description

analysis

• We can use prefix and find out the number of non-adjustable schemes of the original string first.

• For a flip, it must be to flip the interval of [i. [a] [i. [a] [i] [i. [a] [i]] [i. [i]] or [a [i]. [i]] [a [i].. [i] [a [i]. [i]].

• It can also be considered that the point of i+a[i]2{i+a[i]}over 22i+a[i] is the turning center to flip the interval.

• So we got all the nnn turning centers out and saved the turning length with the vector vector vector vector.

• For each turning center, arrange the turning length from small to large.

• Since the current length flip only affects a point from an illegal point to a legitimate point, each scheme increases.

• Left and right endpoints [l,r][l,r][l,r], each lll becomes smaller, rrr becomes larger, the number of alternatives increases, and then add alternatives outside the area to update the answer

• It can be said to be a clever way of thinking.

code

#pragma GCC optimize("O3")
#pragma G++ optimize("O3")
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<vector>
#define MAXN 100005
#define INF 1000000007 
#define ll long long
#define reg register ll
#define fo(i,a,b) for (reg i=a;i<=b;++i)
#define fd(i,a,b) for (reg i=a;i>=b;--i)

using namespace std;

ll a[MAXN],sum[MAXN],pos[MAXN];
vector<ll>v[MAXN<<1];
ll n,ans;

inline ll read()
{
	ll x=0,f=1;char ch=getchar();
	while (ch<'0' || '9'<ch){if (ch=='-')f=-1;ch=getchar();}
	while ('0'<=ch && ch<='9')x=x*10+ch-'0',ch=getchar();
	return x*f;
}
inline ll get(ll x,ll y){return sum[y]-sum[x-1];}
int main()
{
	freopen("rotate.in","r",stdin);
	freopen("rotate.out","w",stdout);
	n=read();
	fo(i,1,n)pos[a[i]=read()]=i,sum[i]=sum[i-1]+(a[i]==i);
	fo(i,1,n)v[i+pos[i]].push_back(abs(i-pos[i])+1);
	ans=sum[n];
	fo(i,1,n<<1)
	{
		if (v[i].size()==0)continue;
		ll mid=i/2,cnt=0;
		sort(v[i].begin(),v[i].end());
		fo(j,0,v[i].size()-1)
		{
			ll len=v[i][j],l,r;
			if (i%2==0)l=mid-len/2,r=mid+len/2;
			else l=mid-len/2+1,r=mid+len/2;
			++cnt;
			ans=max(ans,get(1,l-1)+cnt+get(r+1,n));
		}
	}
	printf("%lld\n",ans);
	return 0;
}