项目目录简单解析：1、AFNetwork--- 通用网络库2、GCDiscreetNotificationView--- 顶部弹出并会自动消失的通知栏3、Thread--- 后台线程对象，处理后台发送带图片的动弹4、SoftwareGroup--- 所有软件索引页以及软件分组页5、Friends--...
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移动开发 时间：
2015-11-04 09:23:41
阅读次数：
264
原文 Jeremy Hampden has a large circle of friends and is very popular at parties. Everybody admires him for his great sense of humor -- everybody, that is, except his six-year-old daughter, Jenny. ...
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其他好文 时间：
2015-10-14 20:05:52
阅读次数：
229
8 Productivity hacks for Data Scientists & Business AnalystsIntroductionI was catching up with one of my friends from a past organization. She had alw...
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其他好文 时间：
2015-10-09 19:59:14
阅读次数：
306
Recommended BooksHere is a list of books which I have read and feel it is worth recommending to friends who are interested in computer science.Machine...
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其他好文 时间：
2015-10-03 10:34:33
阅读次数：
169
计算表达式。只有3种运算符：*，+，- ，*优先级高于后两者，后两者优先级相同。有两种符号：{}，（）。利用递归和堆栈即可解决。首先遇到左括号开始入栈直到遇到右括号，遇到右括号时对括号内的数进行计算。考虑到*优先级较高，因此遇到*直接对其左右集合进行运算。最后得到不含括号和*的表达式，从左往右计算即...
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其他好文 时间：
2015-09-24 00:40:50
阅读次数：
256
题意：给定n个数，从中分别取出0个，1个,2个...n个，并把他们异或起来，求大于m个总的取法。思路：dp，背包思想，考虑第i个数，取或者不取，dp[i][j]表示在第i个数时，异或值为j的所有取法。dp[i][j] = dp[i - 1][j] + dp[i - 1][j ^ a[i]]);其中d...
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移动开发 时间：
2015-09-22 23:19:31
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219
Problem DescriptionMatt has N friends. They are playing a game together.Each of Matt’s friends has a magic number. In the game, Matt selects some (cou...
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移动开发 时间：
2015-09-22 16:32:36
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207
Mohammad has recently visited Switzerland. As he loves his friends very much, he decided to buy some chocolate for them, but as this fine chocolat...
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其他好文 时间：
2015-09-18 21:43:19
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159
C. Bear and PokerLimak is an old brown bear. He often plays poker with his friends. Today they went to a casino. There are n players (including Limak ...
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其他好文 时间：
2015-09-15 23:04:50
阅读次数：
186
Problem DescriptionPrincess Alisha invites her friends to come to her birthday party. Each of her friends will bring a gift of some valuev, and all of...
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其他好文 时间：
2015-09-14 20:52:32
阅读次数：
527
