B - Minimum Inversion Number Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1394 Description The inversion number of a given number sequence a1, a2, ..., an is the number of pairs (ai, aj) that…

typeof可以用来检测给定变量的数据类型,typeof是一个操作符而不是函数,所以圆括号可以省略. Undefined类型只有一个值,即特殊的undefined.在使用var声明变量但未对其加以初始化时,这个变量的值就是undefined. undefined的主要目的是用于比较.第三版引入这个值是为了正式区分空对象指针与未经初始化的变量. Null类型是第二个只有一个值得数据类型,这个特殊的值是null. 从逻辑角度来看,null值表示一个空对象指针,而这也正是使用typeof操作符检测nu…

#include<iostream> #include<cstdio> using namespace std; struct node { int l, r, m; int max; }num[800005]; int val[200005]; int n, m; int init(int l, int r, int k) { num[k].l = l; num[k].r = r; if(l==r) { num[k].m = l; num[k].max=val[l]; retur…

Gold Coins Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 21767   Accepted: 13641 Description The king pays his loyal knight in gold coins. On the first day of his service, the knight receives one gold coin. On each of the next two days…

D. Minimum number of steps time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output We have a string of letters 'a' and 'b'. We want to perform some operations on it. On each step we choose one of s…

Gold Balanced Lineup Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 13215 Accepted: 3873 Description Farmer John's N cows (1 ≤ N ≤ 100,000) share many similarities. In fact, FJ has been able to narrow down the list of features shared by h…

Power Network Time Limit: 2000MS Memory Limit: 32768K Total Submissions: 24867 Accepted: 12958 Description A power network consists of nodes (power stations, consumers and dispatchers) connected by power transport lines. A node u may be supplied with…

Red and Black Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other) Total Submission(s) : 45   Accepted Submission(s) : 34 Font: Times New Roman | Verdana | Georgia Font Size: ← → Problem Description There is a rectangular…

C. Mahmoud and a Message time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Mahmoud wrote a message s of length n. He wants to send it as a birthday present to his friend Moaz who likes strin…

  精选常用: 1.  ctrl+shift+r:打开资源 这可能是所有快捷键组合中最省时间的了.这组快捷键可以让你打开你的工作区中任何一个文件,而你只需要按下文件名或mask名中的前几个字母,比如applic*.xml.美中不足的是这组快捷键并非在所有视图下都能用. 2. ctrl+o:快速outline 如果想要查看当前类的方法或某个特定方法,但又不想把代码拉上拉下,也不想使用查找功能的话,就用ctrl+o吧.它可以列出当前类中的所有方法及属性,你只需输入你想要查询的方法名,点击enter就…

在平常项目中,我们会遇到这样的业务场景: 客户希望把自己的门店绘制在百度地图上,通过省.市.区的选择,然后加载不同区域下的店铺位置. 先看看效果图吧: 实现思路: 第一步:整理行政区域表: 要实现通过地区筛选来动态加载地图,首先要有一套中国行政区域表.哪里来呢?如果你做过淘宝API接入,应该会想到淘宝物流接口提供了一个官方的行政区域代码,这个比较靠谱. 第二步:收集行政区域的经纬度: 这个就有点麻烦了,虽然可以在百度坐标拾取系统一个一个的收集整理,但是3000多条记录,是个不小的体力活.于是经过…

棋盘问题 Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 28474 Accepted: 14084 Description 在一个给定形状的棋盘(形状可能是不规则的)上面摆放棋子,棋子没有区别.要求摆放时任意的两个棋子不能放在棋盘中的同一行或者同一列,请编程求解对于给定形状和大小的棋盘,摆放k个棋子的所有可行的摆放方案C. Input 输入含有多组测试数据. 每组数据的第一行是两个正整数,n k,用一个空格隔开,表示了将在一…

Robot Motion Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 11262 Accepted: 5482 Description A robot has been programmed to follow the instructions in its path. Instructions for the next direction the robot is to move are laid down in a g…

Oracle 11g的dmp备份文件导入到Oracle 10g,出现错误信息: Import: Release 10.2.0.1.0 - Production on 星期四 7月 9 13:47:04 2015 Copyright (c) 1982, 2005, Oracle. All rights reserved. 连接到: Oracle Database 11g Enterprise Edition Release 11.2.0.1.0 - Production With the Part…

Fire Net Problem Description Suppose that we have a square city with straight streets. A map of a city is a square board with n rows and n columns, each representing a street or a piece of wall. A blockhouse is a small castle that has four openings t…

Pie Time Limit : 5000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other) Total Submission(s) : 67   Accepted Submission(s) : 34 Font: Times New Roman | Verdana | Georgia Font Size: ← → Problem Description My birthday is coming up and trad…

Stones Time Limit : 5000/3000ms (Java/Other)   Memory Limit : 65535/32768K (Java/Other) Total Submission(s) : 41   Accepted Submission(s) : 34 Font: Times New Roman | Verdana | Georgia Font Size: ← → Problem Description Because of the wrong status of…

Zipper Problem Description Given three strings, you are to determine whether the third string can be formed by combining the characters in the first two strings. The first two strings can be mixed arbitrarily, but each must stay in its original order…

DFS Problem Description A DFS(digital factorial sum) number is found by summing the factorial of every digit of a positive integer. For example ,consider the positive integer 145 = 1!+4!+5!, so it's a DFS number. Now you should find out all the DFS n…

Oil Deposits Problem Description The GeoSurvComp geologic survey company is responsible for detecting underground oil deposits. GeoSurvComp works with one large rectangular region of land at a time, and creates a grid that divides the land into numer…

简单计算器 Problem Description 读入一个只包含 +, -, *, / 的非负整数计算表达式,计算该表达式的值. Input 测试输入包含若干测试用例,每个测试用例占一行,每行不超过200个字符,整数和运算符之间用一个空格分隔.没有非法表达式.当一行中只有0时输入结束,相应的结果不要输出. Output 对每个测试用例输出1行,即该表达式的值,精确到小数点后2位. Sample Input 1 + 2 4 + 2 * 5 - 7 / 11 0 Sample Output 3.0…

B. Mahmoud and a Triangle time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Mahmoud has n line segments, the i-th of them has length ai. Ehab challenged him to use exactly 3 line segments to…

A Plug for UNIX Description You are in charge of setting up the press room for the inaugural meeting of the United Nations Internet eXecutive (UNIX), which has an international mandate to make the free flow of information and ideas on the Internet as…

Does the following error message looks familiar to you? (When you go to Site Actions –> Site Settings –> [Site Administration] –> Term store management)  "The managed Metadata Service or Connection is currently not available. The Application…

reference: Rabin-Karp and Knuth-Morris-Pratt Algorithms By TheLlama– TopCoder Member https://www.topcoder.com/community/data-science/data-science-tutorials/introduction-to-string-searching-algorithms/ // to be improved #include <cstdio> #include <…

reference: 6.4 knapsack in Algorithms(算法概论), Sanjoy Dasgupta University of California, San Diego Christos Papadimitriou University of California at Berkeley Umesh Vazirani University of California at Berkeley the unbounded knapsack and 0-1 knapsack a…

IMO, version 1 better than version 2, version 2 better than version 3. make some preprocess to make you code simple and efficient. Here divide the input by 2, so you don't have to do dividsion on each loop. version 1 is best thanks to http://www.cnbl…

TinyXML2是simple.small.efficient C++ XML文件解析库!方便易于使用,是对TinyXML的升级改写!源码见本人上传到CSDN的TinyXML2.rar资源:http://download.csdn.net/detail/k346k346/8500915,或者到官网下载:https://github.com/leethomason/tinyxml2. 使用方法:将tinyxml2.cpp和tinyxml2.h拷贝至项目目录,使用时包含#include "tinyx…

#include<iostream> #include<cstdio> #include<algorithm> #define Max 1005 using namespace std; struct line{ double x, y1, y2; int flag; }x_line[Max]; struct node{ int l, r, flag; double x, f; }tree[Max]; double point[Max]; int n, m, xm; i…

#include<iostream> #include<cstring> #include<cstdio> using namespace std; int n, m; int num[100005]; int front(int x) { return x&(-x); } int update(int x,int k) { while(x<=n) { num[x]+=k; x+=front(x); } return 1; } int sum(int x)…