Codeforces 323C Two permutations

题目描述

You are given two permutations pp and qq , consisting of nn elements, and mm queries of the form: l_{1},r_{1},l_{2},r_{2}l1,r1,l2,r2 $ (l_{1}<=r_{1}; l_{2}<=r_{2}) $ . The response for the query is the number of such integers from 11 to nn , that their position in the first permutation is in segment [l_{1},r_{1}][l1,r1] (borders included), and position in the second permutation is in segment [l_{2},r_{2}][l2,r2](borders included too).

A permutation of nn elements is the sequence of nn distinct integers, each not less than 11 and not greater than nn .

Position of number vv (1<=v<=n)(1<=v<=n) in permutation g_{1},g_{2},...,g_{n}g1,g2,...,gn is such number ii , that g_{i}=vgi=v .

输入输出格式

输入格式：

The first line contains one integer n\ (1<=n<=10^{6})n (1<=n<=106) , the number of elements in both permutations. The following line contains nn integers, separated with spaces: p_{1},p_{2},...,p_{n}\ (1<=p_{i}<=n)p1,p2,...,pn (1<=pi<=n) . These are elements of the first permutation. The next line contains the second permutation q_{1},q_{2},...,q_{n}q1,q2,...,qn in same format.

The following line contains an integer m\ (1<=m<=2·10^{5})m (1<=m<=2⋅105) , that is the number of queries.

The following mm lines contain descriptions of queries one in a line. The description of the ii -th query consists of four integers: a,b,c,d\ (1<=a,b,c,d<=n)a,b,c,d (1<=a,b,c,d<=n) . Query parameters l_{1},r_{1},l_{2},r_{2}l1,r1,l2,r2 are obtained from the numbers a,b,c,da,b,c,dusing the following algorithm:

Introduce variable xx . If it is the first query, then the variable equals 00 , else it equals the response for the previous query plus one.
Introduce function f(z)=((z-1+x)\ mod\ n)+1f(z)=((z−1+x) mod n)+1 .
Suppose l_{1}=min(f(a),f(b)),r_{1}=max(f(a),f(b)),l_{2}=min(f(c),f(d)),r_{2}=max(f(c),f(d))l1=min(f(a),f(b)),r1=max(f(a),f(b)),l2=min(f(c),f(d)),r2=max(f(c),f(d)) .

输出格式：

Print a response for each query in a separate line.

输入输出样例

输入样例#1：

输出样例#1：

输入样例#2：

输出样例#2：

1

1

2

把第二个排列的数在第一个排列中对应的位置记一下，主席树跑一跑就行了。

#include<iostream>

#include<cstdio>

#include<cstdlib>

#include<cmath>

#include<algorithm>

#include<cstring>

#define ll long long

#define maxn 1000005

using namespace std;

struct node{

    node *lc,*rc;

    int s;

}nil[maxn*30],*rot[maxn],*cnt;

int a[maxn],n,ky,num[maxn];

int m,le,ri,k,preans=-1,ple,pri;

char ch;

inline int add(int x,int y,const int ha){

  return (x+y)%ha;

}

node *update(node *u,int l,int r){

    node *ret=++cnt;

    *ret=*u;

    ret->s++;

    if(l==r) return ret;

    int mid=l+r>>1;

    if(le<=mid) ret->lc=update(ret->lc,l,mid);

    else ret->rc=update(ret->rc,mid+1,r);

    return ret;

}

int query(node *u,node *v,int l,int r){

    if(l>=le&&r<=ri) return v->s-u->s;

    int mid=l+r>>1,an=0;

    if(le<=mid) an+=query(u->lc,v->lc,l,mid);

    if(ri>mid) an+=query(u->rc,v->rc,mid+1,r);

    return an;

}

inline void prework(){

    cnt=rot[0]=nil->lc=nil->rc=nil;

    nil->s=0;

    for(int i=1;i<=n;i++){

        le=a[i];

        rot[i]=update(rot[i-1],1,n);

    }

}

inline void solve(){

    scanf("%d",&m);

    while(m--){

        scanf("%d%d%d%d",&le,&ri,&ple,&pri);

        le=add(le,preans,n)+1;

        ri=add(ri,preans,n)+1;

        ple=add(ple,preans,n)+1;

        pri=add(pri,preans,n)+1;

        if(le>ri) swap(le,ri);

        if(ple>pri) swap(ple,pri);

        preans=query(rot[ple-1],rot[pri],1,n);

        printf("%d\n",preans);

    }

}

int main(){

    scanf("%d",&n);

    int now;

    for(int i=1;i<=n;i++){

        scanf("%d",&now);

        num[now]=i;

    }

    for(int i=1;i<=n;i++){

        scanf("%d",&now);

        a[i]=num[now];

    }

    prework();

    solve();

    return 0;

}