Problem A. Ascending Rating Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 2057 Accepted Submission(s): 622
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 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 #include
#include
#include
#include 对于一个序列，用单调队列维护滑块区间，
不要用STL尽量，有说不尽的毛病，能用数组就用数组，（还有数组开全局。） 单调队列的性质，它如何降低了复杂度呢？
比如初始化一个长度为k的序列，不难维护一个单调的序列，
由于我们只要最大值和最长单调长度，所以初始化的队列符合要求。
那么如果进了一个新元素呢？
如果比最小值小（咱们是倒着放的，比如9，8，7，现在进个2这样）
那么显然直接添加，如果大的话则贪心删除直至维护单调性质即可。 区间动的时候也删除了一个元素，只要这个元素不是我们区间维护的想要的答案，
就不做操作，如果是单调队列倒退，比如9,8,7，现在尾端下标离开了9的位置，则9退出队列。 倒退的那一端我们利用来判断是否对答案产生影响，前进的那一端我们
用来分析对单调队列整体的影响。
*/
ll n,m,k,p,q,r,mod;///全都要用ll，为啥会爆数据范围呢？
ll a[maxn];
ll lb=1,rb=0,deq[maxn];///双向队列 ll A,B; int main()
{
int t;scanf("%d",&t);
while(t--)
{
scanf("%lld%lld%lld%lld%lld%lld%lld",&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]=( a[i-1]*p+q*i+r )%mod; A=0,B=0;
lb=1,rb=0; for(int i=n;i;i--)
{
while(rb>=lb && deq[rb]<=a[i]) rb--;
deq[++rb]=a[i];
if(i+m-1<=n)
{
if(a[i+m]==deq[lb]) lb++;
B += (rb-lb+1)^(i);
cout< A += (deq[lb]^(i));
}
}
printf("%lld %lld\n",A,B);
}
return 0;
}
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