Codeforces Round #569 (Div. 2) C. Valeriy and Deque

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;
}