Problem A. Ascending Rating

最新推荐文章于 2024-02-19 10:33:47 发布

原创最新推荐文章于 2024-02-19 10:33:47 发布 · 692 阅读

0 ·

CC 4.0 BY-SA版权

单调队列专栏收录该内容

4 篇文章

订阅专栏

本文介绍了一个竞赛编程题目，涉及计算特定区间内元素的最大值和上升序列的长度，通过使用单调队列进行高效求解。

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=0 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<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

Recommend

chendu | We have carefully selected several similar problems for you: 6331 6330 6329 6328 6327

题意：

从1到n-m+1分别为起点的连续的m个区间长度中从起点开始的上升序列的长度和最大值

分析：

窗口问题就需要用到单调队列，只是需要从后往前

代码：

#include<bits/stdc++.h>
#define ll long long
#define inf 0x3f3f3f3f
#define pii pair<ll,ll>
using namespace std;
const int maxn=10000005;
ll nex[maxn],a[maxn],qq[maxn],t;
ll p,q,r,mo,ans,cnt,j,ctt,n,m,k,i,ma,ct,kk;
int main()
{
    ll T;
    scanf("%lld",&T);
    while(T--)
    {
        i=1;
        scanf("%lld%lld%lld%lld%lld%lld%lld",&n,&m,&k,&p,&q,&r,&mo);
        while(i<=n)
        {
            if(i<=k) scanf("%lld",&a[i]);
            else a[i]=(p*a[i-1]+q*i+r)%mo;
            i++;
        }
        ans=cnt=0;
        for(ll h=1,t=0,i=n;i;i--)
        {
            while(h<=t&&a[qq[t]]<=a[i])t--;
            qq[++t]=i;
            if(i+m-1<=n){
               while(qq[h]>=i+m)h++;
               ans+=i^a[qq[h]];
               cnt+=i^(t-h+1);
            }
        }
    printf("%lld %lld\n",ans,cnt);
    }
    return 0;
}