杭电多校第三场 A Ascending Rating

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

 

题意：输入 n,m,k,p,q,r,MOD  输入k个数，总共有n个数，如果输入不够的话就由上面那个公式把剩下的补齐，然后区间长度为m，访问所有的长度m的区间，

我们要求每个区间的最大值变化次数 （比如 3 2 2 1 5 7        那么我们的最大值3-5-7   所以次数是3，最大值是7） 和最大值     ，然后求每个区间的变化次数和最大值分别异或第几个区间号i的累加和

 

思路：一看复杂度，只有o(n)可以过，那么就去想o（n）算法，乍一看最大值就是用一个滑动窗口可以求出来是o(n)的，但是count我们不好记录，我们可以发现count的那些最大值

其实是一个单调递增的序列，我们想怎么去维护呢，我们求最大值那么肯定就要用o(n)了，我们可不可以同时求呢，说明我的单调序列也要能尽量沿用上一个区间的值，避免遍历

我们可以回想下，求滑动窗口的时候那个队列就是存的一个以队列首为最大值的一个最长单调递减序列，那么我们怎么由递减序列变成递增序列呢，很简单，我们倒着求滑动窗口

到时候队列长度即是最大值的变化次数

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<deque>
#include<vector>
#include<algorithm>
using namespace std;
typedef long long ll;
int n,m,t,k,p,q,r,mod;
int a[10000001];
deque<ll> qx;
int cnt;
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%d%d%d%d%d",&n,&m,&k,&p,&q,&r,&mod);
        qx.clear();
        for(int i=1;i<=k;i++)
            scanf("%d",&a[i]);
        for(int i=k+1;i<=n;i++)
             a[i]=(1ll*p*a[i-1]+1ll*q*i+r)%mod;
        ll x=0,y=0;
        for(int i=n;i>=1;i--)
        {
            while(!qx.empty()&&a[i]>=a[qx.front()])
                 qx.pop_front();
            qx.push_front(i);
            if(i-1<=n-m)
            {
                while(!qx.empty()&&qx.back()>=i+m) qx.pop_back();
                x+=i^a[qx.back()];
                y+=i^qx.size();
            }
        }    
        printf("%lld %lld\n",x,y);
    }
    
}

 

 

转载于:https://www.cnblogs.com/Lis-/p/9393797.html