HDU5806 NanoApe Loves Sequence Ⅱ

NanoApe Loves Sequence Ⅱ

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/131072 K (Java/Others)
Total Submission(s): 517    Accepted Submission(s): 250

Problem Description

NanoApe, the Retired Dog, has returned back to prepare for for the National Higher Education Entrance Examination!

In math class, NanoApe picked up sequences once again. He wrote down a sequence with n numbers and a number m on the paper.

Now he wants to know the number of continous subsequences of the sequence in such a manner that the k-th largest number in the subsequence is no less than m.

Note : The length of the subsequence must be no less than k.

 

 

Input

The first line of the input contains an integer T, denoting the number of test cases.

In each test case, the first line of the input contains three integers n,m,k.

The second line of the input contains n integers A1,A2,...,An, denoting the elements of the sequence.

1≤T≤10, 2≤n≤200000, 1≤k≤n/2, 1≤m,Ai≤109

 

 

Output

For each test case, print a line with one integer, denoting the answer.

 

 

Sample Input

1

7 4 2

4 2 7 7 6 5 1

 

Sample Output

18

求有多少个区间，使得区间内的第k大的值>=m.

 

将不小于m的数看作1，剩下的数看作0，那么只要区间内1的个数不小于k则可行，枚举左端点，右端点可以通过two-pointer求出。

/* ***********************************************
Author        :guanjun
Created Time  :2016/8/7 10:02:54
File Name     :hdu5806.cpp
************************************************ */
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <iomanip>
#include <list>
#include <deque>
#include <stack>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 0x3f3f3f3f
#define maxn 10010
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << 61;
const double eps=1e-5;
using namespace std;
priority_queue<int,vector<int>,greater<int> >pq;
struct Node{
    int x,y;
};
struct cmp{
    bool operator()(Node a,Node b){
        if(a.x==b.x) return a.y> b.y;
        return a.x>b.x;
    }
};

bool cmp(int a,int b){
    return a>b;
}
int a[200010];
int sum[200010];
int n,m,k;
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif
    //freopen("out.txt","w",stdout);
    int t;
    cin>>t;
    while(t--){
        scanf("%d%d%d",&n,&m,&k);
        int x;
        sum[0]=0;
        for(int i=1;i<=n;i++){
            scanf("%d",&x);
            if(x>=m)a[i]=1;
            else a[i]=0;
            sum[i]=sum[i-1]+a[i];
        }
        ll ans=0;
        int r=1;
        for(int l=1;l<=n;l++){//枚举左端点
            while(r<=n&&sum[r]-sum[l-1]<k)r++;
            if(r>n)break;
            ans+=(n-r+1);
        }
        printf("%lld\n",ans);
    }
    return 0;
}