题意：

给定一个序列，问在这个序列里有多少区间第k的的数>=m

思路：

在比赛的时候是多想了，开始像区间第k大的问题，赛后想想实在是偏了。

正确的解法是枚举起点然后用尺取法维护一段区间，直到找到k个数>=m的最短终点r，ans便是后边这一小段即n-r+1。

code：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <sstream>
#include <string>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;const int INF=0x3fffffff;
const int inf=-INF;
const int N=200005;
const int M=2005;
const int mod=1000000007;
const double pi=acos(-1.0);#define cls(x,c) memset(x,c,sizeof(x))
#define cpy(x,a) memcpy(x,a,sizeof(a))
#define ft(i,s,n) for (int i=s;i<=n;i++)
#define frt(i,s,n) for (int i=s;i>=n;i--)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lrt  rt<<1
#define rrt  rt<<1|1
#define middle int m=(r+l)>>1
#define lowbit(x) (x&-x)
#define pii pair<int,int>
#define mk make_pair
#define IN freopen("in.txt","r",stdin)
#define OUT freopen("out.txt","w",stdout)int read() {char ch;while (ch = getchar(), !isdigit(ch));int res = ch - '0';while (ch = getchar(), isdigit(ch))res = res * 10 + ch - '0';return res;
}
//++++++++++++密++++++++++++++封++++++++++++++++++++线
int a[N];
int main()
{int T=read();while (T--){int n=read(),m=read(),k=read();for (int i=1;i<=n;i++) a[i]=read();int num=0,r=0;ll ans=0; //num表示大于m的个数,r为最短终点for (int i=1;i<=n;i++){while (num<k&&r<n){r++;num+=(a[r]>=m);}if (num<k) break;ans+=n-r+1;num-=(a[i]>=m);}printf("%I64d\n",ans);}
}