題目連結：

http://acm.hdu.edu.cn/showproblem.php?pid=6319

Problem Description

Before the start of contest, there are n ICPC contestants waiting in a long queue. They are labeled by 1 to n from left to right. It can be easily found that the i-th contestant's QodeForces rating is ai.
Little Q, the coach of Quailty Normal University, is bored to just watch them waiting in the queue. He starts to compare the rating of the contestants. He will pick a continous interval with length m, say [l,l+m−1], and then inspect each contestant from left to right. Initially, he will write down two numbers maxrating=−1and count=0. Everytime he meets a contestant k with strictly higher rating than maxrating, he will change maxrating to ak and count to count+1.
Little T is also a coach waiting for the contest. He knows Little Q is not good at counting, so he is wondering what are the correct final value of maxrating and count. Please write a program to figure out the answer.

Input

The first line of the input contains an integer T(1≤T≤2000), denoting the number of test cases.
In each test case, there are 7 integers n,m,k,p,q,r,MOD(1≤m,k≤n≤107,5≤p,q,r,MOD≤109) in the first line, denoting the number of contestants, the length of interval, and the parameters k,p,q,r,MOD.
In the next line, there are k integers a1,a2,...,ak(0≤ai≤109), denoting the rating of the first k contestants.
To reduce the large input, we will use the following generator. The numbers p,q,r and MOD are given initially. The values ai(k<i≤n) are then produced as follows :ai=(p×ai−1+q×i+r)modMOD。

 Output

Since the output file may be very large, let's denote maxratingi and counti as the result of interval [i,i+m−1].
For each test case, you need to print a single line containing two integers A and B, where :

AB==∑i=1n−m+1(maxratingi⊕i)∑i=1n−m+1(counti⊕i)
Note that ``⊕'' denotes binary XOR operation.

Sample Input

1 10 6 10 5 5 5 5 3 2 2 1 5 7 6 8 2 9

 Sample Output

46 11

Description：

從 1 到 n-m+1 分別為起點的連續的長度為 m 的區間中，以區間起點為開始的上升序列的長度和最大值

Solution：

滑動視窗 + 單調佇列。

如果當前元素比棧頂元素大，則不斷彈出棧頂元素直到當前元素比棧頂元素小，然後把當前元素壓入棧中。如果當前元素比棧頂元素小，則直接入棧。如果當前棧中有元素不在當前區間中，則不斷彈出棧底元素。每段區間最大值就是棧底元素，遞增序列的長度就是棧的大小。

Code：

#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#define mst(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
const int INF = 0x3f3f3f3f;
const double eps = 1e-9;
const int MaxN = 1e7 + 5;

LL a[MaxN];
int pos[MaxN];
int n, m;
LL ans, res;

void solve() {
	for(int i = 0; i <= n; i++) pos[i] = 0;
	ans = res = 0;
	int l = 1, r = 0;
	for(int i = n; i >= 1; i--) {
		while(r >= l && a[pos[r]] <= a[i]) r--;
		pos[++r] = i;
		if(i + m - 1 > n) continue;
		while(r >= l && pos[l] > i + m - 1) l++;
		ans += a[pos[l]] ^ i;
		res += (r - l + 1) ^ i;
	}
}

int main()
{
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    
    int t; cin >> t;
    while(t--) {
		int k, p, q, r, mod;
		cin >> n >> m >> k >> p >> q >> r >> mod;
		for(int i = 1; i <= k; i++) cin >> a[i];
		for(int i = k + 1; i <= n; i++) {
			LL tmp = (p * a[i-1] + (LL)q * i + r) % mod;
			a[i] = tmp;
		}
		solve();
		cout << ans << " " << res << endl;
	}
    return 0;
}