hdu 5806 NanoApe Loves Sequence Ⅱ (尺取法)

NanoApe Loves Sequence Ⅱ
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 262144/131072 K (Java/Others)
Total Submission(s): 286 Accepted Submission(s): 138

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

这是很明显的尺取法，前后追击。统计答案就可以了。

#include
#include
#include
#include
using namespace std;

const int N = 200020;
int a[N];
int n, m, k;

void solve() {
    int i, j;
    long long ans = 0;
    int num = 0;
    j = 0;
    for(i = 1; i <= n; i++) {
        while(num < k && j <= n) {
            j++;
            if(a[j] >= m) num++;
        }
        if(num < k) break;
        ans += n-j+1;
        if(a[i] >= m) num--;
    }
    printf("%I64d\n", ans);
}

int main() {
    int T;
    scanf("%d", &T);
    while(T--) {
        scanf("%d%d%d", &n, &m, &k);
        for(int i = 1; i <= n; i++) {
            scanf("%d", &a[i]);
        }
        solve();
    }
    return 0;
}