E. Copying Data
time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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:

  1. 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 < k). The given operation is correct — both subsegments do not touch unexistent elements.
  2. 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.

Input

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.

Output

For each second type query print the result on a single line.

Examples
input
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
output
0
3
-1
3
2
3
-1

题目链接:点击传送

思路:分块

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
#define bug(x) cout<<"bug"<<x<<endl;
const int N=2e5+,M=1e6+,inf=;
const ll INF=1e18+,mod=1e9+;
/// 数组大小
int n,q,k;
int a[N],b[N],pos[N];
int si[N],st[N];
void update(int L,int R,int l)
{
if(pos[L]==pos[R])
{
if(si[pos[L]])
for(int i=(pos[L]-)*k+,j=st[pos[L]];i<=pos[L]*k;i++,j++)
b[i]=a[j];
si[pos[L]]=;
for(int i=L,j=;i<=R;i++,j++)
b[i]=a[l+j];
return;
}
if(si[pos[L]])
for(int i=(pos[L]-)*k+,j=st[pos[L]];i<=pos[L]*k;i++,j++)
b[i]=a[j];
si[pos[L]]=;
for(int i=L;i<=pos[L]*k;i++)
b[i]=a[l++];
for(int i=pos[L]+;i<=pos[R]-;i++,l+=k)
{
si[i]=;
st[i]=l;
}
if(si[pos[R]])
for(int i=(pos[R]-)*k+,j=st[pos[R]];i<=pos[R]*k;i++,j++)
b[i]=a[j];
si[pos[R]]=;
for(int i=(pos[R]-)*k+;i<=R;i++)
b[i]=a[l++];
}
int query(int x)
{
if(si[pos[x]])
{
int l=x-(pos[x]-)*k-;
return a[st[pos[x]]+l];
}
else
return b[x];
}
int main()
{
scanf("%d%d",&n,&q);
k=sqrt(n);
for(int i=;i<=n;i++)
scanf("%d",&a[i]),pos[i]=(i-)/k+;
for(int i=;i<=n;i++)
scanf("%d",&b[i]);
while(q--)
{
int t;
scanf("%d",&t);
if(t==)
{
int x,y,l;
scanf("%d%d%d",&x,&y,&l);
update(y,y+l-,x);
}
else
{
int x;
scanf("%d",&x);
printf("%d\n",query(x));
}
///for(int i=1;i<=n;i++)
/// printf("%d ",query(i));
/// printf("\n");
}
return ;
}
E. Copying Data
time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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:

  1. 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 < k). The given operation is correct — both subsegments do not touch unexistent elements.
  2. 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.

Input

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.

Output

For each second type query print the result on a single line.

Examples
input
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
output
0
3
-1
3
2
3
-1

Croc Champ 2013 - Round 1 E. Copying Data 分块的更多相关文章

  1. Croc Champ 2013 - Round 1 E. Copying Data 线段树

    题目链接: http://codeforces.com/problemset/problem/292/E E. Copying Data time limit per test2 secondsmem ...

  2. Croc Champ 2013 - Round 2 C. Cube Problem

    问满足a^3 + b^3 + c^3 + n = (a+b+c)^3 的 (a,b,c)的个数 可化简为 n = 3*(a + b) (a + c) (b + c) 于是 n / 3 = (a + b ...

  3. D. Connected Components Croc Champ 2013 - Round 1 (并查集+技巧)

    292D - Connected Components D. Connected Components time limit per test 2 seconds memory limit per t ...

  4. ACM: Copying Data 线段树-成段更新-解题报告

    Copying Data Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u Description W ...

  5. E. Copying Data 解析(線段樹)

    Codeforce 292 E. Copying Data 解析(線段樹) 今天我們來看看CF292E 題目連結 題目 給你兩個陣列\(a,b\),有兩種操作:把\(a\)的一段複製到\(b\),或者 ...

  6. codeforces 292E. Copying Data

    We often have to copy large volumes of information. Such operation can take up many computer resourc ...

  7. CROC 2016 - Final Round [Private, For Onsite Finalists Only] C. Binary Table FWT

    C. Binary Table 题目连接: http://codeforces.com/problemset/problem/662/C Description You are given a tab ...

  8. CROC 2016 - Elimination Round (Rated Unofficial Edition) D. Robot Rapping Results Report 拓扑排序+二分

    题目链接: http://www.codeforces.com/contest/655/problem/D 题意: 题目是要求前k个场次就能确定唯一的拓扑序,求满足条件的最小k. 题解: 二分k的取值 ...

  9. codeforces 292E. Copying Data 线段树

    题目链接 给两个长度为n的数组, 两种操作. 第一种, 给出x, y, k, 将a[x, x+k-1]这一段复制到b[y, y+k-1]. 第二种, 给出x, 输出b[x]的值. 线段树区间更新单点查 ...

随机推荐

  1. [py模块]random&string取随机字符串

    栗子 - 取n位的随机字符串(大小写/数字) def get_random_str(len_str): import string import random letters_nums = strin ...

  2. PAT 1003 Emergency[图论]

    1003 Emergency (25)(25 分) As an emergency rescue team leader of a city, you are given a special map ...

  3. Summary: Binary Search

    Iterative ways: int binarySearch (int[] a, int x) { int low = 0; int high = a.length - 1; int mid; w ...

  4. shell篇(一)

    login shell与non-login shell: login shell:取得shell时,需要完整的登入流程.如:tty1~tty6登入时,需要输入用户名和密码.此时取得的shell就称为l ...

  5. R 简明教程

    R 是一门统计语言.它有很多数据分析和挖掘程序包.可以用来统计.分析和制图. 你也可以在 LaTeX 文档中运行 R 命令. # 注释以 # 开始 # R 语言原生不支持 多行注释 # 但是你可以像这 ...

  6. linux常用命令:split 命令

    split是linux下常用的分割文件命令.Linux下文件分割可以通过split命令来实现,而用cat进行文件合并.而分割可以指定按行数分割和按大小分割两种模式. 1.命令格式: split [OP ...

  7. python3.4学习笔记(十九) 同一台机器同时安装 python2.7 和 python3.4的解决方法

    python3.4学习笔记(十九) 同一台机器同时安装 python2.7 和 python3.4的解决方法 同一台机器同时安装 python2.7 和 python3.4不会冲突.安装在不同目录,然 ...

  8. iOS之第三方库以及XCode插件介绍

    前言 第三方库是现在的程序员离不开的东西 不光是APP开发 基本上所有的商业项目 都会或多或少的使用到第三方库 Github上Star>100的开源库数量如下 可以看到JS以绝对的优势排名第一 ...

  9. oracle创建dblink的脚本

    创建dblink的脚本 create public database link wsbsbb_27(dblink名字) connect to wsbsbb(用户名) IDENTIFIED BY wsb ...

  10. selenium自动化定位方法

    用selenium操作浏览器进行自动化操作其实就是通过元素属性执行相关操作.所以,我们要知道怎样去查找元素,定位元素. 常见的定位属性有: #查找元素的id find_elements_by_id(i ...