Problem A. Ascending Rating
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1569 Accepted Submission(s): 436

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=−1 and 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

#include<bits/stdc++.h>
#define LL long long
#define INF 0x3f3f3f3f
const int maxn = 1e7+4;
using namespace std;
LL mod;
LL n,m,p,q,r,k;
LL a[maxn];
int T;
LL dep[maxn];
LL ans[maxn];
LL cnt[maxn];
int main()
{
    scanf("%d",&T);
    while(T--){
        scanf("%lld%lld%lld%lld%lld%lld%lld
 
",&n,&m,&k,&p,&q,&r,&mod);
        for(int i = 1;i<=k;i++){
            scanf("%lld",&a[i]);
        }
        for(int i =k+1;i<=n;i++){
            a[i] = (p*a[i-1]+q*i+r)%mod;
        }
        int r = n;
        int l = n+1;
        for(int i = n;i>=1;i--){
                while(l<=r&&a[i]>=a[dep[l]]){
                    l++;
                }
                l--;
                dep[l] = i;
            if(n-i+1>=m){
                while(l<=r&&dep[r]-i+1>m){
                    r--;
                }
                ans[i] = a[dep[r]];
                cnt[i] = r-l+1;

            }

        }
        LL A = 0;
        LL B = 0;
        for(int i = 1;i<=n-m+1;i++){
           // cout<<ans[i]<<' '<<cnt[i]<<endl;
            A += ans[i]^i;
            B += cnt[i]^i;
        }
        printf("%lld %lld\n",A,B);
    }
}