Topic

Title Link: https://www.luogu.com.cn/problem/P4428

General idea of the topic

Count Reg n n n 0 / 1 0/1 0 / 1 string requires support

1. Modify a location
2. Find interval [ l , r ] [l,r] [l,r] how many sub interval rearranged binary numbers can be divided by three

1 ≤ n ≤ 1 0 5 1\leq n\leq 10^5 1≤n≤105

Problem solving ideas

First of all 2 2 k % 3 = 1 ( k ∈ Z ) 2^{2k}\%3=1(k\in Z) 22k%3=1(k ∈ Z) and 2 2 k + 1 % 3 = 2 ( k ∈ Z ) 2^{2k+1}\%3=2(k\in Z) 22k+1%3=2(k∈Z).
It is considered in three cases

• have 1 1 1 1 1 So obviously, it can't be divided by three in any case
• have 2 k 2k 2k 1 1 Then it would be nice if we were all at the bottom.
• have 2 k + 1 2k+1 2k+1 1 1 1( k k k cannot be 0 0 0), then there is a scheme to put a certain in an odd position 1 1 1 can be placed in the even position. At this time, the length of the interval needs to be at least 2 k + 3 2k+3 2k+3.

Then analyze it concretely, which is equivalent to an interval 1 1 The number of 1 cannot be 1 1 1 and if it is an odd number, there must be at least two 0 0 0.

It seems very complicated, which can be divided into the following situations

1. The interval is all 1 1 1 and the length is odd
2. There is one in the interval 0 0 0 and even length
3. There is only one interval 1 1 1
4. because 2 2 2 and 3 3 3 will repeat one, only one 1 1 1 and one 0 0 0, so this scheme needs to be added back

The fourth is the best maintenance. Just use the tree array to record

Then the first three we 0 / 1 0/1 One at 0 / 1 position s e t set set to query the predecessor / successor 0 / 1 of a location.

And then in the third case, we have 1 1 1 consider left and right 0 0 The 0 interval is then recorded in the tree array 1 1 Location of 1

For the second case, we consider for each 0 0 0 consider about 1 1 Then record it there 0 0 Location of 0

For the first case, we record the data at the leftmost end of the interval 0 0 0.

Then remember to consider the boundary when counting the answers

It's a little troublesome to write

Time complexity O ( n log ⁡ n ) O(n\log n) O(nlogn)

code

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<set>
#define lowbit(x) (x&-x)
#define ll long long
using namespace std;
const ll N=1e5+10;
ll n,m,a[N],t[N],p[N];
set<ll> s[2];
void Change(ll x,ll val){
	while(x<=n){
		t[x]+=val;
		x+=lowbit(x);
	}
	return;
}
ll Ask(ll x){
	ll ans=0;
	while(x){
		ans+=t[x];
		x-=lowbit(x);
	}
	return ans;
}
ll Left(ll op,ll x)
{return (*--s[op].upper_bound(x));}
ll Right(ll op,ll x)
{return (*s[op].lower_bound(x));}
ll Count(ll n)
{return (n+1)/2*(n+2-(n&1))/2;}
ll Caunt(ll n)
{return n*(n+1)/2;}
ll Calc(ll L,ll R)
{return (L/2+1)*((R+1)/2)+((L+1)/2)*(R/2+1);}
void Updata(ll x){
	if(x<1||x>n)return;
	if(p[x])Change(x,-p[x]);
	if(a[x]){
		ll L=(x-Left(1,x-1)-1),R=(Right(1,x+1)-x-1);
		p[x]=(L+1)*(R+1)-1;
	}
	else{
		ll L=(x-Left(0,x-1)-1),R=(Right(0,x+1)-x-1);
		p[x]=Calc(L,R)+Count(R);
	}
	if(x<n&&a[x]!=a[x+1])p[x]--;
	Change(x,p[x]);
	return;
}
ll Get(ll x,ll l,ll r){
	ll L=max(Left(0,x-1),l-1),R=min(Right(0,x+1),r+1);
	L=x-L-1;R=R-x-1;
	return Calc(L,R);
}
ll Qet(ll x,ll l,ll r){
	ll L=max(Left(1,x-1),l-1),R=min(Right(1,x+1),r+1);
	L=x-L-1;R=R-x-1;
	return (L+1)*(R+1)-1;
}
signed main()
{
	scanf("%lld",&n);
	s[0].insert(0);s[0].insert(n+1);
	s[1].insert(0);s[1].insert(n+1);
	for(ll i=1;i<=n;i++)
		scanf("%lld",&a[i]),s[a[i]].insert(i);
	for(ll i=1;i<=n;i++)
		Updata(i);
	scanf("%lld",&m);
	while(m--){
		ll op,l,r,x;
		scanf("%lld",&op);
		if(op==1){
			scanf("%lld",&x);
			s[a[x]].erase(x);
			a[x]=!a[x];
			s[a[x]].insert(x);
			Updata(x);
			Updata(Left(0,x-1));
			Updata(Left(1,x-1));
			Updata(Right(0,x+1));
			Updata(Right(1,x+1));
		}
		else{
			scanf("%lld%lld",&l,&r);
			ll ans=(r-l+1)*(r-l+2)/2;
			if(Left(1,r)<l){printf("%lld\n",ans);continue;}
			if(Left(0,r)<l){ans-=Count(r-l+1);printf("%lld\n",ans);continue;}
			ans-=Ask(r)-Ask(l-1);
			if(r<n&&a[r]!=a[r+1])ans--;
			ll Ll=Left(0,l-1),Rr=Right(0,r+1),Lr=Left(0,r),Rl=Right(0,l);
			ans=ans+Get(Rl,1,n)-Get(Rl,l,r);
			if(Lr!=Rl)ans=ans+Get(Lr,1,n)-Get(Lr,l,r);
			if(a[r+1])ans=ans+Count(Rr-Lr-1)-Count(r-Lr);
			if(a[l])ans=ans-Count(Rl-l);
			
			Ll=Left(1,l),Rr=Right(1,r),Lr=Left(1,r),Rl=Right(1,l);
			ans=ans+Qet(Rl,1,n)-Qet(Rl,l,r);
			if(Lr!=Rl)ans=ans+Qet(Lr,1,n)-Qet(Lr,l,r);
//			if(!a[r])ans=ans+Caunt(Rr-Rl-1)-Caunt(r-Rl);
//			if(!a[l])ans=ans-Caunt(Lr-l);
			
			printf("%lld\n",ans);
		}
	}
	return 0;
}