NanoApe Loves Sequence Ⅱ

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=5806

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


##题意：

求满足以下条件的连续子串的数目：
子串中第k大的数比m大.


##题解：

要维护第k大的数估计很麻烦，这里转换一下思路：
满足条件的子串中大于m的数的个数大于等于k.
先把数串中大于m的数的位置全部记录下来, pos[i]是第i个比m大的数.
对于i，对必定包含pos[i]~pos[i+k-1]这k个数的连续子串计数即可.


##代码：
``` cpp
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define LL long long
#define eps 1e-8
#define maxn 201000
#define mod 1000000007
#define inf 0x3f3f3f3f
#define mid(a,b) ((a+b)>>1)
#define IN freopen("in.txt","r",stdin);
using namespace std;

int n,m,k;

LL pos[maxn], cnt;

int main(void)

{

//IN;

int t; cin >> t;

while(t--)

{

    cnt = 0;

    cin >> n >> m >> k;

    for(int i=1; i<=n; i++) {

        int x; scanf("%d", &x);

        if(x >= m) pos[++cnt] = (LL)i;

    }

    LL ans = 0;

    for(int i=1; i<=cnt; i++) {

        if(i+k-1 > cnt) break;

        ans += (pos[i]-pos[i-1]) * (n-pos[i+k-1]+1);

    }

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

}

return 0;

}