树状数组区间修改

//c2[n] = (n-1)*c1[n];
//sum(1,k)=k*(c1(1)+c1(2)+c1(3)+…+c1(k))-(0*c1*(1)+1*c1(2)+2*c1(3)+…+(k-1)*c1(k))。
#include <bits/stdc++.h>
#define ll long long
#define LEN 1000000
using namespace std;
ll c1[LEN],c2[LEN],a[LEN];
ll lowbit(ll k)
{
	return k&(-k);
}
void update(ll *q,ll k,ll v)
{
	while(k<=LEN)
	{
		q[k]+=v;
		k+=lowbit(k);
	}
}
ll getsum(ll *q,ll k)//求c1或c2前k项和
{
	ll re = 0;
	while(k)
	{
		re+=q[k];
		k-=lowbit(k);
	}
	return re;
}
ll sum(ll k)//求数列前k项和
{
	ll ans1,ans2;
	ans1 = k*getsum(c1,k);
	ans2 = getsum(c2,k);
	return ans1-ans2;
}
int main()
{
	ll n,q;
	scanf("%lld %lld",&n,&q);
	for(ll i = 1; i<=n ;i++)
		{
			scanf("%lld",&a[i]);
			update(c1,i,a[i]-a[i-1]);
			update(c2,i,(i-1)*(a[i]-a[i-1]));
		}
	while(q--)
	{
		ll tp,l,r,v;
		scanf("%lld",&tp);
		if(tp==0)
		{
			scanf("%lld%lld%lld",&l,&r,&v);
			update(c1,l,v);
			update(c1,r+1,-v);
			update(c2,l,(l-1)*v);
			update(c2,r+1,r*(-v));
		}else
		{
			scanf("%lld %lld",&l,&r);
			printf("%lld\n",sum(r)-sum(l-1));
		}
	}
	
	return 0;
}