感觉没有太大问题，自己也造了几组范围小一点的数据，样例过了，但是0分，求WA。。。

#include<bits/stdc++.h>
#define int long long
using namespace std;
int n,m;
int a[10000009];
int tree[10000009];
int lazy_tag[10000009];
void build(int l,int r,int p){
    if(l==r){
        tree[p]=a[l];
        return;
    }
    int mid=(l+r)/2;
    build(l,mid,2*p);
    build(mid+1,r,2*p+1);
    tree[p]=tree[p*2]+tree[p*2+1];
}
int getsum(int l,int r,int s,int t,int p){
    if(l<=s&&t<=r) return tree[p];
    int mid=(s+t)/2;
    if(lazy_tag[p]){
        tree[p*2]+=lazy_tag[p]*(mid-l+1);
        tree[p*2+1]+=lazy_tag[p]*(r-mid);
        lazy_tag[2*p]+=lazy_tag[p];
        lazy_tag[2*p+1]+=lazy_tag[p];
    }
    lazy_tag[p]=0;
    int sum=0;
    if(l<=mid) sum+=getsum(l,r,s,mid,2*p);
    if(r>mid) sum+=getsum(l,r,mid+1,t,2*p+1);
    return sum;
}
void update(int l,int r,int s,int t,int k,int p){
	if(l<=s&&t<=r){
		tree[p]+=(t-s+1)*k;
		lazy_tag[p]+=k;
		return;
	}
	int mid=(s+t)/2;
	if(lazy_tag[p]){
		tree[2*p]+=lazy_tag[p]*(mid-s+1);
		tree[2*p+1]+=lazy_tag[p]*(t-mid);
		lazy_tag[2*p]+=lazy_tag[p];
		lazy_tag[2*p+1]+=lazy_tag[p];
	}
	lazy_tag[p]=0;
	if(l<=mid) update(l,r,s,mid,k,2*p);
	if(r>mid) update(l,r,mid+1,t,k,2*p+1);
	tree[p]=tree[p*2]+tree[p*2+1];
}

signed main(){
    cin>>n>>m;
    for(int i=1;i<=n;i++) cin>>a[i];
    build(1,n,1);
    while(m--){
        int op;
        cin>>op;
        if(op==1){
            int x,y,k;
			cin>>x>>y>>k;
			update(x,y,1,n,k,1);
        }
        if(op==2){
            int k;
            cin>>k;
            update(1,1,1,n,k,1);
        }
        if(op==3){
        	int k;
            cin>>k;
            update(1,1,1,n,-1*k,1);
		}
		if(op==4){
			int l,r;
			cin>>l>>r;
			cout<<getsum(l,r,1,n,1)<<endl;
		}
		if(op==5){
			cout<<getsum(1,1,1,n,1)<<endl;
		}
    }
}