beautiful tree ~ thinking question

Link: https://www.nowcoder.com/acm/contest/84/B
Source: Niuke.com

There are n trees on the street, numbered 1...n, and the height of the ith tree is a _i .
Define that n trees are beautiful if and only if
    1. for all i, a _i = a _n-i+1 ;
    2. for 1 <= i < n / 2 (not divisible), a _{i + 1} = a _i + 1;
say "2 3 4 5 5 4 3 2" and "1 2 3 2 1" are pretty but "1 3 3 1" and "1 2 3 1" are not.
Now please modify the height of at least a few trees (can be larger or smaller), so that these trees are beautiful.

Enter description:

The first line contains an integer n representing the number of trees ( 1 <= n <= 100,000). 
The second line of n integers represents the height of the tree ( 1 <= a

_i

<= 100,000)。

Output description:

Output an integer representing the number of heights of the least modified tree.

Example 1

enter

3
2 2 2

output

1

Example 2

enter

4
1 2 2 1

output

0 

I haven't done a thinking problem for a long time. 
It feels useless. 
The slope of this problem is known. 
Solving by slope is very fast, 
pay attention to the case of negative array subscripts. 
So I added 10W to the subscript.

 1 #include <cstdio>
 2 #include <algorithm>
 3 #include <vector>
 4 #include <queue>
 5 #include <cstring>
 6 #include <string>
 7 using namespace std;
 8 const int maxn = 2e5 + 10;
 9 int a[maxn], b[maxn];
10 int main() {
11     int n;
12     while(scanf("%d", &n) != EOF) {
13         for (int i = 1 ; i <= n ; i++ )
14             scanf("%d", &a[i]);
15         memset(b, 0, sizeof(b));
16         for (int i = 1 ; i <= n ; i++) {
17             if (i <= n / 2) b[a[i] - i + 100000]++;
18             else  b[a[i] - (n - i + 1) + 100000]++;
19         }
20         int ans = 0;
21         for (int i = 0 ; i < 200000 ; i++ ) {
22             ans = max(ans, b[i]);
23         }
24         printf("%d\n", n - ans);
25     }
26     return 0;
27 }