Codeforces Round #318 [RussianCodeCup Thanks-Round] (Div. 1) B. Bear and Blocks dp

B. 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 hi 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, ..., hn (1 ≤ hi ≤ 109) — sizes of towers.

Output

Print the number of operations needed to destroy all towers.

Sample test(s)

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.

题意，给出一排序列，每次将外层的方块去掉，问最多要多少步。

如果直接模拟，要用n次所以复杂度为o(n * n);我们可以观察每列最大步数是多少。我们可以发现，每列一步要么减1层，要么是min(h[i-1],h[i+1],所以一步，要使得h[i]变为，min(h[i] -1,h[i-1],h[i+1]),r[i]表示用的次数，那么，每i列用的次数，也就是min(h[i],r[i-1],r[i+1]);所 以用dp的思想，从前向后，再从后向前推一遍就可以了。复杂度为O(n);

#define N 100005
#define M 100005
#define maxn 205
#define MOD 1000000000000000007
int n,h[N],maxx;
int main()
{
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
     while(S(n)!=EOF)
    {
        FI(n) S(h[i]);
        h[0] = 1;
        For(i,1,n){
            h[i] = min(h[i],h[i-1] + 1);
        }
        h[n-1] = 1;
        for(int i = n - 2;i>=0;i--){
            h[i] = min(h[i],h[i+1] + 1);
        }
        maxx = 0;
        FI(n) maxx = max(maxx,h[i]);
        printf("%d\n",maxx);
    }
    //fclose(stdin);
    //fclose(stdout);
    return 0;
}