将大于等于m的数改为1，其余的改为0。问题转变成了有多少个区间的区间和>=k。可以枚举起点，二分第一个终点 或者尺取法。
#pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<vector> #include<map> #include<set> #include<queue> #include<stack> #include<iostream> using namespace std; typedef long long LL; const double pi=acos(-1.0),eps=1e-8; void File() { freopen("D:\in.txt","r",stdin); freopen("D:\out.txt","w",stdout); } inline int read() { char c = getchar(); while(!isdigit(c)) c = getchar(); int x = 0; while(isdigit(c)) { x = x * 10 + c - '0'; c = getchar(); } return x; } const int maxn=200000+10; int T,n,m,k,a[maxn],sum[maxn]; int main() { scanf("%d",&T); while(T--) { scanf("%d%d%d",&n,&m,&k); sum[0]=0; for(int i=1;i<=n;i++) { scanf("%d",&a[i]); if(a[i]>=m) a[i]=1; else a[i]=0; sum[i]=sum[i-1]+a[i]; } LL ans=0; for(int i=1;i<=n;i++) { int L=i,R=n,pos=-1; while(L<=R) { int mid=(L+R)/2; if(sum[mid]-sum[i-1]>=k) pos=mid,R=mid-1; else L=mid+1; } if(pos==-1) continue; ans=ans+n-pos+1; } printf("%lld ",ans); } return 0; }