A. Valeriy and Deque
time limit per test
6 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Recently, on the course of algorithms and data structures, Valeriy learned how to use a deque. He built a deque filled with n
elements. The i-th element is ai (i = 1,2,…,n). He gradually takes the first two leftmost elements from the deque (let's call them A and B, respectively), and then does the following: if A>B, he writes A to the beginning and writes B to the end of the deque, otherwise, he writes to the beginning B, and A
writes to the end of the deque. We call this sequence of actions an operation.
For example, if deque was [2,3,4,5,1]
, on the operation he will write B=3 to the beginning and A=2 to the end, so he will get [3,4,5,1,2]
.
The teacher of the course, seeing Valeriy, who was passionate about his work, approached him and gave him q
queries. Each query consists of the singular number mj (j=1,2,…,q). It is required for each query to answer which two elements he will pull out on the mj
-th operation.
Note that the queries are independent and for each query the numbers A
and B
should be printed in the order in which they will be pulled out of the deque.
Deque is a data structure representing a list of elements where insertion of new elements or deletion of existing elements can be made from both sides.
Input
The first line contains two integers n
and q (2≤n≤105, 0≤q≤3⋅105) — the number of elements in the deque and the number of queries. The second line contains n integers a1, a2, ..., an, where ai (0≤ai≤109) — the deque element in i-th position. The next q lines contain one number each, meaning mj (1≤mj≤1018
).
Output
For each teacher's query, output two numbers A
and B — the numbers that Valeriy pulls out of the deque for the mj
-th operation.
Examples
Input
Copy
5 3 1 2 3 4 5 1 2 10
Output
Copy
1 2 2 3 5 2
Input
Copy
2 0 0 0
Output
Copy
Note
- Consider all 10 steps for the first test in detail:
- [1,2,3,4,5]
— on the first operation, A and B are 1 and 2, respectively.
So, 2
we write to the beginning of the deque, and 1
— to the end.
We get the following status of the deque: [2,3,4,5,1]
-
.
- [2,3,4,5,1]⇒A=2,B=3
- .
- [3,4,5,1,2]
- [4,5,1,2,3]
- [5,1,2,3,4]
- [5,2,3,4,1]
- [5,3,4,1,2]
- [5,4,1,2,3]
- [5,1,2,3,4]
- [5,2,3,4,1]⇒A=5,B=2
题意：
给你一个序列，然后每次操作都是从这个序列拿出前2个数，大的放序列头，小的放序列尾，问你第n次操作后，前2个数是什么
思路：
很明显，当我们的序列头为最大时，剩下的操作只是一个循环，我们用一个双向队列就能解决了。
#include<iostream>
#include<deque>
#include<memory.h>
#include<stdio.h>
#include<map>
#include<string.h>
#include<algorithm>
#include<vector>
#include<math.h>
#include<stack>
#include<queue>
#include<set>
using namespace std;
typedef long long LL;
deque<LL>q;
LL n,m,Max=0,temp,ci=0,a,b,aa[200100],bb[200100],arr[200100];
int main()
{
cin>>n>>m;
for(int i=1;i<=n;i++)
{
scanf("%lld",&temp);
q.push_back(temp);
Max = max(Max,temp);
if(i == 1)
aa[0]=temp;
else if(i ==2)
bb[0]=temp;
}
while(1)
{
a = q.front();
q.pop_front();
b = q.front();
q.pop_front();
if(a == Max)
{
q.push_front(b);
q.push_front(a);
break;
}
else
{
ci++;
aa[ci] = a;
bb[ci] = b;
if(a>b)
{
q.push_front(a);
q.push_back(b);
}
else
{
q.push_back(a);
q.push_front(b);
}
}
}
for(int i=1;i<=n;i++)
{
arr[i]=q.front();
q.pop_front();
}
for(int i=1;i<=m;i++)
{
cin>>temp;
if(temp<=ci)
cout<<aa[temp]<<' '<<bb[temp]<<endl;
else
{
temp-=ci;
temp%=(n-1);
if(temp!=0)
cout<<Max<<' '<<arr[temp+1]<<endl;
else
cout<<Max<<' '<<arr[n]<<endl;
}
}
return 0;
}