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?<?k). The given operation
is correct — both subsegments do not touch unexistent elements. - 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.
Sample test(s)
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
这道题可以用线段树成段跟新做，属于染色一类线段树题目，可以定义两个变量stra,strb。stra表示这一线段是否被a[]数组覆盖，如果没有覆盖，那么stra=0,如果覆盖，那么被a[]覆盖的区间范围是a[stra]~a[stra+k-1],strb同理,只不过记录的是b[]的开始位置。
对于操作1，输入x1,y1,k,只要使得区间[y1,y1+k-1]的stra变成x1,strb变成y1.
对于操作2，输入x1,在线段树中寻找，如果这一点(即[x1,x1])的stra是0，就输出c[x1]，否则输出a[x1+strb-stra].
#include<stdio.h>
#include<string.h>
#define maxn 100006
int a[maxn],c[maxn];
int st1,st2,x1,y1,k;
struct node
{
int l,r,stra,strb;
}b[4*maxn];
void build(int l,int r,int i)
{
int mid;
b[i].l=l;b[i].r=r;b[i].stra=b[i].strb=0;
if(l==r)return;
mid=(l+r)/2;
build(l,mid,i*2);
build(mid+1,r,i*2+1);
}
void update(int l,int r,int i) //stra=x1,strb=y1;
{
int mid;
if(b[i].stra==x1 && b[i].strb==y1)return;
if(b[i].l==l && b[i].r==r){
b[i].stra=x1;b[i].strb=y1;return;
}
if(b[i].stra!=-1){
b[i*2].stra=b[i*2+1].stra=b[i].stra;
b[i*2].strb=b[i*2+1].strb=b[i].strb;
b[i].stra=b[i].strb=-1;
}
mid=(b[i].l+b[i].r)/2;
if(r<=mid)update(l,r,i*2);
else if(l>mid) update(l,r,i*2+1);
else {
update(l,mid,i*2);update(mid+1,r,i*2+1);
}
}
void question(int id,int i)
{
int mid;
if(b[i].stra!=-1){
st1=b[i].stra;st2=b[i].strb;return;
}
mid=(b[i].l+b[i].r)/2;
if(id<=mid) question(id,i*2);
else question(id,i*2+1);
}
int main()
{
int n,m,i,j,h,d,e,f,x;
while(scanf("%d%d",&n,&m)!=EOF)
{
for(i=1;i<=n;i++){
scanf("%d",&a[i]);
}
for(i=1;i<=n;i++){
scanf("%d",&c[i]);
}
build(1,n,1);
while(m--){
scanf("%d",&h);
if(h==1){
scanf("%d%d%d",&x1,&y1,&k);
update(y1,y1+k-1,1);
}
else if(h==2){
scanf("%d",&x);
st1=st2=0;
question(x,1);
if(st1==0){
printf("%d\n",c[x]);continue;
}
else {
printf("%d\n",a[x-st2+st1]);continue;
}
}
}
}
return 0;
}