Kanade's sum

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 505 Accepted Submission(s): 176

Problem Description
Give you an array A[1..n]of length n.

Let f(l,r,k) be the k-th largest element of A[l..r].

Specially , f(l,r,k)=0 if r−l+1<k.

Give you k , you need to calculate ∑nl=1∑nr=lf(l,r,k)

There are T test cases.

1≤T≤10

k≤min(n,80)

A[1..n] is a permutation of [1..n]

∑n≤5∗105

Input
There is only one integer T on first line.

For each test case,there are only two integers n,k on first line,and the second line consists of n integers which means the array A[1..n]

Output
For each test case,output an integer, which means the answer.

Sample Input

5 2

1 2 3 4 5

Sample Output

Source
2017 Multi-University Training Contest - Team 3

思路：
　　很容易想到按公式算是不可行的（O（n^2）的时间复杂度），然后想到枚举计算每个数的贡献：即第k大为x的区间个数乘以x
　　然后只要考虑怎么快速求出第k大为x的区间个数。如果能知道左边大于x的80个数的位置和右边大于x的80个数的位置就可以计算区间个数了。
　　之后想到用从小到大枚举或者从大到小，用链表维护即可。
　　比赛时用的set模拟的，结果被卡了，T的连妈都不认识。还是觉得时限卡太紧了。

 #include <bits/stdc++.h>

 using namespace std;

 #define MP make_pair

 #define PB push_back

 typedef long long LL;

 typedef pair<int,int> PII;

 const double eps=1e-;

 const double pi=acos(-1.0);

 const int K=1e6+;

 const int mod=1e9+;

 int n,k,p[K],pre[K],nxt[K],pos[K],tl[],tr[];

 LL ans;

 int main(void)

 {

     int t;cin>>t;

     while(t--)

     {

         ans=;

         scanf("%d%d",&n,&k);

         for(int i=;i<=n;i++)

             scanf("%d",p+i),pos[p[i]]=i;

         for(int i=;i<=n;i++)

             pre[i]=i-,nxt[i]=i+;

         pre[]=,nxt[n]=n+;

         for(int i=;i<=n;i++)

         {

             int la=,lb=;

             for(int j=pos[i];j>&&la<=k;j=pre[j])

                 tl[la++]=j-pre[j];

             for(int j=pos[i];j<=n&&lb<=k;j=nxt[j])

                 tr[lb++]=nxt[j]-j;

             for(int j=;j<la;j++)

             if(k-j-<lb)

                 ans+=i*1LL*tl[j]*tr[k-j-];

             pre[nxt[pos[i]]]=pre[pos[i]];

             nxt[pre[pos[i]]]=nxt[pos[i]];

         }

         printf("%lld\n",ans);

     }

     return ;

 }