Codeforces Round #465 (Div. 2) F. Fafa and Array-优快云博客

本文介绍了一种处理数组查询的高效算法，该算法能在给定的时间和内存限制内处理大量不同类型的操作，包括查找最大可能值及更新数组元素。通过构建特定的数据结构并采用分段函数分析的方法，实现了快速响应查询请求。

Fafa has an array A of n positive integers, the function f(A) is defined as $\text{[math]}$ . He wants to do q queries of two types:

1 l r x — find the maximum possible value of f(A), if x is to be added to one element in the range [l, r]. You can choose to which element to add x.
2 l r x — increase all the elements in the range [l, r] by value x.

Note that queries of type 1 don't affect the array elements.

Input

The first line contains one integer n (3 ≤ n ≤ 105) — the length of the array.

The second line contains n positive integers a1, a2, ..., an (0 < ai ≤ 109) — the array elements.

The third line contains an integer q (1 ≤ q ≤ 105) — the number of queries.

Then q lines follow, line i describes the i-th query and contains four integers ti li ri xi $\text{[math]}$ .

It is guaranteed that at least one of the queries is of type 1.

Output

For each query of type 1, print the answer to the query.

Examples

input

Copy

output

2
8

input

output

6
10

(没有qls 灵活的数学分析能力弱鸡只能用分类来讨论分段函数不过跑的飞快哦

#include <bits/stdc++.h>
#define ll long long
#define N 100005 
using namespace std;
ll readll(){
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int readint(){
    int x=0,f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
    return f*x;
}
typedef struct node{
	int l,r;ll minn;
}node;
node d[N<<2];
int n,q;
ll a[N],ans,b[N];
void up(int x){
	d[x].minn=min(d[x<<1].minn,d[x<<1|1].minn);
}
void built(int root,int l,int r){
	if(l==r){
		d[root].l=d[root].r=l;d[root].minn=b[l];
		return ;
	}
	int mid=(l+r)>>1;
	built(root<<1,l,mid);
	built(root<<1|1,mid+1,r);
	d[root].l=d[root<<1].l;d[root].r=d[root<<1|1].r;
	up(root);
}
void update(int root,int t,ll vul){
	if(d[root].l==d[root].r){
		d[root].minn=vul;
		return ;
	}
	int mid=(d[root].l+d[root].r)>>1;
	if(t<=mid) update(root<<1,t,vul);
	if(t>mid) update(root<<1|1,t,vul);
	up(root);
}
ll ans1;
void querty(int root,int l,int r){
	if(l<=d[root].l&&d[root].r<=r){
		ans1=min(d[root].minn,ans1);
		return ;
	}
	int mid=(d[root].l+d[root].r)>>1;
	if(l<=mid) querty(root<<1,l,r);
	if(r>mid) querty(root<<1|1,l,r);
}
ll jue(ll x){
	if(x<0) return -1*x;
	else return x;
}
ll chulil(ll x){
	ll sum1=ans;
	ll tt=max(0ll,-1*a[1]);
	//sum1+=(x-2*tt);
	if(tt==0) sum1+=x;
	else{
		if(x>=2*tt) sum1+=(x-2*tt);
		else if(x>=tt&&x<2*tt) sum1-=(2*tt-x);
		else sum1-=x;
	}
	return sum1;
}
ll chulin(ll x){
	ll sum1=ans;
	ll tt=max(0ll,a[n-1]);
	//sum1+=(x-2*tt);
	if(tt==0) sum1+=x;
	else{
		if(x>=2*tt) sum1+=(x-2*tt);
		else if(x>=tt&&x<2*tt) sum1-=(2*tt-x);
		else sum1-=x;
	}
	return sum1;
}
ll chulie(int l,ll x){
	ll sum1=ans;
	ll tt=max(0ll,a[l-1])+max(0ll,-1*a[l]);
	ll ttt=min(max(0ll,a[l-1]),max(0ll,-1*a[l]));
	ll tttt=max(max(0ll,a[l-1]),max(0ll,-1*a[l]));
//	cout<<tt<<" "<<ttt<<" "<<tttt<<endl;
	if(x>=tt) sum1+=(x-tt)*2;
	else if(x>=tttt&&x<tt) sum1+=(x-tt)*2;
	else if(x>=ttt&&x<tttt) sum1+=(-ttt)*2;
	else sum1+=(-x)*2;
	return sum1;
}
ll slove(int l,int r,ll x){
	ll sum1=ans;
	if(l==1&&r!=n){
		if(r-l==1){
			return max(chulil(x),chulie(r,x));
		}
		ans1=1e18;querty(1,l+1,r);
		ans1=min(ans1,x);sum1+=(x-ans1)*2;
		return max(chulil(x),sum1);
	}
	else if(l!=1&&r==n){
		if(r-l==1) return max(chulin(x),chulie(l,x));
		ans1=1e18;querty(1,l,r-1);
		ans1=min(ans1,x);sum1+=(x-ans1)*2;
		return max(sum1,chulin(x));
	}
	else if(l!=1&&r!=n){
		ans1=1e18;querty(1,l,r);
		//cout<<ans1<<endl;
		ans1=min(ans1,x);sum1+=(x-ans1)*2;
		return sum1;
	}
	else{
		if(n==3) return max(max(chulil(x),chulin(x)),chulie(2,x));
		ans1=1e18;querty(1,l+1,r-1);
		ans1=min(ans1,x);sum1+=(x-ans1)*2;
		return max(max(chulil(x),chulin(x)),sum1);
	}
}
void work(int l,int r,ll x){
	if(l==1&&r!=n){
		b[r]-=max(0ll,-1*a[r]);b[r+1]-=max(0ll,a[r]);
		ans-=jue(a[r]);a[r]+=x;ans+=jue(a[r]);
		b[r]+=max(0ll,-1*a[r]);b[r+1]+=max(0ll,a[r]); 
		update(1,r,b[r]);
		if(r<n+1) update(1,r+1,b[r+1]);
	}
	else if(l!=1&&r==n){
		b[l]-=max(0ll,a[l-1]);ans-=jue(a[l-1]);b[l-1]-=max(0ll,-1*a[l-1]); 
		a[l-1]-=x;ans+=jue(a[l-1]);b[l]+=max(0ll,a[l-1]);b[l-1]+=max(0ll,-1*a[l-1]); 
		update(1,l,b[l]);
		if(l>2) update(1,l-1,b[l-1]);
	}
	else if(l!=1&&r!=n){
		b[l]-=max(0ll,a[l-1]);ans-=jue(a[l-1]);b[l-1]-=max(0ll,-1*a[l-1]);
		a[l-1]-=x;ans+=jue(a[l-1]);b[l]+=max(0ll,a[l-1]);b[l-1]+=max(0ll,-1*a[l-1]); 
		update(1,l,b[l]);
		if(l>2) update(1,l-1,b[l-1]);
		b[r]-=max(0ll,-1*a[r]);b[r+1]-=max(0ll,a[r]);
		ans-=jue(a[r]);a[r]+=x;ans+=jue(a[r]);
		b[r]+=max(0ll,-1*a[r]);b[r+1]+=max(0ll,a[r]);
		update(1,r,b[r]);
		if(r<n-1) update(1,r+1,b[r+1]); 
	}
}
void print(){
	for(int i=2;i<n;i++) cout<<b[i]<<" ";
	cout<<endl;
}
int main(){
	ios::sync_with_stdio(false);
	n=readint();ans=0;
	for(int i=1;i<=n;i++) a[i]=readll();a[n+1]=a[n];
	for(int i=1;i<=n;i++) a[i]-=a[i+1],ans+=jue(a[i]);
	for(int i=2;i<n;i++) b[i]=max(0ll,a[i-1])+max(0ll,-1*a[i]);
	//print();
	built(1,1,n);
	q=readint();int op,l,r;ll x;
	for(int i=1;i<=q;i++){
		op=readint();l=readint();r=readint();x=readll();
		if(op==1){
			//cout<<ans<<endl;
			if(l==r){
				if(l==1){
					printf("%lld\n",chulil(x));
					continue;
				}
				else if(l==n){
					printf("%lld\n",chulin(x));
					continue;
				}
				else{
					printf("%lld\n",chulie(l,x));
					continue;
				}
			}
			printf("%lld\n",slove(l,r,x));
		}
		else work(l,r,x);
		//cout<<ans<<endl;
		//print();
	}
	return 0;
}

转载于:https://www.cnblogs.com/wang9897/p/8457901.html