HDU 5806 NanoApe Loves Sequence Ⅱ two pointers

NanoApe Loves Sequence Ⅱ

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/131072 K (Java/Others)
Total Submission(s): 425    Accepted Submission(s): 207

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

题意：给n,m,k,n个数的数列，询问这个数列中有多少个区间里的第 k 大的数不小于 m。

two pointers可做，统计l到r区间>=m的数字个数，当等于k个时候就进行计数。

CODE

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <string.h>
#include <stdlib.h>

using namespace std;
typedef long long LL;
const int N = 2e5+10;

int a[N];

int main()
{
    int T;
    scanf("%d",&T);
    while(T--){
        int n,m,k;
        scanf("%d%d%d",&n,&m,&k);
        for(int i = 1;i <= n;i++)
            scanf("%d",&a[i]);
        LL ans = 0;
        int tmp = 0;
        for(int l = 1,r = 1;l <= n;l++){
            while(r <= n && tmp < k){
                if(a[r++] >= m) tmp++;
            }
            if(tmp == k){
                ans += n-r+2;
            }
            if(a[l] >= m) tmp--;
        }
        printf("%I64d\n",ans);
    }
    return 0;
}