Problem A. Ascending Rating

单调队列模板链接
https://blog.csdn.net/qq_41268947/article/details/81483268

Problem A. Ascending Rating

Problem Description

Before the start of contest, there are n$n$ ICPC contestants waiting in a long queue. They are labeled by 1$1$ to n$n$ from left to right. It can be easily found that the i$i$ -th contestant’s QodeForces rating is ai$a_i$ .
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$m$ , say [l,l+m−1]$[l,l+m-1]$ , and then inspect each contestant from left to right. Initially, he will write down two numbers maxrating=−1$maxrating=-1$ and count=0$count=0$ . Everytime he meets a contestant k$k$ with strictly higher rating than maxrating$maxrating$ , he will change maxrating$maxrating$ to ak$a_k$ and count$count$ to count+1$count+1$ .
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$maxrating$ and count$count$ . Please write a program to figure out the answer.

Input

The first line of the input contains an integer T(1≤T≤2000)$T(1\le T\le 2000)$ , denoting the number of test cases. In each test case, there are 7$7$ integers n,m,k,p,q,r,MOD(1≤m,k≤n≤107,5≤p,q,r,MOD≤109)$n,m,k,p,q,r,MOD(1\le m,k\le n\le {10}^7,5\le p,q,r,MOD\le {10}^9)$ in the first line, denoting the number of contestants, the length of interval, and the parameters k,p,q,r,MOD$k,p,q,r,MOD$ . In the next line, there are k$k$ integers a1,a2,...,ak(0≤ai≤109)$a_1,a_2,...,a_k(0\le a_i\le {10}^9)$ , denoting the rating of the first k$k$ contestants. To reduce the large input, we will use the following generator. The numbers p,q,r$p,q,r$ and MOD$MOD$ are given initially. The values ai(k<i≤n)$a_i(k<i\le n)$ are then produced as follows :

ai=(p×ai−1+q×i+r)modMOD$$\begin{array}{rcl}{a}_i& =& (p\times a_{i-1}+q\times i+r)modMOD\end{array}$$

It is guaranteed that ∑n≤7×107$\sum n\le 7\times {10}^7$ and ∑k≤2×106$\sum k\le 2\times {10}^6$ .

Output

Since the output file may be very large, let’s denote maxratingi$maxratin{g}_i$ and counti$coun{t}_i$ as the result of interval [i,i+m−1]$[i,i+m-1]$ . For each test case, you need to print a single line containing two integers A$A$ and B$B$ , where :

AB==∑i=1n−m+1(maxratingi⊕i)∑i=1n−m+1(counti⊕i)$$\begin{array}{rcl}A& =& \sum _{i=1}^{n-m+1}(maxratin{g}_i\oplus i)\\ B& =& \sum _{i=1}^{n-m+1}(coun{t}_i\oplus i)\end{array}$$

Note that “⊕$\oplus$ ” 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

有序列a1~an，对于每个连续的长为m的子区间[ax,ax+m-1]，求两个值：maxrating和count，两者初始值均为0，从左到右扫描每个元素，如果有元素大于maxrating，则更新maxrating且让count++。求出每个区间的maxrating和count后，再按题给公式求出A和B。

思路

这道题运用单调队列，但是我们不能从首到尾判断，因为单调队列只能存每个区间的最大值，但count求的是最大值更新了多少次，那么，区间变换了，count值也会不同。还有就是如果遇到那种最小值不在该区域内，需要队首后移这时也需要重算count
所以倒过来，从尾到首进行单调队列

代码

#include<cstdio>
const int N=10000010;

int i,a[N],v[N],h,t; //v是单调队列,h是队首，t是队尾
long long A,B;

int main()
{
    int T,n,m,k,p,q,r,mod;
    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;
        A=0,B=0;
        h=1,t=0;
        for(i=n;i;i--)
        {
            while(h<=t&&a[v[t]]<=a[i]) t--;
            //队列递减所以队尾上小于要插入值的元素都删除
            v[++t]=i;
            if(i+m-1<=n)//i可以作为区间起点了
            {
                while(v[h]>=i+m) h++;//去掉区间外的元素
                A+=i^a[v[h]];
                B+=i^(t-h+1);
            }
        }
        printf("%lld %lld\n",A,B);
    }
    return 0;
}