Hdu-5806 NanoApe Loves Sequence(尺取法)

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 n numbers and a number  m m on the paper. 

Now he wants to know the number of continous subsequences of the sequence in such a manner that the  k k-th largest number in the subsequence is no less than  m m. 

Note : The length of the subsequence must be no less than  k k.

Input

The first line of the input contains an integer  T T, denoting the number of test cases. 

In each test case, the first line of the input contains three integers  n,m,k n,m,k. 

The second line of the input contains  n n integers  A1,A2,...,An A1,A2,...,An, denoting the elements of the sequence. 

1≤T≤10, 2≤n≤200000, 1≤k≤n/2, 1≤m,Ai≤109 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

題意：問數列中有多少區間滿足區間第k大數不小于m。

分析：尺取法求出以每個下标結尾的所有區間個數。

#include<iostream>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<cstdio>
#include<set>
#include<map>
#include<vector>
#include<cstring>
#include<stack>
#include<queue>
#define INF 2147483640
#define eps 1e-9
#define N 200005
#define MOD 1000000007
typedef long long ll;  
using namespace std;
int t,n,m,k,tot,a[N];
int main()
{
	scanf("%d",&t);
	for(int T = 1;T <= t;T++)
	{
		ll ans = 0;
		int i,j,sum = 0,tot = 0;
		scanf("%d%d%d",&n,&m,&k);
		for(int i = 1;i <= n;i++) scanf("%d",&a[i]);
		for(j = 1;j <= n && sum != k;j++) 
		{
			if(a[j] >= m) 
			{
				sum++;
				i = j;
			}
			tot++;
		}
		if(sum != k)
		{
			cout<<0<<endl;
			continue;
		}
		tot--,sum--;
		for(;i <= n;i++)
		{
			tot++;
			if(a[i] >= m) sum++;
			while(sum >= k)
			{
				if(a[i - tot + 1] >= m) sum--;
				tot--; 
			}
			tot++;sum++;
			ans += i - tot + 1ll; 
		}
		cout<<ans<<endl;
	}
}