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
It is guaranteed that ∑n≤7×107 and ∑k≤2×106.
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 :
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
题意：maxrating(i)就是从i开始m个数中的最大值。count(i)就是从i开始m个数中最大值变换的次数。然后按要求求出A，B。题目中只给前k项，后面到n个数要自己算出来。
思路：就是倒着用单调队列维护一个滑窗的最大值。
具体操作看代码吧。
代码：（标程）
const int N=10000010;
int T,n,m,k,P,Q,R,MOD,i,a[N],q[N],h,t;
long long A,B;
int main()
{
scanf("%d",&T);
while(T--)
{
scanf("%d%d%d%d%d%d%d",&n,&m,&k,&P,&Q,&R,&MOD);
for(i=1;i<=k;i++)scanf("%d",&a[i]);
for(i=k+1;i<=n;i++) a[i]=(1LL*P*a[i-1]+1LL*Q*i+R)%MOD;
h=1;
t=0;
A=0;
B=0;
i=n;
for(;i;i--)//倒着放进去
{
while(h<=t&&a[q[t]]<=a[i]) t--;将元素弹出
q[++t]=i;放入
if(i+m-1<=n)
{
while(q[h]>=i+m) h++;//保证是维护的元素是滑窗的大小
A+=(i^a[q[h]]);//队列首的元素就是最大值
B+=(i^(t-h+1));//队列中元素的个数就是此滑窗中最大值变换次数
}
}
printf("%lld %lld\n",A,B);
}
}