Hdu 5806 NanoApe Loves Sequence Ⅱ(双指针) （C++,Java）

Hdu 5806 NanoApe Loves Sequence Ⅱ(双指针)

Hdu 5806

题意：给出一个数组，求区间第k大的数大于等于m的区间个数

#include<queue>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define ll long long
#define inf 0x3f3f3f3f
using namespace std;
inline int read()
{
    int x=0,f=1;
    char ch=getchar();
    while(ch<'0'||ch>'9')
    {
        if(ch=='-')f=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9')
    {
        x=x*10+ch-'0';
        ch=getchar();
    }
    return x*f;
}
const int N=2e5+10;
int a[N];
int main()
{
    int T=read();
    while(T--){
        int n=read(),m=read(),k=read();
        for(int i=0;i<n;i++) a[i]=read();
        int l=0,r=0,cnt=0;
        ll ans=0;
        while(l<=r&&r<n){
            while(r<n&&cnt<k){
                if(a[r++]>=m) cnt++;
            }
            if(r>n||cnt<k) break;
            if(cnt==k) ans+=n-r+1;
            while(cnt==k){
                if(a[l++]>=m) cnt--;
                if(cnt==k) ans+=n-r+1;
            }
        }
        printf("%lld
",ans);
    }
    return 0;
}

Java版本实现，输入可能会导致超时，需要使用BufferedInputStream加速

import java.io.*;
import java.util.*;
class MyInputStream extends InputStream {  
    public BufferedInputStream bis = new BufferedInputStream(System.in);  
    public int read() throws IOException {  
        int i;  
        while ((i = bis.read()) < 48)  
            if (i == -1)  
                return -1;  
        int temp = 0;  
        while (i > 47) {  
            temp = temp * 10 + i - 48;  
            i = bis.read();  
        }  
        return temp;  
    }  
}  
public class Main
{
	static final int N=200005;
	static int a[]=new int[N];
    public static void main(String[] args) throws IOException{
    	MyInputStream sc=new MyInputStream();
    	int T=sc.read();
    	Integer zero=new Integer("0");
        while(!zero.equals(T--)){
        	int n=sc.read(),m=sc.read(),k=sc.read();
        	for(int i=0;i<n;i++) a[i]=sc.read();
        	int l=0,r=0,cnt=0;
            long ans=0;
            while(l<=r&&r<n){
                while(r<n&&cnt<k){
                    if(a[r++]>=m) cnt++;
                }
                if(r>n||cnt<k) break;
                if(cnt==k) ans+=n-r+1;
                while(cnt==k){
                    if(a[l++]>=m) cnt--;
                    if(cnt==k) ans+=n-r+1;
                }
            }
            System.out.println(ans);
        }
    	sc.close();
    }
}