Acwing week 57 longest continuous subsequence - (binary or tree array)

The longest continuous subsequence

The question ：
Just give you an array , Ask you the longest continuous subsequence , Make the sum of this interval > Interval length *100.

reflection ：
At that time, seeing this question was a glance question , Subtract all 100, Then seek >0 Subinterval of . For the longest , Just use a tree array . It is the same as the question I have done before ： Xiao Yang buys fruit .
Then this question card nlongn How to do it , All the time . Then I thought it was not two points , Does the binary answer satisfy monotonicity , I think of the old one , When the answer is No , Not necessarily the answer on the right is no , So it feels wrong . Actually, this question , For when the length is 5 When there is a plan to meet , So the length is 6 You can do it when you are . That is to say 5 When it doesn't work ,6 Definitely not . Because this view is not just the kind of fixed continuity , The answer in two is at least >=mid.

Code ：

 Two points ：

#include<bits/stdc++.h>
#define fi first
#define se second
#define pb push_back
#define db double
#define int long long
#define PII pair<int,int >
#define mem(a,b) memset(a,b,sizeof(a))
#define IOS std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
		
using namespace std;
const int mod = 1e9+7,inf = 1e18;
const int N = 1e6+10,M = 2010;

int T,n,m,k;
int va[N];
int sum[N];
int minn[N];

bool check(int mid)
{
    
	for(int i=mid;i<=n;i++)
	{
    
		if(sum[i]>minn[i-mid]) return true;
	}
	return false;
}

signed main()
{
    
	IOS;
	cin>>n;
	for(int i=1;i<=n;i++) cin>>va[i];
	for(int i=1;i<=n;i++)
	{
    
		va[i] -= 100;
		sum[i] = sum[i-1]+va[i];
	}
	for(int i=1;i<=n;i++) minn[i] = inf;
	for(int i=1;i<=n;i++) minn[i] = min(minn[i-1],sum[i]);
	int l = 0,r = n;
	while(l<r)
	{
    
		int mid = (l+r+1)>>1;
		if(check(mid)) l = mid;
		else r = mid-1;
	}
	cout<<l;
	return 0;
}

 Tree array timeout :
#include<iostream>
#include<cstring>
#include<string>
#include<math.h>
#include<algorithm>
#include<vector>
#include<queue>
#include<deque>
#include<stack>
#include<map>
#include<set>
#include<unordered_map>
//#pragma GCC optimize(2)
#define eps 1e-9
#define fi first
#define se second
#define pb push_back
#define db double
#define ll long long
#define PII pair<int,int >
#define cin(x) scanf("%d",&x)
#define cout(x) printf("%d ",x)

#define mem(a,b) memset(a,b,sizeof(a)) 
#define IOS std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
#define int ll

using namespace std;

const int N = 5e6,M = 2000;
const int mod = 1e9+7;
const int inf = 0x3f3f3f3f;
const double pai = acos(-1);

int T,n,m,k;
int va[N];
int sum[N];
int tr[N],R = 2e6+5;

vector<int > v;

int get(int x)
{
    
	return lower_bound(v.begin(),v.end(),x)-v.begin()+1;
}

int bit(int x)
{
    
	return x&(-x);
}

void update(int x,int value)
{
    
	while(x<=R)
	{
    
		tr[x] = min(tr[x],value);
		x += bit(x);
	}	
}

int query(int x)
{
    
	int minn = inf;
	while(x)
	{
    
		minn = min(minn,tr[x]);
		x -= bit(x);
	}
	return minn;
}

signed main()
{
    
	IOS;
	cin>>n;
	for(int i=1;i<=R;i++) tr[i] = inf;
	for(int i=1;i<=n;i++)
	{
    
		cin>>va[i];
		va[i] -= 100;
	}
	for(int i=1;i<=n;i++)
	{
    
		sum[i] = sum[i-1]+va[i];
		v.pb(sum[i]);
	}
	sort(v.begin(),v.end());
	v.erase(unique(v.begin(),v.end()),v.end());
	int maxn = 0;
	for(int i=1;i<=n;i++)
	{
    
		int now = get(sum[i]);
		if(sum[i]>0) maxn = max(maxn,i);
		else maxn = max(maxn,i-query(now-1));
		update(now,i);
	}
	cout<<maxn;
	return 0;
}

summary ：
Think a lot , Don't take it for granted .