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Counting
git tag查看、创建与删除
20150525 # 创建tag $ git tag -a dev_20150525_16 -m 'xxxx' 把tag push到远程 $ git push origin dev_20150525_16
Counting
·
2015-07-14 16:00
git
OC与Java的 runtime
所以目前Objective-C没有GC(只有ref-
counting
的内存管理策略)。JavaVM有ge
liutianshx2012
·
2015-07-13 15:00
【十分不错】【离线+树状数组】【TOJ4105】【Lines
Counting
】
Onthenumberaxis,thereareNlines.ThetwoendpointsLandRofeachlineareinteger.GiveyouMqueries,eachquerycontainstwointervals:[L1,R1]and[L2,R2],canyoucounthowmanylinessatisfythisproperty:L1≤L≤R1andL2≤R≤R2?Inp
zy691357966
·
2015-07-11 02:00
Java 多线程之信号量-Semaphore
技术信号量(
Counting
Semaphore)用来控制同时访问某个特定资源的操作数量,或者执行某个特定操作的数量。
zhangwei_david
·
2015-07-05 10:00
Semaphore
[USACO Open10]数三角形Triangle
Counting
解题报告
题目http://cojs.tk/cogs/problem/problem.php?pid=543分析这道题的要点在于:含原点的三角形不好数,我们数不含原点的三角形,最后用C(N,3)扣掉它就是答案了。怎么数“不含原点的三角形”呢?画出我们所在的坐标系。我们拿一条过原点的直线l,按极角序去扫描:直线l分成两条射线,一个是a,另一个是b。它绕着原点(Bessie所在点)旋转,请自行想象在孙悟空手中旋
wmdcstdio
·
2015-07-03 21:00
数学
USACO
解题报告
计算几何
在Xcode4.5中禁用ARC(Automatic Referencing
Counting
)
转自: http://blog.csdn.net/aking21alinjuju/article/details/8089134最近升级了xcode4.5,用上了ios6的SDK。但用着用着发现一个比较烦的问题,以前很多代码提示错误,发现原来因为xcode启用了ARC,当ARC启用后会自动在代码中加入retain、release,原先需要手动添加的用来处理内存管理的引用计数的代码可以自动地由编译器
hemeinvyiqiluoben
·
2015-07-02 19:00
在iOS 中创建自定义Delegates properties with ARC错误的解决方法
错误提示:error: Automatic Reference
Counting
Issue: Existing ivar 'delegate' forunsafe_unretained property
robinson_911
·
2015-06-28 12:00
Big Data
Counting
: How to count a billion distinct objects using only 1.5KB of Memory
ThisisaguestpostbyMattAbrams(@abramsm),fromClearspring,discussinghowtheyareabletoaccuratelyestimatethecardinalityofsetswithbillionsofdistinctelementsusingsurprisinglysmalldatastructures.Theirserversre
doctor_who2004
·
2015-06-26 23:00
Data
big
Counting
hdu 2952
Counting
Sheep
#include #include #include #include #include #include #include #include #include #include #include #include #include usingnamespacestd; intn,m,vis[104][104]; intdx[]={0,0,1,-1}; intdy[]={-1,1,0,0}; c
xinag578
·
2015-06-22 17:00
UVa1225 Digit
Counting
#include <stdio.h>#include <string.h>int main(){ int T, N, i, j; int a[10]; scanf("%d", &T); while (T-
·
2015-06-21 00:00
count
JVM调优总结
一、相关概念 基本回收算法 引用计数(Reference
Counting
) 比较古老的回收算法。
lzyuwenbo
·
2015-06-12 11:00
JVM调优
hoj2430
Counting
the algorithms
MyTags (Edit) Source : mostleg Timelimit :1sec Memorylimit :64MSubmitted :725, Accepted :286AsmostoftheACMers,wy'snexttargetisalgorithms,too.wyisclever,sohecanlearnmostofthealgorithmsquickly.Afterash
Kirito_Acmer
·
2015-06-10 18:00
树状数组
HDOJ 3664 Permutation
Counting
PermutationCountingTimeLimit:2000/1000MS(Java/Others) MemoryLimit:32768/32768K(Java/Others)TotalSubmission(s):1407 AcceptedSubmission(s):711ProblemDescriptionGivenapermutationa1,a2,…aNof{1,2,…,N
zwj1452267376
·
2015-06-09 22:00
hdoj 3664 Permutation
Counting
【递推 有点类似杨辉三角】
PermutationCountingTimeLimit:2000/1000MS(Java/Others) MemoryLimit:32768/32768K(Java/Others)TotalSubmission(s):1404 AcceptedSubmission(s):709ProblemDescriptionGivenapermutationa1,a2,…aNof{1,2,…,N
chenzhenyu123456
·
2015-06-09 21:00
hdu 3887
Counting
Offspring
题目大意:给你一棵树,求每个父节点的子树中序号小于父节点序号的节点个数。#include #include #include #include #include #include #include #include #include #include #include #include #include usingnamespacestd; constintN=100000+5; intn,s
xinag578
·
2015-06-07 11:00
HDU ACM 3887
Counting
Offspring 树状数组+DFS
分析:一棵树,求每个结点的子树中有几个数是小于这个数的。dfs会进入一个节点一次,出一个节点一次,中间经过的点都是它的子树中的点,因此,进入的时候统计一遍,出来的时候统计一遍,两个结果相减即可,使用树状数组。纯dfs会爆栈,需要重设栈的大小。#pragmacomment(linker,"/STACK:100000000,100000000") #include #include usingname
a809146548
·
2015-05-31 21:00
编程
C++
c
算法
ACM
Counting
Sort
Letusunderstanditwiththehelpofanexample.Forsimplicity,considerthedataintherange0to9. Inputdata:1,4,1,2,7,5,2 1)Takeacountarraytostorethecountofeachuniqueobject. Index:0123456789 Count:0220110100
yuanhisn
·
2015-05-28 09:00
Counting
inversions in an array
Countthenumberofinversionpairsinanarrayofnnumbers. An inversionpair isdefinedasfollowing:- Ifiarr[j]then(arr[i],arr[j])iscalled inversionpair Solution1:Thisisalmostnormalmergesort,thewholemagi
yuanhisn
·
2015-05-28 09:00
Counting
Boolean Parenthesizations
Youaregivenabooleanexpressionconsistingofastringofthesymbols'true','false','and','or',and'xor'.Countthenumberofwaystoparenthesizetheexpressionsuchthatitwillevaluatetotrue.Forexample,thereare2waystopar
yuanhisn
·
2015-05-28 09:00
JS Caffeine Script & Add property to every object in array &
Counting
power sets&Singleton Patter
CompletethefunctioncaffeineBuzz,whichtakesanon-zerointegerasit'soneargument. Iftheintegerisdivisibleby3,returnthestring"Java". Iftheintegerisdivisibleby3anddivisibleby4,returnthestring"Coffee" Ifth
power0405hf
·
2015-05-27 21:00
JavaScript
数据流基本问题--确定频繁元素(二)lossy
counting
和sticky sampling
我们之前在数据流基本问题--确定频繁元素(一)中提到了频繁元素的一个计算问题(找出出现次数超过m/k的元素),里面的算法返回的结果里肯定包含出现次数超过m/k的元素,但是也可能包含不超过m/k的元素(falsepositive)。对于这个缺点,必须得进行额外一次的重新扫描,以确定最终答案。我们只允许进行一次的扫描,那么该怎么去做呢?这里我们简单讨论下lossycounting算法和stickysa
dm_ustc
·
2015-05-27 19:00
数据流算法
iOS另类的内存管理
从开始的MRR(manual retain-release)到后来ARC(Automatic Reference
Counting
),包括CoreFoundation的内存管理都遵守引用计数的基本原则。
·
2015-05-25 09:00
内存管理
Pymongo Tutorial & Pymongo入门教
获取一个数据库获取一个Collection文件(Documents)文件插入操作单个文件获取find_one()按照ObjectId查询关于Unicode字符串的一点说明批量插入查询多个文件
Counting
鸭脖
·
2015-05-24 17:37
MongoDB
SGU117
Counting
SGU117
Counting
题目大意输入N个数,输出有多少个数的M次方可以被K整除算法思路快速幂取模求出每个数的M次方模K的结果,统计结果为0的个数时间复杂度:O(NlogM)代码/***Copyright
u010929036
·
2015-05-23 20:00
sgu
快速幂取模
UVa 1225 - Digit
Counting
1225-DigitCountingTrungisboredwithhismathematicshomeworks.Hetakesapieceofchalkandstartswritingasequenceofconsecutiveintegersstartingwith1toN(1 #include #include usingnamespacestd; intmain() { //freo
a995549572
·
2015-05-15 22:00
ACM
uva
1225
HDU 3240
Counting
Binary Trees
ProblemDescriptionThereare5distinctbinarytreesof3nodes:LetT(n)bethenumberofdistinctnon-emptybinarytreesofnomorethan n nodes,yourtaskistocalculateT(n)mod m.InputTheinputcontainsatmost10testcases.Eachca
jtjy568805874
·
2015-05-13 23:00
HDU
Counting
Leaves (30)
题目如下:Afamilyhierarchyisusuallypresentedbyapedigreetree.Yourjobistocountthosefamilymemberswhohavenochild.InputEachinputfilecontainsonetestcase.Eachcasestartswithalinecontaining0 usingnamespacestd; #de
xyt8023y
·
2015-05-08 18:00
算法
bfs
家谱树
HDU 3518 Boring
counting
后缀自动机
题目大意:就是对于给出的字符串S,长度不超过1000,求其中本质不同的子串的数量,这些子串满足在字符串S中出现了至少不重合的2次大致思路:如果将S建立后缀自动机的话,记录每个节点处的串最早一次和最后一次出现的位置,然后状态t处的字符串长度为[Min(t),Max(t)]比较其和rightmost-leftmost的大小即可知道这个状态下有多少满足条件的串时间复杂度O(n)代码如下:Result :
u013738743
·
2015-05-05 09:00
HDU
后缀自动机
Counting
boring
3518
理解ARC在Objective-C中的应用
原文:http://longweekendmobile.com/2011/09/07/objc-automatic-reference-
counting
-in-xcode-explained/名词解释:
billfanggs
·
2015-05-04 14:00
内存管理
自动化
内存泄露
compiler
arc
计数排序
转自: http://blog.csdn.net/pi9nc/article/details/12220851 1)算法简介 计数排序(
Counting
sort)是一种稳定的排序算法。
北极的。鱼
·
2015-04-30 15:00
排序
Counting
Leaves (30)
题目链接:http://www.patest.cn/contests/pat-a-practise/1004题目: 时间限制400ms 内存限制65536kB 代码长度限制16000B 判题程序Standard 作者 CHEN,Yue Afamilyhierarchyisusuallypresentedbyapedigreetree.Yourjob
Apie_CZX
·
2015-04-28 13:00
UVa 269
Counting
Patterns(dfs+构造)
题意:求出所有的n个绝对值不大于k的数的组成排列,排列的和为0。且对于一个排列,全部正负号翻转、排列顺序翻转、循环左移、循环右移所构成的新排列视为同一个排列,并要求按照字典序打印所有的排列。解析:参考了学长的解题报告,其思路是按照字典序构造出和为0的排列,且该排列的字典序是当前排列所有字典序最大的。然后记录下来,最后打印。AC代码#include #include #include #includ
HelloWorld10086
·
2015-04-22 20:00
uva
269
Lake
Counting
-- DFS(深搜)
TimeLimit:1000MSMemoryLimit:65536KTotalSubmissions:22424Accepted:11300DescriptionDuetorecentrains,waterhaspooledinvariousplacesinFarmerJohn'sfield,whichisrepresentedbyarectangleofNxM(1 #include #defin
u012965373
·
2015-04-19 11:00
SCU 4426
Counting
_3 (dp)
题目链接4426:
Counting
_3Submityoursolution Discussthisproblem BestsolutionsTimeLimit:1000msDescriptionDefinef
madaidao
·
2015-04-09 21:00
dp
ACM
hdu3518---Boring
counting
(后缀数组,对后缀分组)
ProblemDescription035nowfacedatoughproblem,hisenglishteachergiveshimastring,whichconsistswithnlowercaseletter,hemustfigureouthowmanysubstringsappearatleasttwice,moreover,suchapearancescannotoverlapeac
Guard_Mine
·
2015-04-09 14:00
后缀数组
UPC:2224 Boring
Counting
(二分+划分树)
题意:给一串数字,每次查询区间内大于等于A小于等于B的数字个数。思路:二分+划分树。划分树可以求区间内第k小的数字,配合二分可以求到某个数字在区间内是第几小数字。#include #include #include #include #include usingnamespacestd; constintmaxn=50005; typedeflonglongLL; structDivideTree
kkkwjx
·
2015-04-07 21:00
数据结构
二分法
划分树
sdut2610---Boring
Counting
(离线+树状数组+离散化)
BoringCountingTimeLimit:3000msMemorylimit:65536K有疑问?点这里^_^题目描述InthisproblemyouaregivenanumbersequencePconsistingofNintegerandPiistheithelementinthesequence.Nowyoutaskistoansweralistofqueries,foreachqu
Guard_Mine
·
2015-04-06 20:00
树状数组
Lake
Counting
-DFS
LakeCountingTimeLimit: 1000MS MemoryLimit: 65536KTotalSubmissions: 22269 Accepted: 11226DescriptionDuetorecentrains,waterhaspooledinva
bao_libra
·
2015-04-06 17:00
codeforces 372A
Counting
Kangaroos is Fun
Thereare n kangarooswithpockets.Eachkangaroohasasize(integernumber).Akangaroocangointoanotherkangaroo'spocketifandonlyifthesizeofkangaroowhoholdthekangarooisatleasttwiceaslargeasthesizeofkangaroowhois
页永仲
·
2015-04-04 15:00
sdut2610--Boring
Counting
(二分+划分树)
BoringCountingTimeLimit:3000ms Memorylimit:65536K 有疑问?点这里^_^题目描述 InthisproblemyouaregivenanumbersequencePconsistingofNintegerandPi istheith elementinthesequence.Nowyoutaskistoansweralistofqueries,f
u013015642
·
2015-04-03 20:00
poj 2386 Lake
counting
只需要用到DFS就行了,算搜索中比较直接的,前面也写过一体类似的,DFS和BFS只是一种策略,只是搜索。 #include usingnamespacestd; intN,M; charmaze[101][101]; voiddfs(intx,inty) { //将现在所在的位置替换为‘.’ maze[x][y]='.'; //循环遍历可以移动的八个方向 intdx,dy; for(dx=-1;
hujian_
·
2015-03-30 21:00
Algorithm
ACM
poj
DFS
Counting
Sort
Let us understand it with the help of an example. For simplicity, consider the data in the range 0 to 9. Input data: 1, 4, 1, 2, 7, 5, 2 1) Take a count array to store the count of each unique o
yuanhsh
·
2015-03-29 13:00
count
Counting
Sort
Let us understand it with the help of an example. For simplicity, consider the data in the range 0 to 9. Input data: 1, 4, 1, 2, 7, 5, 2 1) Take a count array to store the count of each unique o
yuanhsh
·
2015-03-29 13:00
count
Counting
Kangaroos is Fun
Thereare n kangarooswithpockets.Eachkangaroohasasize(integernumber).Akangaroocangointoanotherkangaroo'spocketifandonlyifthesizeofkangaroowhoholdthekangarooisatleasttwiceaslargeasthesizeofkangaroowhois
viphong
·
2015-03-27 23:00
[CF 335E]
Counting
Skyscrapers题解翻译
题目翻译:见http://cojs.tk/cogs/problem/problem.php?pid=1921题解翻译:这道题中的摩天大楼描述了被称作跳跃表(SkipList)的数据结构。跳跃表在某种意义上和AVL、红黑树相似,因为它支持O(logN)的插入,操作和查找(包括上下界查询),同时支持常数时间的在排序顺序中前后移动。跳跃表和任何树结构都不同,因为它有实用的,不需要线程锁的(所谓‘无锁’)
wmdcstdio
·
2015-03-26 10:00
数学
概率
codeforces
iOS开发--关闭ARC
Build settings -> Apple LLVM complier 3.0 - Language -> objective-C Automatic Reference
Counting
·
2015-03-24 19:00
ios开发
SGU 117
Counting
n个数,求其M次幂能够被k整除的数的个数。(n,k,m #include usingnamespacestd; #defineLL__int64 intpow(inta,intb,intk) { intans=1; while(b) { if(b&1)ans=ans*a%k; if(!ans)returnans; a=a*a%k; b>>=1; } returnans; } intmain() {
u014679804
·
2015-03-24 19:00
Counting
inversions in an array
Count the number of inversion pairs in an array of n numbers. An inversion pair is defined as following: - If i < j a
yuanhsh
·
2015-03-24 11:00
version
Counting
inversions in an array
Count the number of inversion pairs in an array of n numbers. An inversion pair is defined as following: - If i < j a
yuanhsh
·
2015-03-24 11:00
version
D - Lake
Counting
-dfs
#include#include#include#include#include#includeusingnamespacestd;charmap[101][101];intfloorr,k;intdx[]={0,-1,-1,0,1,1,1,0,-1};intdy[]={0,0,-1,-1,-1,0,1,1,1};intn,m;intmain(){inti,j;intbfs(intr,intc);
viphong
·
2015-03-20 15:00
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