回档|带标记的线段树

果然还是递归线段树好理解啊，写了个模板题，用了黄学长的模板。codevs的线段树练习3.
代码：

#include"iostream"
#include"cstdio"
using namespace std;

struct node{
    int s,t,sum,lab;
}tr[800001];
int n,q;
int a[200001];

int read()
{
    char c=getchar();
    int a=0;
    while (c<‘0‘||c>‘9‘) c=getchar();
    while (c>=‘0‘&&c<=‘9‘)
    {
          a=a*10+c-‘0‘;
          c=getchar();
    }
    return a;
}

void build(int k,int x,int y)
{
    tr[k].s=x;tr[k].t=y;
    if (x==y) { tr[k].sum=a[x]; return; }
    int mid=(x+y)>>1;
    build(k<<1,x,mid);
    build(k<<1|1,mid+1,y);
    tr[k].sum=tr[k<<1].sum+tr[k<<1|1].sum;
}

void pushdown(int k)
{
    int x=tr[k].t-tr[k].s+1;
    tr[k<<1].lab+=tr[k].lab;
    tr[k<<1|1].lab+=tr[k].lab;
    tr[k<<1].sum+=(x-(x>>1))*tr[k].lab;
    tr[k<<1|1].sum+=(x>>1)*tr[k].lab;
    tr[k].lab=0;
}

void change(int now,int x,int y,int k)
{
    int l=tr[now].s,r=tr[now].t;
    if (x==l && y==r)
    {
        tr[now].lab+=k;
        tr[now].sum+=(y-x+1)*k;
        return;
    }
    if (tr[now].lab) pushdown(now);
    int mid=(l+r)>>1;
    if (x>mid) change(now<<1|1,x,y,k);
        else if (y<=mid) change(now<<1,x,y,k);
            else { change(now<<1,x,mid,k); change(now<<1|1,mid+1,y,k);        }
    tr[now].sum=tr[now<<1].sum+tr[now<<1|1].sum;
    return;
}

long long ask(int now,int x,int y)
{
    int l=tr[now].s,r=tr[now].t;
    if (x==l && y==r) return tr[now].sum;
    if (tr[now].lab) pushdown(now);
    int mid=(l+r)>>1;
    if (x>mid) return ask(now<<1|1,x,y);
        else if (y<=mid) return ask(now<<1,x,y);
            else return (ask(now<<1,x,mid)+ask(now<<1|1,mid+1,y));
}

int main()
{
        n=read();
        for (int i=1; i<=n; i++) a[i]=read();
        build(1,1,n);
        q=read();
        for (int i=1; i<=q; i++)
        {
            int now=read();
            if (now==1)
            {
                int x=read(),y=read(),k=read();
                change(1,x,y,k);
            }
            else if (now==2)
            {
                int x=read(),y=read();
                printf("%lld\n",ask(1,x,y));
            }
        }
        return 0;
}