HDU 5806 - NanoApe Loves Sequence Ⅱ

[align=left]Problem Description[/align]
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.
 

[align=left]Input[/align]
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
 

[align=left]Output[/align]
For each test case, print a line with one integer, denoting the answer.
 

[align=left]Sample Input[/align]

1
7 4 2
4 2 7 7 6 5 1

 
题意：给出三个数，n，m，k，有一个n个数组成的数列，在这个数列中的某个区间中的第k个数大于等于m，这样的区间一共有多少个。

我的解法就是每找到一个大于等于m的数就记下他的位置，然后从开始找到第k个开始，计算出能有多少个空间，具体公式在代码中。

#include <cstdio>

long long a[200000 + 5] = {0};
int main()
{
int T;
int m, n, k;
scanf("%d", &T);
while (T--)
{
scanf("%d%d%d", &n, &m, &k);
int cur = 0;
long long ans = 0;
for (int i = 1; i <= n; ++i)
{
int num;
scanf("%d", &num);
if (num >= m)
{
a[++cur] = i;
if (cur == k)
ans += a[1]*(n-i+1);
else if (cur > k)
ans += (a[cur-k+1]-a[cur-k])*(n-i+1);
}
}
printf("%I64d\n", ans);
}
return 0;
}