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解一元一次方程移项我解一元一次方程的时候,会解方程,但是出一些应用题让我列方程我就不会列方程了,我该怎么办?
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1.75亿学生的选择
应用题要注意以下几点题目需要求什么?将要求得设为未知数x根据题目的意思列出处于x的方程解出未知数（这个你会）作答当然还是熟能生巧,多做做练习,问问老师.、希望能帮到你
为您推荐：
其他类似问题
应用题找里面的关键字 是 比 以之类的 然后设未知数就是这些字之后的 根据题意列方程即可
扫描下载二维码第一资料 算法
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HTML：&!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"&
&html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en"&
题意：求左下角到右上角的最短路，按照方向输出 解题思路:优先方向的广搜，格式错误判WA解题代码：// File Name: 2594.c
// Author: darkdream
// Created Time: 日 星期六 22时18分29秒#include&stdio.h&
#include&string.h&
#include&stdlib.h&
#include&ti
private ArrayList RemoveSame(ArrayList list)
for (int i = 0; i & list.Count - 1; i++)
for (int j = i + 1; j & list.C j++)
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C#网络通信编程
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题目描述:http://acm./showproblem.php?pid=2544
#include &stdio.h&
#define MAX_N 10003
double dp[MAX_N][MAX_N];
void floyd(void)
for(k = 0; k & n
&问题分析&
与数塔问题相似，状态转移方程,s[i][j]=max{s[i+1][k]} j-2&=k&=j+2
1 #include &stdio.h&
4 int main()
long v[201][201];
int i,j,k,m,n;
scanf(\"%d %d\",&m,&n);
题目： Implement atoi to convert a string to an integer.Hint: Carefully consider all possible input cases. If you wanta challenge, please do not see below and ask yourself what are the possible input ca
Time Limit:
MS (Java/Others)
Memory Limit:
K (Java/Others) Total Submission(s): 2122
Accepted Submission(s): 449
Problem Description
Coco is a clever boy, who
The sequence of triangle numbers is generated by adding the natural numbers. So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be:
1, 3, 6, 10, 1
回到l2fwd的main函数中 int
MAIN(int argc, char **argv)
struct lcore_queue_conf *
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There was once a 3 by 3 by 3 cube built of 27 smaller cubes. It has fallen apart into seven pieces:
Figure 1: The seven pieces that once forme……
http://acm./sdutoj/problem.php?action=showproblem&problemid=2718
题意：如果没有障碍就按原方向直走，否则就右转直走，若右边走不通就左转直走，若左边也走不通就后转直走，直到走出去。
1 #include &stdio.h&
2 #include &string.h&
3 const in
Windows CE
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Windows CE is……
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题意：在一个平面上，有若干个球，给出球的坐标，每次可以将一个球朝另一个球打过去（只有上下左右），碰到下一个球之后原先的球停下来，然后被撞的球朝这个方向移动，直到有一个球再也撞不到下一个球后，这个球飞出，球只能是朝上下左右四个方向打，并且要一个球四个方向都没有球了，这个球就不能打了。问说最少平面上剩几个球，并且给出打球的方案。
思路：把四个方向可以连成一块的球用一个并查集连起来，有多少个并查集，最
Uncle Tom's Inherited Land*Time Limit:
MS (Java/Others)
Memory Limit:
K (Java/Others)Total Submission(s): 1249
Accepted Submission(s): 544Special JudgeProblem DescriptionYo……
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' get the number of rows in the tablerowCount=Browser(\"Browser\").FlexApplication(\"App\").FlexApplication(\"Fl exClient\").FlexPanel(\"FlexPanel\").FlexCanvas(\"App Home\").FlexPanel(\"Search Results\").Flex
using System.IO;
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/// &/summary&
public class JSMin
This Technical Bulletin discusses the troubleshooter architecture and what is available and shipped as a part of Exchange 2010 SP1 as well as troubleshooter that can be provided as a add on the produc……
今天帮忙修了一个bug， 在拖动TreeViewItem时，需要滚动TreeView向前翻页，或向后翻页。
1.找到TreeView控件里的ItemsControl
2.找到ItemsControl里的ScrollViewer
3.判断当前每个Item的高度
4.通过GetCursorPos获取屏幕绝对坐标
5.通过ItemsControl的PointFromS
题目描述有些坑。。题意：有一条高速公路在x轴上，从（0,0）到（L,0）。周围有一些村庄，希望能够在高速公路上开通几个出口，使得每个村庄到最近的出口距离小于D，求出最少需要开通多少个出口。解题思路：典型的区间问题，将每个点化为区间（x-sqrt(D^2-y^2),x+sqrt(D^2+y^2)），然后按区间右端点排序，依次对每个区间进行操作，如果该区间中已有出口则无需增加，若该区间没有出口则取右
总时间限制: 2000ms 内存限制: 65536kB描述John is going on a fishing trip. He has h hours available (1 &= h &= 16), and there are n lakes in the area (2 &= n &= 25) all reachable along a single, one-way road. John
The following table lists limits on representation of data by array variables.
Maximum or range
Array storage size
Limited by available memory
Number of dimensions
努力总会有收获的，虽然一直在刷水题，但感觉代码的正确率越来越高了。坚持是最重要的，看到几个大神的博客，hdu，poj题目数量上百道，真心佩服，希望我也能坚持下来。What Is Your Grade?Time Limit:
MS (Java/Others)
Memory Limit:
K (Java/Others) Total Submissio
see also:/files/pdf/techpaper/vmware-multipathing-configuration-software-iSCSI-port-binding.pdf Multipathing between a server and storage array provides the ability to load-balanc……
1.在VBA中使用以下方法，即可破解Excel工作表密码。
（1）在Excel文档中,选择&视图&选项卡，选择&宏&，录制宏，然后停止录制。
（2）然后，按Alt + f11，调出VBA界面，发现有一个模块1，下面就是刚才录制宏的方法，把下面的代码Copy进去，替换掉，最后点工具栏的绿色三角形，执行方法，即可破解Excel工作表的密码。
由于马里奥的飞行距离有限，因此为了方便处理，我们首先用floyd预处理出马里奥可以飞行的两点间的最短路，然后再将图分成K+1层用SPFA求最短路即可。
#include&stdio.h&#include&string.h&#define MAXD 130#define MAXN 20#define INF int A, B, M, L, K,
01 March 2012 Women exercise in a gym in Bogot, Colombia, part of the fitness chain started by Gigiola Aycardi.Policymakers and business experts have pitched entrepreneurship to women as a way to real
1、典型的Point结构体
1 struct point {
point(double _x = 0, double _y = 0): x(_x), y(_y) {
void input() {
scanf("%lf%lf", &x, &y);
题目：401 - Palindromes
#include &cstdio&
#include &iostream&
#include &algorithm&
#include &cmath&
#include &cstring&
#include &vector&
#include &map&
map&char,c
Description
Problems involving the computation of exact values of very large magnitude and precision are common. For example, the computation of the national debt is a taxing experience for many comp
时间限制：3000 ms
内存限制：65535 KB
As is known to all,if you throw a coin up and let it droped on the desk there are usually three results. Yes,just believe what I say ~it can
题目链接：http://poj.org/problem?id=2253题目意思：找出从Freddys stone
Fionas stone
最短路中的最长路。 很拗口是吧，举个例子。对于 i 到 j 的一条路径，如果有一个点k， i 到 k 的距离 && k 到 j 的距离都小于 i 到 j 的距离，那么就用这两条中较大的一条来更新 i 到 j 的距离 。每两点之间都这样求出路径。最后输
The code i wrote a while ago recently caused a disaster and as I reviewed it I found it is the silliest code I've ever written,
1 static int BadMaximalMatch(TListRef list1, int start1, int count1,
/***************************参考自****************************/
/newpanderking/archive//2726757.html
上有详细注解，我是看了之后写的，我的代码有一咪咪的改动
//这是一道典型的拓扑排序的题，刚开始时没有理解，//上网
Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2.
For example,Given:s1 = "aabcc",s2 = "dbbca",
When s3 = "aadbbcbcac", return true.When s3 = "aadbbbaccc", return false
* Flexihash - A simple consistent hashing implementation for PHP.
* The MIT License
* Copyright (c) 2008 Paul Annesley
linux/mm/swap.c * *
Copyright (C)
Linus Torvalds *//* * This file should contain most things doing the swapping from/to disk. * Started 18.12.91 */#include &linux/mm.h&#include &lin
Point 1: Whether a filtering law should be obeyed is determined by how it is made.
Point 2: Businesses should not have to decide what is bad or immoral. Whose standards would they apply?
Point 3: Th
由于计算顺序已经给定了，那么便只需要依次枚举符号和符号后面ai的值即可，其中符号是可以重复使用的，但每个ai只能使用一次。
当递归到所有ai都已使用时就检查计算结果是否为23即可。
#include&stdio.h&#include&string.h&int a[10],vis[10];int dfs(int cur,int tot){
int i,j,t;
Time Limit:
MS (Java/Others)
Memory Limit:
K (Java/Others)Total Submission(s): 11566
Accepted Submission(s): 4205
Problem Description
Angel was caught by the MOLI
题目：/problem.php?pid=1553
分别算出时针和分针的度数，做差之后分几种情况讨论，(-360,-180),[-180,0),[0,180],(180,360).
#include &stdio.h&
int main(){
while (~scanf("%d:%d",
1.链接地址：
/practice/1979
http://poj.org/problem?id=1979
总时间限制:1000ms内存限制:65536kB描述There is a rectangular room, covered with square tiles. Each tile is colored either
struct Point
double x,y;
Point(double tx = 0,double ty = 0) : x(tx),y(ty){}
typedef Point V
//向量的加减乘除
Vtor operator + (Vtor A,Vtor B) { return Vtor(A.x + B.x,A
~题目链接~
http://poj.org/problem?id=2506
class Point{
\tPoint();
\tPoint(int x ,int y) ;
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\tPoint operator +(Point A) ;
\tint get_X() ;
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os.listdir()
os.path.isdir()
os.path.join()
os.mkdir()
1 # -*- coding:utf-8 -*-
2 import os,sys
5 def mkdir(path = os.getcwd()):
if 'img' in os.listdir(path) and os.
If it's a set type,then turn the set into list before serializing, or use a custom default handler to do so:
1 def set_default(obj):
if isinstance(obj, set):
return list(obj)
题目链接：http://pat./contests/pat-a-practise/1009
分析：简单题。相乘时指数相加，系数相乘即可，输出时按指数从高到低的顺序。注意点：多项式相乘后指数最高可达2000。
题目描述：
This time, you are supposed to find A*B where A and B are two polynomial
真是酷毙了。
Double Dealing
时间限制(普通/Java):20000MS/60000MS
运行内存限制:65536KByte 总提交: 14
测试通过: 5
Take a deck of n unique cards. Deal the entire deck out to k players in the usual way: the t
值得一提的是，如果我们用KM算法求最小权完美匹配时，要把边的权值初始化成MAX-G[i][j]然后去求最大权的完美匹配，计算结果的时候再转化回来即可。
如果把G[i][j]初始化成-G[i][j]去求最大权的完美匹配的话，我写的程序会超时，暂时没有明白是这个想法有问题还是我的写法有问题。
#include&stdio.h&#include&string.h&#defin
Given a collection of numbers that might contain duplicates, return all possible unique permutations.
For example, [1,1,2] have the following unique permutations: [1,1,2], [1,2,1], and [2,1,1].
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subroutine basis_function_b_val ( tdata, tval, yval )
!*******************************************************************************
!! BASIS_FUNCTION_B_VAL evaluates the B spline basis fu
我们考虑一种特殊的图：
1. 有向图
2. 只有一个连通分量
3. 不存在环
那么这样的图里，必然可以找到一种排序方式，来确定谁在谁的&前面&。
简单的来说可以这么理解：如果存在一条边a-&b，那么a顶点就在b的前面。
下面我们通过例子来看看拓扑排序的过程，确定所有的顶点中，谁……
agents v changement des sacs du cr&ateurs main duplicata de marque sur la prix beaucoup dans bas tarifs.Donc, ce dernier ya encore de public femelle se trouvent ici pour fa&en&age
#include&stdio.h&
int main()
long long a[40][70];//二维矩阵用来存放数据
int i,j,n;
for(i=0;i&70;i++)//设定第一行的初始值
a[0][i]=0;
if(i==35||i==37)
a[0][i]=1;
Good afternoon everyone, I'm so glad to join this English Hour again with you after my two weeks leave. I miss you all very much.
As we all know, it&s harmful when you take a long site
Part of the reason why the Jordan 11 Bred worked so well is blank slate that the upper of the shoe offered. Following the same logic, the almost lineless Air Jordan XX8 is shown here with a similar ……
http://acm./showproblem.php?pid=3787
1 #include &stdio.h&
2 #include &string.h&
4 char s1[20],s2[20];
5 int len1,len2;
7 int main()
There are N gas stations along a circular route, where the amount of gas at station i is gas[i].
You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from station i to
//重新整理的比较清楚的opencv框架#include "stdafx.h" #include &cv.h&#include &highgui.h&#include &iostream&
int _tmain(int argc, _TCHAR* argv[]){
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#include &stdio.h&
#define MAX_N 100
//最大总人数
void main()
SPOJ - BALNUM
Balanced Numbers
Time Limit: 1000MS
Memory Limit: Unknown
64bit IO Format: %lld & %llu
Description
Balanced numbers hav
题目很水，不多说.........
#include&stdio.h&
int main()
long t,n,m,a,i,j,dp[10005],vol[505],jizhi[505],sum,w;
scanf(\"%ld\",&t);
while(t--)
scanf(\"%ld%ld\",&n,&
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Math Resources
If you dont have a basic understanding of 2D and 3D math for computer graphics your head will probably explode once you try learning OpenGL.
To prevent that from happening you should
http://nodeschool.io/
第一部分：Learn You The Node.js For Much Win!
5.filtered ls：
题目及提示
#####################################################################
FILTERED LS
IMMEDIATE DECODABILITY
Time Limit: 1000MS
Memory Limit: 10000K
Total Submissions: 9630
Accepted: 4555
Description
An encoding of a set of symbols is said to be immediat
1 #include&stdio.h&
2 #include&algorithm&
3 #include&functional&
4 #include&string.h&
5 #include&math.h&
7 struct mark
int Mark,cl,
10 }mark1[1
Super Jumping! Jumping! Jumping!
Problem Description
Nowadays, a kind of chess game called &Super Jumping! Jumping! Jumping!& is very popular in HDU. Maybe you are a good boy, and know l……
1. 表格的重置
/****************************************************************
* 機 能： クリア処理
* 引 数： なし
* 戻り値： なし
*************************************************************
A + B for you again
Time Limit:
MS (Java/Others)
Memory Limit:
K (Java/Others)Total Submission(s): 3623
Accepted Submission(s): 917
Problem Description
Generally speakin
这个题目很简单，虽然看上去高大上，只要找规律就行了，多看看，找找例子。-不过我花了蛮久才看出来，惭愧啊，其实就是看n.m中最小的数是奇数还是偶数。After winning gold and silver in IOI 2014, Akshat and Malvika want to have some fun. Now they are playing a game on a grid made
最近学习了AC自动机，做了notonlysuccess大牛里面的题，也该来个总结了。
AC自动机(Aho-Corasick Automaton)在1975年产生于贝尔实验室，是著名的多模匹配算法之一。
至于算法的讲解，notonlysuucess大牛极力推荐
http://www.cs.uku.fi/~kilpelai/BSA05/lectures/slides04.pdfhttp://
Oracle基本语法集锦
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摘要：本文给出了ORACLE的基本语法建立的源代码，供大家参考！标签：语法
Oracle帮您准确洞察各个物流环节
create table test (names varch
题目大意：贪吃蛇的简化版，给出一串操作命令，求蛇的最终状态是死是活。
解法：这条蛇一共20格的长度，所以用一个20个元素的队列表示，队列的每个元素是平面的坐标。每读入一条指令，判断其是否越界，是否咬到了自己。若都没有，把改点压入队列，front元素弹出。
参考代码：
#include&iostream&
#include&queue&
#include&string&
u-boot第二阶段初始化内容的入口函数是_main,_main位于arch/arm/lib/crt0.S文件中：
_main函数中先为调用board_init_f准备初始化环境（设置栈指针sp和并给gd_t结构分配空间）：
.global _main
* Set up initial C runtime environment and call boa
Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.
For example, given n = 3, a solution set is:
"((()))", "(()())", "(())()", "()(())", "()()()"
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/problem/1049/这题我之前写没想到迭代加深，看了题解，然后学习了这种搜索（之前我写的某题也用过，，但是不懂专业名词 囧。）迭代加深搜索就是限制搜索深度，一旦有可行解立即跳出，优化了深搜一直搜下去的毛病。（囧，这题搜索题写了我一下午，我搜索的确很弱啊！！！）第一次写出来的版本我没有注意到，应该是从多个点拓展下去，而不是从某个点。第二次写出来的版本
DECLARE @SOInfoList TABLE
SONumber INT,
SODate datetime,
Status char(1)
INSERT INTO @SOInfoList
T.c.value('(SONumber/te
C#网络编程技术总结
C#网络编程技术总结
/album/view/2d7aaf5acfce3c/album/view/c5d2158ecc22bcd126ff0c3c/album/view/bd1cc734eefdc8d376ee323f
通过Ajax操作Dom：
1 if(xmlHttp.readyState == 4){
if(xmlHttp.status == 200){
var sobj = document.getElementById("suggest");
var str = xmlHttp.resp
Jordan 3 Powder Blue 2014 Sale Online Now! Which pairs are your must-have Jordans In January 2014? If you love the blue colors, the January 18 will really a busy day for you. The Powder Blue 3 2014 is……
一次通过：
1 public class Solution {
public int romanToInt(String s) {
int sum = 0;
int[] num = new int[s.length()];
if(s == null) return 0;
man upstart
nit(8) init(8)
init - Upstart process management daemon
init [OPTION]...
DESCRIPTION
the parent of all processes on the system, it is execu
Problem 1607 Greedy division
http://acm./problem.php?pid=1607
Accept: 402
Submit: 1463Time Limit: 1000 mSec
Memory Limit : 32768 KB
Problem Description
Oaiei has inherited……
#include&stdio.h&
#include&math.h&
int main(void)
int count,i,m,n;
for(m=2;m&=100;m++){
n=sqrt(m);
for(i=2;i&=n;i++)
if(m%i==0)
中共领袖毛泽东把它概括为三句话八个字。三句话是：坚定正确的政治方向，艰苦朴素的工作作风，灵活机动的战略战术；八个字是：团结，紧张，严肃，活泼。通称“三八作风”。、 这三句话是毛泽东于日为延安中国人民抗日军政大学（即抗大）成立3周年所发表的文章中归纳的，加上之前为抗大确定的校训“团结紧张严肃活泼”8个字，合成为“三八作风”。 毛泽东在题为《抗大三周年纪念》的文章中提出，“抗大
1.这个插件挺好用的，可以用它来代替微软的gridview，前提是您用了asp.net mvc模式开发下面是找到的一些英文资料，大家可以参考下
网址如下：
/category/plugins/templates/
jQuery.tmpl( template, [data,] [options] ) Returns: jQuery
planification de la h&te permettrait d'&conomiser nombre d'argent si le
incident survient succinct ce transfert d'une soci&t& d'un partenariat
d'une soci&t&
On July 29, 2014, Thomson Reuters announced the release of the 2013 Journal Citation Reports, which covers 11,471 journals. The impact factor for World Journal of Gastroenterology is 2.433 [5-year imp
1 #include &iostream&
2 #include &cstdio&
3 #include &math.h&
6 double A,B,H;
7 double V;
9 double F(double x){
double ……
摘自MSDN：#include &windows.h&#include &tchar.h&#include &stdio.h&#define BUFSIZE 80typedef void (WINAPI *PGNSI)(LPSYSTEM_INFO);int _tmain(){
OSVERSIONINFOEX
SYSTEM_INFO
PGNSI pGNSI;
uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1058
半平面交求面积最值。直接枚举C(20,8)的所有情况即可。
代码如下：
1 #include &cstdio&
2 #include &cstring&
3 #include &i
1.To determine the installed edition, run:
DISM /online /Get-CurrentEdition
2.To check the possible target editions, run:
DISM /online /Get-TargetEditions
3. Finally, to initiate an upgrade,
The Famous Clock
时间限制：1000 ms
内存限制：65535 KB
Mr. B, Mr. G and Mr. M are now in Warsaw, Poland, for the 2012&s ACM-ICPC World Finals Contest. They&ve decided to take a
http://acm./showproblem.php?pid=1690
坑爹的题，必须用__int64 %I64d(以前没用过）
1 #include &iostream&
2 #include &stdio.h&
3 #include &string.h&
4 #include &stdlib.h&
5 #include &math.h&
这题说的是n个人进行石头剪刀布 然后给了m个 指令
1 A B A与B相等
2 A B A大于B
然 后 计 算 给 出的 点 是 否 存 在 错 误 并 计 算 错 误 出 现 的 次
每 个 点 设 置 一 个 吃 与 被 吃 不 断 的 去 维 护 他 们 之 间 的 关 系 例 如
例如 A&B&C
D&E&F 然后 将 相 应 的 赋 值 祖 先 ，赋值完后就……
Rio de Janeiro is the perfect place to begin a Brazil vacation. Read about top Rio tours爄n this article, written by a specialist in Brazil travel at Brazil For Less. Many travelers, faced with the pos
USE [msdb]EXEC msdb.dbo.sp_grant_login_to_proxy @proxy_name=NSSISProxyAgentV1, @login_name=NWTC\Ebw.Admin GO
Given n non-negative integers a1, a2, ..., an, where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Find two
如果i==j&&j-1&=0时候，f[i][j]=f[i][j-1];
如果j==0时候，f[i][j]=1;
其他 f[i][j]=f[i-1][j]+f[i][j-1];
#include &stdio.h&
int main(){
__int64 gird[40][40];
for(i=0;i&=35;++i)
题目大意是求给出的N个数分成连续的M组， 求组内数之和的最小值。
发现二分循环里最后输出mid比较保险。。一开始输出了l总是错
Monthly Expense
Time Limit: 2000MS
Memory Limit: 65536K
Total Submissions: 11653
Accepted: 4769
Sometimes, internal user or customer might request you to change the Title of the SAP Transaction code to a more meaningful one and SAP allows this to be done painlessly.
The steps to change the Titl
题目描述：
FatMouse prepared M pounds of cat food, ready to trade with the cats guarding the warehouse containing his favorite food, JavaBean. The warehouse has N rooms. The i-th room
function quickSort(&$arr, $l, $r)
if (count($arr)&2 || $l&$r)
$tmp_l = $l;
$tmp_r = $r;
$privot = $arr[$r];
while($tmp_l&$tmp_r) {
while($arr[$t
for (int i = 0 ; i& tempArr. i++) {
MessageDetailInfo *info = [tempArr objectAtIndex:i];
for (int j = 0; j&self.dataArray. j++) {
MessageDetailInfo *mess
//php版插入排序
$arr=array('','5','3','7','6','4','8','2');
for($i=2;$i&count($arr);$i++)
if($arr[$i]&$arr[$i-1])
$arr[0]=$arr[$i];
for($j=$i-1;$arr[0]&$arr
增量法的最小包围圈算法，不会&&
#include &cstdio&
#include &cstring&
#include &iostream&
#include &cmath&
#include &algorithm&
const double EPS = 1e-10;
inline int sgn(doub
纯动态规划。
注意初始化为-INF
#include&stdio.h&
#include&algorithm&
#include&iostream&
#include&string.h&
#define INF
#define max3(a,b,c) max(max(a,b),c)
int n,m,i,j;
function IsWin(IsBlack:boolean):
label exit1;
if IsBlack then
wtag:=1 else
fuck #include &stdio.h&#include &string.h&#include &stdlib.h&#include &iostream&#include &algorithm&#include &math.h&#define MAXX 105#define eps 1e-8typedef struct point{
Counting Black
Time Limit: 1000MS
Memory Limit: 10000K
Total Submissions: 9772
Accepted: 6307
Description
There is a board with 100 * 100 grids as shown below. The left……
#include &stdio.h&
#define N 10000 + 10
char s[N];
int main()
int n, m, i, j, t, p, q, test = 1;
while(scanf(\"%c%*c\", &ch) != EOF && ch != '#')
Search in Rotated Sorted Array I
Suppose a sorted array is rotated at some pivot unknown to you beforehand.
(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).
You are given a target value to search
假设有两种微生物 X 和 Y
X出生后每隔3分钟分裂一次（数目加倍），Y出生后每隔2分钟分裂一次（数目加倍）。
一个新出生的X，半分钟之后吃掉1个Y，并且，从此开始，每隔1分钟吃1个Y。
现在已知有新出生的 X=10, Y=89，求60分钟后Y的数目。
如果X=10，Y=90
本题的要求就是写出这两种初始条件下，60分钟后Y的数目。
Dexter Strickland and Marcus Paige took to Instagram to show off their new gifts from the Jordan Brand. Looks as if players of this years UNC basketball team received some very Player Exclusive Air ……
&!--#include file=\"Conn.asp\"--&&!DOCTYPE html PUBLIC \"-//W3C//DTD XHTML 1.0 Transitional//EN\" \"http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd\"&&html xmlns=\"http://www.w3.org/1999/xhtml\"&&head
The Sultan's Successors
The Sultan of Nubia has no children, so she has decided that the country will be split into up to k separate parts on her death and each part will be
In the theory of formal languages in computability theory, a pumping lemma or pumping argument states that, for a particular language to be a member of a language class, any sufficiently long string i
我们不妨用f[i][j]表示到第i个加油站油量为j时这个状态所需的最小花费，那么首先有一部分状态是从第i-1个加油站继承过来的，即f[i][j]=f[i-1][j+d[i]-d[i-1]]，之后就是考虑在第i个加油站买多少油会更划算，那么状态转移方程为f[i][j]=min{min{f[i][k]+(j-k)*m[i]},f[i][j]}，这样乍看上去是三个for，然
1 package main
3 import (
7 type Graphic struct {
edges [][]int
colors int
color []int
14 func (g *Graphic) ch
A: Fine Year
DescriptionThis time, Alice will begin her amazing journey from Harbin. Just lick most of the participants taking part in this contest, Alice is born after 1990. Do you notice that 201
#include&stdio.h&
#include&iostream&
bool map[];//1 P 0 N;
int main()
int i,j,k;
map[1][1]=1;
for(i=2;i&=2000;i++)
we are a professional stone crushing machinery manufacturer in China. We primarily produce ring hammer crushers. Our hammer crushers are widely to crush medium-hard friable materials copper crusher we
# -*- coding: utf-8 -*-
import pygame
from sys import exit
import random
pygame.init()
screen = pygame.display.set_mode((450, 600), 0, 32)
background = pygame.image.load(\"back.jpg\").conver
Download bitnami installer: bitnami-redmine-2.4.1-1-linux-installer.run
$ chmod 755 bitnami...installer.run (run as a common user)
$ ./bitnami...install.run
you choose a installation destinati
上一篇教程中，我们学习了如何计算轮廓的凸包，其实对一个轮廓而言，可能它的凸包和它本身是重合的，也有可能不是重合的。比如下面左边图像的轮廓本身就是凸包，而右边图像的轮廓则不是。我们可以通过函数bool isContourConvex(InputArray contour)，来判定一个轮廓是否是凸包，是的话返回true，否则false[注意测试的轮廓必须是简单轮廓，没有自交叉之类的]。
Description
Triathlon is an athletic contest consisting of three consecutive sections that should be completed as fast as possible as a whole. The first section is swimming, the second section is rid
Basic terminology used in the workflow:
The workflow definition is the set of rules that determine the path that the process takes. For example, how a purchase requisition is processed, from the init
1.选取最左下边的点p0
2.根据剩余的m点集Q相对p0的极角排序，如果极角相同，只考虑最远的点
3.对堆栈进行初始化
\tpush(p0,S)
\tpush(p1,S)
\tpush(p2,S)
\tfor i=3 to m
\t\tdo the angle formed by points next_to_top(s) top(s) and pi make a
/article/7795
/TUTORIALS/LibraryArchives-StaticAndDynamic.html!!!!!!!!!!!!!!!!!!!!!!!!!!1
http://www.tldp.org/HOWTO/Program-Library-HOWTO/
Oil Deposits
The GeoSurvComp geologic survey company is responsible for detecting underground oil deposits. GeoSurvComp works with one large rectangular region of land a……
一副牌中发五张扑克牌给你，判断是四条，三带二、三带一加一、两对、一对、顺子、还是什么都不是。
控制台输入：
示例输出：
Java方法的代码：
static String func(String str){
String[] parts = str.split(",");
网上看到的一个题，在某个网页上发现了关于这个题目的应用，是用js写的，分享下源代码（稍作了修改）：
1 function tdisoper(f0,f1,f2,f3)
3 this[0]=f0;
4 this[1]=f1;
5 this[2]=f2;
6 this[3]=f3;
8 disoper=new tdisoper(\"-\",\"+\",\"/\",
圆桌骑士。有的骑士之间是相互憎恨的，不能连坐，需要安排奇数个骑士围着桌子坐着，大于3个，求哪些骑士不可能安排到座位。根据给定的关系，如果两个骑士之间没有憎恨关系，那么连边。最终就是求有多少个点无法位于奇圈之内。首先求所有联通分量，对于每个连通分量二分图染色，看看是否存在一个奇圈，如果有一个，那么这个联通分量里面的所有点都可以在至少一个奇圈之内。（详细的见白书）下面重点说说如何找联通分量的。方法是用
1.1 Kernels Let \(E\) be a set and
\(\mathscr{E}\)
a \(\sigma\)-algebra of subsets of
E. We assume that the a-algebra
countably generated, i.e. generated by a countable collection of subsets
简单的判断互质
Uniform Generator
Computer simulations often require random numbers. One way to generate pseudo-random numbers is via a function of the form
\" is t……
扁平化设计的流行，源于它简约的设计态度。扁平设计去除所有多余的元素和功能，在网页设计，如图形，声音，渐变，3D元素，阴影效果等，可以使设计看起来紧凑和充满吸引力。在这篇文章中，我收集了30个免费的扁平化设计 PSD 素材，这些对设计师很有用。
您可能感兴趣的相关文章
20套很精美的扁平化设计素材免费下载
24款扁平风格 PSD 格式图标免费下载
35套用于扁平化设计的图标和网页……
原题： HDU 3362 http://acm./showproblem.php?pid=3362
开始准备贪心搞，结果发现太难了，一直都没做出来。后来才知道要用状压DP。
题意：题目给出n(n &= 18)个点的二维坐标，并说明某些点是被固定了的，其余则没固定，要求添加一些边，使得还没被固定的点变成固定的，可见题目中的图形sample。
由于n很小，而且固定点的顺序没有……
Oracle Assets Additions API Documentation Supplement, Jan 2003
Modified Date :
21-APR-2005
@ font-face的CSS规则，使得它比以往任何时候都更容易实现独特的字体在您的网站上，而不需要依赖于谷歌的字体或任何溢价选项。您可能不希望使用自定义的字体为您的整个网站，但他们可以有很大的影响，使标题站了出来，同时保留了能够被搜索引擎读取。
这些著名的字体，距离著名的科幻小说展示
字体：亚历克斯&墨菲固 - 电影：机械战警
本字体：异形联赛 - 电影：外国……
#include &stdio.h&
#include &math.h&
mmin(x,y) (x)&(y)?(x):(y)
int main(){
double dp[10000+10];
c[10000+10];
double w[10000+10];
int n,m,i,j;
原题地址：/problems/triangle/
Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below.
For example, given t
$ emacs /etc/profile
export PATH=/opt/Qt/bin:$PATH
保存,重启或是
$ source /etc/profile
centos下安装qt时出现/usr/lib/libstdc++.so.6: version `GLIBCXX_3.4.9' not found执行ls -l /usr/lib/libstdc
C:\\Program Files\\NUnit 2.5.3\\bin\\net-2.0\\nunit.exe/run $(ProjectDir)/$(ProjectFileName)$(ProjectDir)
run nuit from menu:
ref:http://www.dijksterhuis.org/setting-up-nunit-for-c-unit-testing-with-……
走进纪念馆，看见周恩来！铁肩担道义，袖手撑乾坤！西花厅中坐，艰苦奋斗立!竹椅衬西花，清廉挺正气！寒梅傲春雪，风骨永不朽！水中若明镜，照出铁骨铮！死而方后已，鞠躬才尽瘁！年轻怀梦想，伴侣同终身！威武而不屈，敌国皆心惊！骨灰撒江河，灵柩落山青！
这道题做了好久，题并不难，一开始的思路，和做法都是对的。但是要求最小字典序必须从右向左求，这样找最小的时候可以从左向右，保证最小字典序。以后做题特别要记住，越是情况多的时候越要总结归纳，最小样例，最大样例都要试一试，保证输出正确。
1 #include&stdio.h&
2 #include&string.h&
3 int main()
I just create test app and test. So for devices with out retina: ImageName.png - For iPhone/iPod ImageName~ipad.png -- For iPad For devices with retina display: ImageName@2x~ipad.png -- For
Description某石油公司计划建造一条由东向西的主输油管道。该管道要穿过一个有n 口油井的油田。从每口油井都要有一条输油管道沿最短路经(或南或北)与主管道相连。如果给定n口油井的位置,即它们的x 坐标（东西向）和y 坐标（南北向）,应如何确定主管道的最优位置,即使各油井到主管道之间的输油管道长度总和最小的位置?证明可在线性时间内确定主管道的最优位置。
给定n 口油井的位置,编程计算各油
A
先考虑单栈排序。
显然出现231就跪了。
其他情况？
更一般的，若在一个数列中找不到i,j,k满足i&j&k && a[i] & a[j] && a[i] & a[k]，则肯定能够单栈排序。
首先满足231就肯定不可以。
其他情况呢？
把每个a[i]看成二维的点(i,a[i]),则原来的条件就转化成：&&（自己想想吧&&
强连通+缩点+搜索特殊处理，求单向连通
这一题的符合条件的图经过缩点重构后为拓扑图。
对于重新构造出的新图，判断这个图是否单向连通
若结点的后继大于等于2个，则需判断这些所有的后继是之间
否为单向连通，然后搜索所有点。
从中找到的规律为：
入度为0的点有且仅有一个，删除这个点后的图中，入度为0的点依然有且仅有一个
因此只需判断入度为0的结点个数即可判断是否单向连通
void bubbleSort( int[] values )
for ( int i = 0; i & values. i++ )
for ( int j = 0; j & values.length
Coin Change
Time Limit:
MS (Java/Others)
Memory Limit:
K (Java/Others)Total Submission(s): 10590
Accepted Submission(s): 3535
Problem Description
Suppose there are 5 typ
Problem Description
Have you ever played quoit in a playground? Quoit is a game in which flat rings are pitched at some toys, with all the toys encircled awarded.In the field of Cyberground,
A cache coherence protocol facilitates a distributed cache coherency conflict resolution in a multi-node system to resolve conflicts at a home node.FIELDThe invention relates to high-speed point-to-po
点击打开链接
给多个矩形和其被顺时针旋转的角度
求矩形面积和凸包面积的比
验证模板题
//大白p263
#include &cmath&
#include &cstdio&
#include &cstring&
#include &string&
#include &queue&
#include &functional&
#include &set&
SD Relevant Steps:
1.Check if all delivery in due list have been processedVL06G
(Could set background jobs to automatically generate
the billing due list for posting)
2.Check if all picking in due
转载请注明出处一块努力的牛皮糖：/yuxc/
新手上路，翻译不恰之处，恳请指出，不胜感谢
Updated 1st
2.2 Conditional Frequency Distributions 条件频率分布
We introduced frequency distributions in Section 1.3. W……
Knight Moves题意:骑士巡游到某位置的最少步数#include &stdio.h&#include &string.h&int s[20][20];int rear,char a,int b,struct{int x,}Susake[300000];void bfs(int n, int m){rear++;Susa
Description Question 1: Is Bigger Smarter?The ProblemSome people think that the bigger an elephant is, the smarter it is. To disprove this, you want to take the data on a collection of elephants and p
题目链接：http://acm./showproblem.php?pid=4421
思路：枚举32位bit,然后2-sat判断可行性，这里给出2-sat矛盾关系构图：
a-&~b,b-&~a;
~a-&a,~b-&b;
a-&~a,b-&~b;
~a-&b,~b-&
题目本身不难 但是要用滚动数组，否则的integer会爆128M
考虑记sqrr[i,j] 为以[i,j]为右下角向左上扩展的正方形的最大值，up[i,j]为以[i,j]为最后一格能向上扩展的最大值,left[i,j]同理
则sqrr[i,j]=min(sqrr[i-1,j-1]+1,up[i,j],left[i,j]);
可以看到第i行状态只与i-1行有关
Is It A Tree?Time Limit:
MS (Java/Others)
Memory Limit:
K (Java/Others)Total Submission(s): 10139
Accepted Submission(s): 2323Problem DescriptionA tree is a well-known data……
USE [tablename]GO/****** Object:
StoredProcedure [dbo].[paginate]
Script Date: 05/12/:10 ******/SET ANSI_NULLS ONGOSET QUOTED_IDENTIFIER ONGO-- ==========================================
Application Developer
If you would like to...
In this position you will be responsible for technical leadership contributing to the successful delivery of application development projects. Your resp
That's Why ( You Go away)
Baby won't you tell me why there is sadness in your eyes I don't wanna say goodbye to you Love is one big illusion I should try to forgetbut there is something left in my he
升级版不用栈，用递归来列举出所有的拓扑序列，循环内嵌套递归的思想值得好好领悟！
若干士兵站队，已知一些约束的关系，比如A必须站在B前面，B必须站在F前面等。求所有可能的站队方式
输入：所有的约束关系，比如AB（表示A必须站在B前面）HJ
输出：列出所有的站队方式
测试用例：
1 #include &stdio.h&
2 #include &string.……
QUESTION 10The RVPC user can do which of the following? (Choose all that apply.)A. Register databases if granted the register database privilegeB. See all databases in the recovery-catalog schemaC. S
# Doxyfile 1.6.3
# This file describes the settings to be used by the documentation system
# doxygen (www.doxygen.org) for a project
# All text after a hash (#) is considered a comment and wil
private static int[] pyValue = new int[]
-2,-2,-2,-2,-2,-2,
-2,-2,-1,-1,-19784,-1
Q: Say you have an array for which the ith element is the price of a given stock on day i.
Design an algorithm to find the maximum profit. You may complete as many transactions as you like (ie, buy o
ROUTE-COUNT（G）
DFS(G)
count = 0
return COUNT(G,s,t)
COUNT(G,s,t)
for each v in G:Adj[s]
if v.d == t.d
count = count + 1
else if v.d &t.d
return count + COUNT(G,
前面几篇已经建立游戏框架和元素，现在还差游戏的主要逻辑：滑动时对卡片的合并计分以及游戏结算检查。现在就来完成卡片合并的逻辑。卡片的合并需要检查各个卡片的数值，这里使用一个二维数组来存储卡片。
首先在GameLayer.h中声明一个二维数组： //卡片数组
CardSprite* cardArr[4][4];
卡片数组声明后需要初始化……
The New Villa
Mr. Black recently bought a villa in the countryside. Only one thing bothers him: although there are light switches in most rooms, the lights they control are
题目链接：http://acm./onlinejudge/showProblem.do?problemId=3490
题意：给出两组数，问区间[L,R]内有多少数字满足能被第一组数的至少一个整除且不能被第二组数的至少一个整除。
思路：设第一组的条件为P，第二组的条件为Q，
vector&i64& V;
#include &stdio.h&
#include &malloc.h&
#define INFINITY 999
#define MAX 20
int visited[1000] = {0};
int sum = 0;
typedef struct
int vexs[MAX];
int arcs[MAX][MAX];
题目意思就是通过一系列翻煎饼的动作让整个堆保持递增顺序
在学习STL 从网上复制了别人的代码学习一下。。非常简洁
Stacks of Flapjacks
Background
Stacks and Queues are often considered the bread and butter of data structure
昨天正画荷花，突见一群麻雀从窗前飞过，清脆之鸣破窗而来，心里突然一动，麻雀常常立于莲蓬上面，似在歇脚，又似在赏花闻香，于是添了一只麻雀在画上，卧在莲蓬上，压弯了莲蓬，花依然开放，叶依然舒展，花骨朵依然含羞，但有了麻雀，整个画面有了动感，有了趣味，画完，一边擦着手上的墨，一边想，大自然本应如此生动，只是我们在艺术表现的时候，人为的取舍，把一些应该留的弃了，于是就单薄了，空洞了，趣的流淌和味的
Emag eht htiw Em Pleh
Time Limit: 1000MS
Memory Limit: 65536K
Total Submissions: 2312
Accepted: 1548
Description
This problem is a reverse case of the problem 2996. You
Young people from less-privileged homes are more likely to graduate from college and earn more if raised by two married parents.
privileged：特权的，特殊的
Young adults are 44 percent more likely to hav
题目：Given an array of non-negative integers, you are initially positioned at the first index of the array.Each element in the array represents your maximum jump length at that position.Determine if you
Help is needed for Dexter
Time Limit: 3 Second
Dexter is tired of Dee Dee. So he decided to keep Dee Dee busy in a game. The game he planned for her is quite easy to play but not easy to win at l
时间限制：100 ms
内存限制：65535 KB
描述Little A is one member of ACM team. He had just won the gold in World Final. To celebrate, he decided to invite all to have one meal. As bowl, kni
一、Digital Roots
HDOJ地址： http://acm./showproblem.php?pid=1013
不过数学规律是大神！
代码如下：
/***** Digital Roots********/
/******** written by C_Shit_Hu ***********
Fiddler：在PC和移动设备上抓取HTTPS数据包 /5575368/tour/a2i12783.html /5575368/tour/a2ie1tb9.html /5575368/tour/a2hbolo5.html
思路：将每对夫妻看成是对立状态，每个不正常关系都是一个矛盾，按2-SAT的方式建边。最后建一条新娘到新郎的边。具体看注释
#include&iostream&
#include&cstdio&
#include&algorithm&
#include&cstring&
#include&queue&
#define Maxn 62
#define Maxm Maxn*Maxn
题目：http://acm./showproblem.php?pid=1874
/************************************************************************/
畅通工程续
dijkstra求起始点到目标点最短距离
Super Jumping! Jumping! Jumping!
Time Limit : ms (Java/Other)
Memory Limit : K (Java/Other)
Total Submission(s) : 5
Accepted Submission(s) : 1
Problem Description
Nowadays,……
The next release of BizTalk Server will be called BizTalk Server 2010 R2 and will focus on three primary themes: Platform Support
Helping customers stay up-to-date with the latest platforms. Imp
Given n non-negative integers a1, a2, ..., an, where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Find two
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