Codeforces Round #569 (Div. 2) C. Valeriy and Deque
作者:互联网
链接:
https://codeforces.com/contest/1180/problem/C
题意:
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.
思路:
考虑当第一个值为最大值时,后面的顺序就是数组排列的顺序。
先记录第一个不是最大值,然后队列进出队,当第一个值最大时,将剩余元素出队到数组。找的时候取模计算即可。
代码:
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int MAXN = 1e5+10;
int a[MAXN];
int n, q;
int main()
{
int v, mmax = -1;
cin >> n >> q;
deque<int> que;
for (int i = 1;i <= n;i++)
{
cin >> v;
mmax = max(mmax, v);
que.push_back(v);
}
vector<pair<int, int> > par;
while (que.front() != mmax)
{
int fi = que.front();
que.pop_front();
int se = que.front();
que.pop_front();
par.emplace_back(fi, se);
if (fi < se)
swap(fi, se);
que.push_front(fi);
que.push_back(se);
}
que.pop_front();
int cnt = -1;
while (!que.empty())
{
a[++cnt] = que.front();
que.pop_front();
}
LL w;
while (q--)
{
cin >> w;
if (w <= par.size())
cout << par[w-1].first << ' ' << par[w-1].second << endl;
else
{
w -= par.size()+1;
cout << mmax << ' ' << a[w%(cnt+1)] << endl;
}
}
return 0;
}
标签:569,elements,Deque,int,Codeforces,deque,que,front,he 来源: https://www.cnblogs.com/YDDDD/p/11155479.html