翔特卡洛和解题报告

题目 ：
定义：

$$A_i=(1023^i \mod 10^9) xor (1025^i \mod 10^9)$$
求 ：
$$Sum=\sum_{i=l_1}^{r_1} \sum_{i=max(i,l_2)}^{r_2} \max_{k=i}^{j} A[k]-\min_{k=i}^{j} A[k]$$
多组询问
Limit:
$$1\le t \le 40000,1\le l1 \le r1 \le 5*10^4,1\le l2 \le r2 \le 5*10^4$$
题解：一开始以为这是一个数论题目，后来膜拜了正解之后才发现$A_i$ 其实是随机的= =
Q：什么数据结构是引入随机化的
A：Treap
然而Treap不能做序列操作。。。等等真的不行么，fhq Treap不就可以了么，那么我们把A_i当fhq Treap的键值，然后向Maintain时计入键值对答案的贡献$fix*(lsize+1)*(risze+1)+leftans+rightans$ 即可
然而我们需要分类讨论l1,r1,l2,r2的关系，这里懒得讨论了，详见代码。code：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define P 1000000000LL
#define n 100000
using namespace std;
struct treap_node{
    treap_node *left,*right;
    int size,wgt;
    long long fix,ans;
    treap_node(long long fix): fix(fix) {ans=fix; wgt=size=1; left=right=NULL;}
    int lsize()
      {
        if (left)
          return left->size;
        else
          return 0;
      }
    int rsize()
      {
        if (right)
          return right->size;
        else
          return 0;
      }
    void Maintain()
      {
        size=wgt;
        size+=lsize()+rsize();
        ans=(long long)fix*(long long)(lsize()+1)*(long long)(rsize()+1);
        if (left) ans=(long long)(ans+left->ans);
        if (right) ans=(long long)(ans+right->ans);
      }
};
treap_node *rootlarge,*rootsmall;
long long a[100001];
int t,l1,r1,l2,r2;
typedef pair<treap_node*,treap_node*> droot;
void print(treap_node *p)
{
    if (p->left)
      print(p->left);
    printf("%d\n",p->fix);
    if (p->right)
      print(p->right);
}
treap_node *mergesmall(treap_node *a,treap_node *b)
{
    if (!b) return a; if (!a) return b;
    if (a->fix<b->fix)
      {
        a->right=mergesmall(a->right,b);
        a->Maintain();
        return a;
      } 
    else
      {
        b->left=mergesmall(a,b->left);
        b->Maintain();
        return b;
      }
}
treap_node *mergelarge(treap_node *a,treap_node *b)
{
    if (!b) return a; if (!a) return b;
    if (a->fix>b->fix)
      {
        a->right=mergelarge(a->right,b);
        a->Maintain();
        return a;
      }
    else
      {
        b->left=mergelarge(a,b->left);
        b->Maintain();
        return b;
      }
}
droot split(treap_node *x,int k)
{
    if (!x) return droot(NULL,NULL);
    droot y;
    if (k<=x->lsize())
      {
        y=split(x->left,k);
        x->left=y.second;
        x->Maintain();
        y.second=x;
      }
    else
      {
        y=split(x->right,k-x->lsize()-1);
        x->right=y.first;
        x->Maintain();
        y.first=x;
      }
    return y;
}
long long qlarge(int a,int b)
{
    droot x,y;
    long long t;
    x=split(rootlarge,a-1);
    y=split(x.second,b-a+1);
    t=y.first->ans;
    x.second=mergelarge(y.first,y.second);
    rootlarge=mergelarge(x.first,x.second);
    return t;
}
long long qsmall(int a,int b)
{
    droot x,y;
    long long t;
    x=split(rootsmall,a-1);
    y=split(x.second,b-a+1);
    t=y.first->ans;
    x.second=mergesmall(y.first,y.second);
    rootsmall=mergesmall(x.first,x.second);
    return t;
}
int main()
{
    int i;
    long long ai=1,bi=1,maxn,minn,sum;
    treap_node *temp;
    freopen("sum.in","r",stdin);
    freopen("sum.out","w",stdout);
    a[0]=0;
    for (i=1;i<=100000;++i)
      {
        ai=(long long)((long long)ai*(long long)1023)%P;
        bi=(long long)((long long)bi*(long long)1025)%P;
        a[i]=ai^bi; 
        temp=new treap_node(a[i]);
        rootlarge=mergelarge(rootlarge,temp);
        temp=new treap_node(a[i]);
        rootsmall=mergesmall(rootsmall,temp);
      }
    scanf("%d",&t);
    while (t--)
      {
        scanf("%d%d%d%d",&l1,&r1,&l2,&r2);
        if (r2<l1)
          sum=0;
        else
        {
        if (l1>l2)
          l2=l1;
        if (r1<l2)
          {
            maxn=qlarge(l1,r2)-qlarge(l1,l2-1)-qlarge(r1+1,r2);
            if (r1+1<=l2-1)
              maxn+=qlarge(r1+1,l2-1);
            minn=qsmall(l1,r2)-qsmall(l1,l2-1)-qsmall(r1+1,r2);
            if (r1+1<=l2-1)
              minn+=qsmall(r1+1,l2-1);
            sum=maxn-minn;
          }
        else
          {
            if (r2>r1)
              {
                maxn=qlarge(l1,r2);
                if (l2>l1) maxn-=qlarge(l1,l2-1);
                if (r2>r1) maxn-=qlarge(r1+1,r2);
                minn=qsmall(l1,r2);
                if (l2>l1) minn-=qsmall(l1,l2-1);
                if (r2>r1) minn-=qsmall(r1+1,r2);
                sum=maxn-minn;
              }
            else
              {
                maxn=qlarge(l1,r2);
                if (l2>l1) maxn-=qlarge(l1,l2-1);
                minn=qsmall(l1,r2);
                if (l2>l1) minn-=qsmall(l1,l2-1);
                sum=maxn-minn;  
              }
          }
        }
        printf("%I64d\n",sum);
      }
}