题目链接:
Time Limit: 4000/2000 MS (Java/Others)
 Memory Limit: 262144/131072 K (Java/Others)
/************************************************
┆  ┏┓   ┏┓ ┆   
┆┏┛┻━━━┛┻┓ ┆
┆┃       ┃ ┆
┆┃   ━   ┃ ┆
┆┃ ┳┛ ┗┳ ┃ ┆
┆┃       ┃ ┆ 
┆┃   ┻   ┃ ┆
┆┗━┓    ┏━┛ ┆
┆  ┃    ┃  ┆      
┆  ┃    ┗━━━┓ ┆
┆  ┃  AC代马   ┣┓┆
┆  ┃           ┏┛┆
┆  ┗┓┓┏━┳┓┏┛ ┆
┆   ┃┫┫ ┃┫┫ ┆
┆   ┗┻┛ ┗┻┛ ┆      
************************************************ */ 
 
 
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
//#include <bits/stdc++.h>
#include <stack>
#include <map>
 
using namespace std;
 
#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
 
typedef  long long LL;
 
template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<‘0‘||CH>‘9‘;F= CH==‘-‘,CH=getchar());
    for(num=0;CH>=‘0‘&&CH<=‘9‘;num=num*10+CH-‘0‘,CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + ‘0‘);
    putchar(‘\n‘);
}
 
const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=2e5+10;
const int maxn=2e3+14;
const double eps=1e-12;
int n,m,a[N],sum[N],k;
int main()
{      
        int t;
        read(t);
        while(t--)
        {
            read(n);read(m);read(k);
            For(i,1,n)
            {
                read(a[i]);
                if(a[i]>=m)sum[i]=sum[i-1]+1;
                else sum[i]=sum[i-1];
            }
            LL ans=0;
            int r=1;
            For(i,1,n-k+1)
            {
                r=max(i+k-1,r);
                while(sum[r]-sum[i-1]<k&&r<=n)r++;
                if(r<=n&&r-i+1>=k)ans=ans+(n-r+1);
            }
            cout<<ans<<"\n";
        } 
        return 0;
}
hdu-5806 NanoApe Loves Sequence Ⅱ(尺取法)
原文:http://www.cnblogs.com/zhangchengc919/p/5745171.html