I have a sequence!

I have a sequence! Time Limit: 20000/10000ms (java/others) Memory Limit: 128000/64000kb (java/others) SubmitStatusProblem Description

Little Sunny Day: "I have a series!" ”

Small sunny Day: "I also have a lot of series!" ”

So a small sunny day to the sequence of all successive sub-series of the write out.

Then a sunny day writes the largest number in each successive sub-series.

So, how many are bigger than k?

Input

Multiple sets of data, first a positive integer t (t<=100), representing the number of groups of data

For each set of data, the first is two integers n (1<=n<=200000), K (0<=k<=10^9), but the sum of n in all data does not exceed 1000000.

Next is n integers a[i] (1<=a[i]<=10^9)

Output for each set of data, outputs an integer that represents the number of consecutive sub-sequences with the largest element greater than K. Sample Input

23 21 2 3 3 1 1 2 3

Sample Output

35

Hint

For sample one, there are 6 consecutive subsequence {1}{2}{3}{1,2}{2,3}{1,2,3} (note {1,3} does not meet test instructions because it is discontinuous)

A total of 3 {3}{2,3}{1,2,3} with the largest element greater than 2

For example two, there are 5 consecutive sub-sequences greater than 1, {2}{3}{1,2}{2,3}{1,2,3}

1#include <iostream>2#include <stdio.h>3#include <stdlib.h>4#include <math.h>5#include <string.h>6 using namespacestd;7typedefLong Longll;8ll t,n,k,a[200005],d[200005],cou;9 /**Ten after scanning from the go, for the value of a[i]>k, the number of sequences satisfying the condition =i+1.  One for the value of a[i]<=k, it will be equal to d[i-1]. The sequence range is at least the distance from I to the previous maximum value. The number of sequences will be the maximum number of values.  A Of course, this will be all the scope of the past, there will be no leakage, the value of all added up.  - */ - intMain () the { -scanf"%lld",&t); -      while(t--) -     { +scanf"%lld%lld",&n,&k); -cou=0; +memset (D,0,sizeof(d)); A          for(intI=0; i<n;i++) at         { -scanf"%lld",&a[i]); -                 if(a[i]>k) -                 { -d[i]+=i+1; -                 } in                 Else -                 { to                     if(i>0) d[i]=d[i-1]; +                     Elsed[i]=0; -                 } thecou+=D[i]; *         } $printf"%d\n", cou);Panax Notoginseng     } -     return 0; the}

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I have a sequence!