校内选拔I题题解 构造题 Codeforces Round #318 [RussianCodeCup Thanks-Round] (Div. 2) ——D

http://codeforces.com/contest/574/problem/D

Bear and Blocks

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Limak is a little bear who loves to play. Today he is playing by destroying block towers. He built n towers in a row. The i-th tower is made of h_i identical blocks. For clarification see picture for the first sample.

Limak will repeat the following operation till everything is destroyed.

Block is called internal if it has all four neighbors, i.e. it has each side (top, left, down and right) adjacent to other block or to the floor. Otherwise, block is boundary. In one operation Limak destroys all boundary blocks. His paws are very fast and he destroys all those blocks at the same time.

Limak is ready to start. You task is to count how many operations will it take him to destroy all towers.

Input

The first line contains single integer n (1 ≤ n ≤ 10⁵).

The second line contains n space-separated integers h₁, h₂, ..., h_n (1 ≤ h_i ≤ 10⁹) — sizes of towers.

Output

Print the number of operations needed to destroy all towers.

Examples

Input

6
2 1 4 6 2 2

Output

3

Input

7
3 3 3 1 3 3 3

Output

2

Note

The picture below shows all three operations for the first sample test. Each time boundary blocks are marked with red color.

After first operation there are four blocks left and only one remains after second operation. This last block is destroyed in third operation

题目大意 每次只能消最外层的砖 问多少次能消完

看hint图吧 等价与从右方看 从左到右峰依次为  2 1 4 3 2 1 从左方看  从左到右峰依次为 1 1 2 3 2 2

比较每个位置需要消去的最少次数 依次为 1 1 2 3 2 2的峰

看到这里是不是就明白了呢

我们只需要把山峰等效为 突起 如 1 2 2 3 4 3这样的形式就ok了

具体操作见代码 tw菊苣的dp写法还不是很理解 再研究一下

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
inline void ri(int &num){
    num=;int f=;char ch=getchar();
    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
    while(ch>=''&&ch<='')num=num*+ch-'',ch=getchar();
    num*=f;
}
const int N=1e5+;
int l[N],r[N],a[N];
int main()
{
    int n;
    ri(n);
    for(int i=;i<=n;i++)ri(a[i]);
    for(int i=;i<=n;i++) l[i]=min(l[i-]+,a[i]);
    for(int i=n;i>=;i--) r[i]=min(r[i+]+,a[i]);
    int mx=-;
    for(int i=;i<=n;i++) mx=max(mx,min(l[i],r[i]));
    printf("%d\n",mx);
    return ;
}

AC代码