题目链接：

http://codeforces.com/problemset/problem/292/E

E. Copying Data

time limit per test2 secondsmemory limit per test256 megabytes
#### 问题描述
> We often have to copy large volumes of information. Such operation can take up many computer resources. Therefore, in this problem you are advised to come up with a way to copy some part of a number array into another one, quickly.
>
> More formally, you've got two arrays of integers a1, a2, ..., an and b1, b2, ..., bn of length n. Also, you've got m queries of two types:
>
> Copy the subsegment of array a of length k, starting from position x, into array b, starting from position y, that is, execute by + q = ax + q for all integer q (0 ≤ q  Determine the value in position x of array b, that is, find value bx.
> For each query of the second type print the result — the value of the corresponding element of array b.
#### 输入
> The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 105) — the number of elements in the arrays and the number of queries, correspondingly. The second line contains an array of integers a1, a2, ..., an (|ai| ≤ 109). The third line contains an array of integers b1, b2, ..., bn (|bi| ≤ 109).
>
> Next m lines contain the descriptions of the queries. The i-th line first contains integer ti — the type of the i-th query (1 ≤ ti ≤ 2). If ti = 1, then the i-th query means the copying operation. If ti = 2, then the i-th query means taking the value in array b. If ti = 1, then the query type is followed by three integers xi, yi, ki (1 ≤ xi, yi, ki ≤ n) — the parameters of the copying query. If ti = 2, then the query type is followed by integer xi (1 ≤ xi ≤ n) — the position in array b.
>
> All numbers in the lines are separated with single spaces. It is guaranteed that all the queries are correct, that is, the copying borders fit into the borders of arrays a and b.
#### 输出
> For each second type query print the result on a single line.
####样例输入
> 5 10
> 1 2 0 -1 3
> 3 1 5 -2 0
> 2 5
> 1 3 3 3
> 2 5
> 2 4
> 2 1
> 1 2 1 4
> 2 1
> 2 4
> 1 4 2 1
> 2 2

样例输出

0

3

-1

3

2

3

-1

题意

给你大小为n的两个数组，a,b;

执行两个操作：

1,x,y,k，把a数组中以x开头长度为k的串覆盖到b数组中以y开头的长度为k的位置。

2,x, 查询b[x];

题解

对于查询1在b数组上做成段覆盖更新：(y,y+k-1)，值为查询的id。然后对于查询2做单点查询，查到对应的查询再去a数组中找对应的位置。

代码

#include<map>

#include<set>

#include<cmath>

#include<queue>

#include<stack>

#include<ctime>

#include<vector>

#include<cstdio>

#include<string>

#include<bitset>

#include<cstdlib>

#include<cstring>

#include<iostream>

#include<algorithm>

#include<functional>

using namespace std;

#define X first

#define Y second

#define mkp make_pair

#define lson (o<<1)

#define rson ((o<<1)|1)

#define mid (l+(r-l)/2)

#define sz() size()

#define pb(v) push_back(v)

#define all(o) (o).begin(),(o).end()

#define clr(a,v) memset(a,v,sizeof(a))

#define bug(a) cout<<#a<<" = "<<a<<endl

#define rep(i,a,b) for(int i=a;i<(b);i++)

#define scf scanf

#define prf printf

typedef long long LL;

typedef vector<int> VI;

typedef pair<int,int> PII;

typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;

const LL INFL=0x3f3f3f3f3f3f3f3fLL;

const double eps=1e-8;

const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=1e5+10;

int setv[maxn<<2];

void pushdown(int o){

    if(setv[o]!=-1){

        setv[lson]=setv[rson]=setv[o];

        setv[o]=-1;

    }

}

int qp,qv;

void query(int o,int l,int r){

    if(l==r||setv[o]!=-1){

        qv=setv[o];

    }else{

        if(qp<=mid) query(lson,l,mid);

        else query(rson,mid+1,r);

    }

}

int ul,ur,uv;

void update(int o,int l,int r){

    if(ul<=l&&r<=ur){

        setv[o]=uv;

    }else{

        pushdown(o);

        if(ul<=mid) update(lson,l,mid);

        if(ur>mid) update(rson,mid+1,r);

    }

}

struct Node{

    int x,y,k;

    Node(int x,int y,int k):x(x),y(y),k(k){}

    Node(){}

}que[maxn];

int a[maxn],b[maxn];

int n,m;

void init(){

    clr(setv,-1);

}

int main() {

    scf("%d%d",&n,&m);

    init();

    for(int i=1;i<=n;i++) scf("%d",&a[i]);

    for(int i=1;i<=n;i++) scf("%d",&b[i]);

    ul=1,ur=n,uv=0;

    update(1,1,n);

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

        int cmd; scf("%d",&cmd);

        if(cmd==1){

            int x,y,k;

            scf("%d%d%d",&x,&y,&k);

            que[i]=Node(x,y,k);

            ul=y,ur=y+k-1,uv=i;

            update(1,1,n);

        }else{

            int x; scf("%d",&x);

            qp=x;

            query(1,1,n);

            int ans;

            if(qv==0) ans=b[x];

            else ans=a[x-que[qv].y+que[qv].x];

            prf("%d\n",ans);

        }

    }

    return 0;

}

//end-----------------------------------------------------------------------