线段树0分，求调

#include <iostream> 
#define now (st[u])
#define left (st[u<<1])
#define right (st[(u<<1)+1])
using namespace std;
int n,m,opt,x,y,k,a[100001];
struct STNode{
	int l,r,lazy;
	long long sum;
} st[400001];
void build(int u,int l,int r){
	now.l=l;now.r=r;
	if (l!=r){
		int mid=(l+r)>>1;
		build(u<<1,l,mid);
		build((u<<1)+1,mid+1,r);
		now.sum=left.sum+right.sum;
	}else{
		now.sum=a[l];
	}
}
void pushdown(int u){
	if (now.lazy){
		left.lazy+=now.lazy;
		right.lazy+=now.lazy;
		now.sum+=now.lazy*(now.r-now.l+1);
		now.lazy=0;
	}
}
void modify(int u,int l,int r,int k){
	if (l<=now.l&&now.r<=r) now.lazy+=k;
	else if (now.r<l||now.l>r) return;
	else{
		int mid=(now.l+now.r)/2;
		if (l<=mid) modify(u<<1,l,r,k);
		if (r>mid) modify((u<<1)+1,l,r,k);
	}
}
int query(int u,int l,int r){
	if (now.r<l||now.l>r) return 0;
	pushdown(u);
	if (l<=now.l&&now.r<=r) return now.sum;
	int ans=0,mid=(now.l+now.r)/2;
	if (l<=mid) ans+=query(u<<1,l,r);
	if (r>mid) ans+=query((u<<1)+1,l,r);
	return ans;
}
int main(){
	//ios::sync_with_stdio(false);
	//cin.tie(0);
	cin>>n>>m;
	for (int i=1;i<=n;i++) cin>>a[i];
	build(1,1,n);
	for (int i=1;i<=m;i++){
		cin>>opt;
		if (opt==1) cin>>x>>y>>k;
		else cin>>x>>y;
		if (opt==1) modify(1,x,y,k);
		else cout<<query(1,x,y)<<'\n';
	}
	return 0;
}