hdu-6319-单调队列

Problem A. Ascending Rating

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 2884    Accepted Submission(s): 905

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 :

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

 

Source

2018 Multi-University Training Contest 3

 

　　如果是正向推过去的话发现很难维护这个变化次数，倒着来的话，由于单调队列的特性，大的值进来会把之前小于它的值全部pop出去，所以单调队列内的元素个数就是变化次数了，逆着跑一边deque就好了。

　　

 #include<bits/stdc++.h>
 using namespace std;
 #define LL long long
 #define inf 0x3f3f3f3f
 #define pb push_back
 #define pii pair<int,int>
 deque<int>q;
 int a[];
 int main(){
   LL n,i,j,k,m,P,Q,R,t,MOD;
   scanf("%d",&t);
   while(t--){
     scanf("%lld%lld%lld%lld%lld%lld%lld",&n,&m,&k,&P,&Q,&R,&MOD);
     q.clear();
     for(i=;i<=k;++i) scanf("%d",a+i);
     for(i=k+;i<=n;++i)a[i]=(P*a[i-]+Q*i+R)%MOD;
     LL ans1=,ans2=;
     for(i=n;i>=;--i){
       while(!q.empty()&&a[i]>=a[q.back()])q.pop_back();
       while(!q.empty()&&q.front()>i+m-) q.pop_front();
       q.push_back(i);
       if(i>n-m+) continue;
       ans1+=(a[q.front()]^i);
       ans2+=(q.size()^i);
     }
     cout<<ans1<<' '<<ans2<<endl;
   }
   return ;
 }