CF58C Trees(逆向思维)

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题目描述

On Bertown's main street 

Changing the height of a tree is a very expensive operation, using advanced technologies invented by Berland scientists. In one operation you can choose any tree and change its height to any number, either increase or decrease. Note that even after the change the height should remain a positive integer, i. e, it can't be less than or equal to zero. Identify the smallest number of changes of the trees' height needed for the sequence of their heights to become beautiful.

输入格式

The first line contains integer 

输出格式

Print a single number which is the minimal number of trees whose heights will have to be changed for the sequence to become beautiful.

Examples

Input

3
2 2 2

Output

1

Input

4
1 2 2 1

Output

0

题意：美丽序列：回文，左边一半是连续递增，增加值都一样，右边和左边一一对应。操作是选取一个数，改成任意值。要想把给出的序列变成美丽序列，求最少改动的数的数量。解析：想了一下，要想求最少改动次数，的确不容易。逆向思维一下，最少改动次数=n-最多不需要改动数。对于序列 2 4 6 6 4 2，有：

 　　　　　　　可以看出，美丽序列的每一个a[]-idx都相等(idx也要对称)，而且a[]-idx>=0。既然都相等，所以我们统计一下每个差值出现的次数，出现最多的，就不需要改动，n-maxx就是最少改动次数了。

#include<iostream>
#include<vector>
#include<set>
#include<map>
#include<cmath>
using namespace std;
const int maxn=1e5+10;
int a[maxn];
int main()
{
    int n;
    cin>>n;
    map<int,int>mp;
    int maxx=0;
    for(int i=1;i<=n;i++)
    {
        cin>>a[i];
        int md=a[i]-min(i,n-i+1);
        if(md>=0)
        {
            mp[md]++;
            maxx=max(maxx,mp[md]);
        }
    }
    cout<<n-maxx<<endl;
}