poj 3468 整理一下线段树的写法

// 对于延迟更新，我们在updata 和query的时候 pushdown和pushup两个东西都要存在 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
typedef long long ll;
struct node
{
    ll l,r,sum,add;
} tree[10000*4];
ll a[100001];
//ll x,y;
void pushup(int id)
{
    tree[id].sum=tree[id*2].sum+tree[id*2+1].sum;
}
void build(ll l,ll r,int id)
{
    if(l==r)
    {
        tree[id].sum=a[l];
        return;
    }
    tree[id].l=l;
    tree[id].r=r;
    tree[id].add=0;
    ll mid=(l+r)/2;
    build(l,mid,id*2);
    build(mid+1,r,id*2+1);
    pushup(id);
}
void pushdown(int id)
{
    if(tree[id].add)
    {
        tree[id*2].add +=tree[id].add;
        tree[id*2+1].add += tree[id].add;

        tree[id*2].sum += (tree[id*2].r-tree[id*2].l+1)*(tree[id*2].add);
        tree[id*2+1].sum += (tree[id*2+1].r-tree[id*2+1].l+1)*(tree[id*2+1].add);
        tree[id].add=0;
    }
}

void updata(ll x,ll y,ll l,ll r,int id,ll k)
{
    if(x<=l && r<=y)// 在区间里面 就可以做一定操作
    {
        tree[id].sum += (r-l+1)*k;
        tree[id].add += k;
        return;
    }
    pushdown(id);//
    ll mid=(l+r)/2;
    if(x <= mid) updata(x,y,l,mid,id*2,k);
    if(y > mid)  updata(x,y,mid+1,r,id*2+1,k);
    pushup(id);
}

ll query(ll x,ll y,ll l,ll r,int id)
{
    if(x<=l && r<=y)
    {
        return tree[id].sum;
    }
    pushdown(id);
    ll mid=(l+r)/2;
    ll sum=0;
    if(x<=mid) sum+=query(x,y,l,mid,id*2);
    if(y>mid)  sum+=query(x,y,mid+1,r,id*2+1);
    pushup(id);//
    return sum;
}
int main()
{
    int n,q;
    ll x,y,k;
    while(scanf("%d %d",&n,&q)!=EOF)
    {
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&a[i]);
        }
        build(1,n,1);
        while(q--)
        {
            char s[3];
            scanf("%s",s);
            if(s[0]=='Q')
            {
                scanf("%lld %lld",&x,&y);
                printf("%lld\n",query(x,y,1,n,1));
            }
            else if(s[0]=='C')
            {
                scanf("%lld %lld %lld",&x,&y,&k);
                updata(x,y,1,n,1,k);
            }
        }
    }


    return 0;
}