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NanoApe Loves Sequence Ⅱ

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/131072 K (Java/Others)
Total Submission(s): 514    Accepted Submission(s): 248

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

Source

BestCoder Round #86

题目大意:

退役狗 NanoApe 滚回去学文化课啦！

在数学课上，NanoApe 心痒痒又玩起了数列。他在纸上随便写了一个长度为 n 的数列，他又根据心情写下了一个数 m。

他想知道这个数列中有多少个区间里的第 k 大的数不小于 m，当然首先这个区间必须至少要有 k 个数啦。

解题思路：

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（尺取）求出。

时间复杂度O(n)。

MyCode：

/**
2016 - 08 - 07 上午
Author: ITAK

Motto:

今日的我要超越昨日的我，明日的我要胜过今日的我，
以创作出更好的代码为目标，不断地超越自己。
**/

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <set>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const LL INF = 1e9+5;
const int MAXN = 1e6+5;
const int MOD = 1e9+7;
const double eps = 1e-7;
const double PI = acos(-1);
using namespace std;
int a[MAXN], sum[MAXN];
inline bool cmp(LL a, LL b)
{
    return a > b;
}

int main()
{
    int T, n, m, k;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d",&n,&m,&k);
        memset(a, 0, sizeof(a));
        for(int i=1; i<=n; i++)
        {
            int x;
            scanf("%d",&x);
            if(x >= m)
                a[i] = 1;
        }
        int num = 0, r = 0;
        LL ans = 0;
        for(int i=1; i<=n; i++)
        {
            while(num<k && r<n)
            {
                r++;
                num += a[r];
            }
            if(num < k)
                break;
            ans += n-r+1;
            ///cout<<"ans = "<<ans<<endl;
            num -= a[i];
        }
        printf("%I64d\n",ans);
    }
    return 0;
}