HDU6319：Problem A. Ascending Rating（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): 587    Accepted Submission(s): 150

 

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

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

 题意：生成N个数，对于每个M长度的区间，设为[L, L+M-1]，贡献1为区间最大值异或x，贡献2为以以L为起点的最长上升子序列异或X，输出所有M长度区间的贡献之和。

思路：首先按照公式生成N个数，正着做很难维护，考虑反着做维护一个单调递减的序列，任意时刻队列的大小就是上升子序列的长度，注意边界条件就行，然后得用数组模拟双端队列。

# include 
using namespace std;
typedef long long LL;
const LL mod = 1e9+7;
const int maxn = 1e7+30;
int a[maxn], d[maxn];
int T, n, m, k, p, q, r, MOD;
void scan(int &ret)
{
    char c; ret = 0;
    while((c=getchar()) < '0' || c > '9');
    while(c>='0' && c<='9') ret = ret*10 + (c-'0'), c = getchar();
}
int main()
{
    for(scan(T);T;--T)
    {
        LL ans = 0, ans2=0;
        scan(n);scan(m);scan(k);scan(p);scan(q);scan(r);scan(MOD);
        p %= MOD;q %= MOD;r %= MOD;
        for(int i=1; i<=k; ++i) scan(a[i]);
        for(int i=k+1; i<=n; ++i)
        {
            a[i] = 1LL*p*a[i-1]%MOD+(1LL*q*i%MOD+r)%MOD;
            if(a[i] >= MOD) a[i] -= MOD;
        }
        int tail = 10000001, head = tail+1;
        for(int i=n; i>=n-m+1; --i)
        {
            while(tail+1 != head && a[i] >= a[d[head]]) ++head;
            d[--head] = i;
        }
        ans = (tail-head+1)^(n-m+1);
        ans2 = (a[d[tail]])^(n-m+1);
        for(int i=n-m; i>0; --i)
        {
            if(d[tail] == i+m) --tail;;
            while(tail+1!= head && a[i] >= a[d[head]]) ++head;
            d[--head] = i;
            ans += (tail-head+1)^i;
            ans2 += a[d[tail]]^i;
        }
        printf("%lld %lld\n",ans2,ans);
    }
    return 0;
}