n

　　for j = 1 to w[i]

　　　　b[j] = max (b[j] , a[i]);

　　end

end

这样最坏情况下O(n^n),显然tle.

通过yy可以发现，区间大的可以跟新区间小的，so:

for i = 1 to n 

　　b[w[i]] = max (b[w[i]] , a[i]);

end

for i = n to 2

　　if (b[i - 1] < b[i] )   b[i - 1] = b[i] ;

end

这样就变为O(n)了。

　　2.第二个要解决的问题时我们怎样在O(n)内求出w[i] ?

令w[i] = r[i] - l[i] ;

令r[i] , l[i]分别保存a[i]所能影响大的最右 和 最左。

这样在已知l[i]时，我们便可以在O(1)时间内求出l[i + 1]了。