HDU 5806/BC 86C NanoApe Loves Sequence Ⅱ

题解：满足第K大数大于等于m，也就存在k个大于等于m的数，所以滑动窗口即可，对于每个位置作为右端点求其最右的左端点。累加即可。

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
#include<list>
#include<set>
#include<map>
#include<stack>
#include<queue>
#include<deque>
#define mem(x,y) memset(x,y,sizeof(x))
#define pb push_back
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
#define bug puts("===========");
#define zjc puts("");
const double pi=(acos(-1.0));
const double eps=1e-8;
const ll INF=1e18+10;
const ll inf=1e9+10;
const int mod=1e9+7;
const int maxn=2e5+10;
/*=======================================*/

int a[maxn];
int main()
{
int T_T;
scanf("%d",&T_T);
while(T_T--){
int n,m,k;
scanf("%d%d%d",&n,&m,&k);
for(int i=1;i<=n;i++){
scanf("%d",a+i); a[i]-=m;
}
int l=1,cnt=0;
ll ans=0;
for(int i=1;i<=n;i++){
if(a[i]>=0) cnt++;
if(cnt<k) continue;
while(1){
if(cnt==k&&a[l]>=0) break;
if(a[l++]>=0) cnt--;
}
ans+=l;
// cout<<i<<" "<<l<<endl;
}
printf("%I64d\n",ans);

}
return 0;
}