Problem A. Ascending Rating

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 3867 Accepted Submission(s): 1275

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=−1 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 :

Note that ``⊕ '' denotes binary XOR operation.

Sample Input

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

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题意：给定一个长度为 n 的序列，给出前 k 个值，后面根据公式自己算 n-k 个数。现在问你

A == 每个区间长度为m的子序列最大值异或上 i 的和；

B == 最大值变化次数异或上 i 的和。

题解：从后到前遍历一遍，开一个数组模拟队列，记录最长递减子序列，在 i <= n-m+1 期间，队列的的 l（队头），r（队尾）需要不断地更新区间。

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stack>
#include<list>
#include<algorithm>
using namespace std;
typedef long long ll;
const int M=70000005;
ll a[M];
ll qq[M];
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(int i=1;i<=k;i++){
            scanf("%lld",&a[i]);
        }
        for(int i=k+1;i<=n;i++){
            a[i]=(1ll*p*a[i-1]%mod+1ll*q*i%mod+r%mod)%mod;  // 注意
            // a[i]=(1ll*p*a[i-1]+q*i+r)%mod;  会错
        }
        ll A=0,B=0;
        int l=0,r=0;  // l 队头指针；；r 队尾指针
        for(int i=n;i>=1;i--){       
            while(r>l&&a[qq[r-1]]<=a[i]) r--;  // 逆序求单调递减序列
            qq[r++]=i;

            if(i>(n-m+1))  continue;    // (i 取值 在 1 - n-m+1)
            while(qq[l]>(i+m-1))  l++;   // 区间不断向前移动
            B+=1ll*(r-l)^i;        // 队尾减队头 是变化的次数
            A+=1ll*a[qq[l]]^i;    // 队头存放的是最大元素
        }
        printf("%lld %lld\n",A,B);
    }
    return 0;
}

HDU 6319 Problem A. Ascending Rating---数组模拟单调队列

Problem A. Ascending Rating

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