题解

肥肠煎蛋，直接使用区改区查，主墓碑单独计算。

呆码

#include<bits/stdc++.h>
#define int long long

using namespace std;
int n,a[1000001],b[1000001],trb[1000001],big;
int trib[1000001];
int lowbit(int x){return x&-x;}
int getArea(int l,int r){
	l--;
	int temp;
	int sum1;
	temp=l,sum1=0;
	while(temp){
		sum1+=trb[temp]*(l+1);
		sum1-=trib[temp];
		temp-=lowbit(temp);
	}
	int sum2;
	temp=r,sum2=0;
	while(temp){
		sum2+=trb[temp]*(r+1);
		sum2-=trib[temp];
		temp-=lowbit(temp);
	}
	return sum2-sum1;
}
void change(int p,int k){
	int temp=p;
	while(temp<=n){
		trb[temp]+=k;
		trib[temp]+=k*p;
		temp+=lowbit(temp); 
	}
}
void changeArea(int l,int r,int k){
	r++;
	int temp=l;
	while(temp<=n){
		trb[temp]+=k;
		trib[temp]+=k*l;
		temp+=lowbit(temp);
	}
	temp=r;
	k=-k;
	while(temp<=n){
		trb[temp]+=k;
		trib[temp]+=k*r;
		temp+=lowbit(temp);
	}
}
signed main(){
	cin.tie(0);cout.tie(0);ios::sync_with_stdio(false);
	cin>>n;
	int x;
	cin>>x;
	for(int i=1;i<=n;i++){
		cin>>a[i];
		change(i,a[i]-a[i-1]);
	}
	while(x--){
		int ctrl=0;
		int l,r,k;
		cin>>ctrl;
		switch(ctrl){
		case 1:
			cin>>l>>r>>k;
			changeArea(l,r,k);
			break;
		case 2:
			cin>>k;
			big+=k;
			//主墓碑单独计算
			break;
		case 3:
			cin>>k;
			big-=k;
			break;
		case 4:
			cin>>l>>r;
			if(l==1){
				cout<<getArea(l,r)+big<<endl;
				//若含有主墓碑，加上主墓碑
			}
			else cout<<getArea(l,r)<<endl;
			break;
		case 5:
			cout<<big+getArea(1,1)<<endl;
			//记得算上更改的值
			break;
		}
	}
}