E. Breaking the Wall

原题链接
Monocarp plays “Rage of Empires II: Definitive Edition” — a strategic computer game. Right now he’s planning to attack his opponent in the game, but Monocarp’s forces cannot enter the opponent’s territory since the opponent has built a wall.

The wall consists of n sections, aligned in a row. The i-th section initially has durability ai. If durability of some section becomes 0 or less, this section is considered broken.

To attack the opponent, Monocarp needs to break at least two sections of the wall (any two sections: possibly adjacent, possibly not). To do this, he plans to use an onager — a special siege weapon. The onager can be used to shoot any section of the wall; the shot deals 2 damage to the target section and 1 damage to adjacent sections. In other words, if the onager shoots at the section x, then the durability of the section x decreases by 2, and the durability of the sections x−1 and x+1 (if they exist) decreases by 1 each.

Monocarp can shoot at any sections any number of times, he can even shoot at broken sections.

Monocarp wants to calculate the minimum number of onager shots needed to break at least two sections. Help him!

Input
The first line contains one integer n (2≤n≤2⋅105) — the number of sections.

The second line contains the sequence of integers a1,a2,…,an (1≤ai≤106), where ai is the initial durability of the i-th section.

Output
Print one integer — the minimum number of onager shots needed to break at least two sections of the wall.

Examples

inputCopy
5
20 10 30 10 20
outputCopy
10

inputCopy
3
1 8 1
outputCopy
1

inputCopy
6
7 6 6 8 5 8
outputCopy
4

inputCopy
6
14 3 8 10 15 4
outputCopy
4

inputCopy
4
1 100 100 1
outputCopy
2

inputCopy
3
40 10 10
outputCopy
7

Note
In the first example, it is possible to break the 2-nd and the 4-th section in 10 shots, for example, by shooting the third section 10 times. After that, the durabilities become [20,0,10,0,20]. Another way of doing it is firing 5 shots at the 2-nd section, and another 5 shots at the 4-th section. After that, the durabilities become [15,0,20,0,15].

In the second example, it is enough to shoot the 2-nd section once. Then the 1-st and the 3-rd section will be broken.

In the third example, it is enough to shoot the 2-nd section twice (then the durabilities become [5,2,4,8,5,8]), and then shoot the 3-rd section twice (then the durabilities become [5,0,0,6,5,8]). So, four shots are enough to break the 2-nd and the 3-rd section

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思路很简单,就是分情况讨论(3种情况,详见代码)
为什么比赛时候写不出来呜呜呜呜呜

#include<bits/stdc++.h>
using namespace std;
int const N=2e5+10;
int a[N];
int res=1e9;
int main()
{
    int n;
    cin>>n;
    for(int i=1;i<=n;i++)
        cin>>a[i];
        
      //击破相邻的
    for(int i=1;i<n;i++)
    {
        res=min(res,max(max((a[i]+1)/2,(a[i+1]+1)/2),(a[i]+a[i+1]+2)/3));
    }

  	//击破该值左右两边的
    for(int i=2;i<n;i++)
    {
        res=min(res,min(a[i+1],a[i-1])+(abs(a[i-1]-a[i+1])+1)/2);
    }

    sort(a+1,a+1+n);
    //击破互不相干的;
    res=min(res,(a[1]+1)/2+(a[2]+1)/2);
    cout<<res<<endl;
    return 0;
}