HDU 5806 NanoApe Loves Sequence Ⅱ

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

就是相当于找到合适的队列（符合要求的）以后，然后不管其他的，大的就去替代大的，小的就去替代小的，逐个替代，只要满足的第k大的数字比m不小就可以

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<time.h>
using namespace std;
#define ll long long
const int N=200005;
int t,n,m,k,l,r,a
,tmp;
ll ans;
//移动是相对于tmp来说的
void left(){
if (a[l++]>=m)
tmp--;
}
void right(){
if (a[++r]>=m)
tmp++;
}
int main()
{
int i;
scanf("%d",&t);
while (t--)
{
scanf("%d%d%d",&n,&m,&k);
for (i=1;i<=n;i++)
scanf("%d",&a[i]);
l=1;
r=0;
ans=0;
tmp=0;
for (i=1;i<=n-k+1;i++)
{
while (tmp<k&&r<n)
right();
if (tmp>=k)
ans+=n-r+1;
//printf("n-r:%d ans:%d\n",n-r,ans);
left();
}
printf("%lld\n",ans);
}
return 0;
}

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
using namespace std;
#define ll long long
const int N=100010;
int main()
{
int n,m,i,k,t;
scanf("%d",&t);
while (t--)
{
ll ans=0ll;
scanf("%d%d%d",&n,&m,&k);
queue<int> Q;
for (i=0;i<n;i++)
{
int x;
scanf("%d",&x);
if (x>=m)
Q.push(i);  //重点是这里
if (Q.size()<k)
continue;
while (Q.size()>k)
Q.pop();
ans+=Q.front()+1;
}
printf("%lld\n",ans);
}
return 0;
}