july

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: If I have a string which represents a date, like "2011/07/01" (which is 1st July 2011) , how would I output that in more readable forms, like: 1 July 2011 1 Jul 2011 (month as three letters) And also, how could I make it intelligently show date ranges like "2011/07/01" to "2011/07/11" as 1 - 11 July 2001 (without repeating the 'July' and '2011' in this case) 回答1: As NullUserException mentioned, you can use strtotime to convert the date strings to timestamps. You can output 'intelligent' ranges by using a different date format for the first

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I have the following code in a JUnit test, which seemed to work last week is failing this week: Calendar cal = Calendar.getInstance(); cal.set(2011, Calendar.JULY, 12); cal.set(Calendar.DAY_OF_WEEK, Calendar.FRIDAY); // push the date to 15 System.out.println(cal.get(Calendar.DATE)); As you could probably infer from my comment, since the 12th is a Tuesday, I expect Date to be 15 after setting the DAY_OF_WEEK to Friday. However, the value that is printed is 22, and causes the test to fail. If I, however change the code as follows, and add an

If we just open the door a crack, the light comes pouring in. 只要把门开个小口，光就会倾泻而入。 Don't be so dishearted, life won't be hard all the time. If it would, then why not think that it isn't the first time we fail and it may be the chance we can grow up. Be a little more optimistic. If fear about the future, why not take actions now to make it better? I pray, please let some lights in, and I will reflect more to you. There comes a point in time that you got to take control of your own life. 总有一天你得要去掌控自己的人生。 From Shameless. Since last year, I have kept doing exercises under the instructions from the

费用流。 把每个方格拆成 $T$ 个点，$t$ 时刻一个方格向周围四个方格的 $t + 1$ 的点连一条容量为 $1$ 费用为 $0$ 的边，向自身的 $t + 1$ 连一条容量为 $1$ 费用为该方格最大幸福值的边。 源点向方格为 'S' 的0时刻连一条容量为 $1$ 费用为 $0$ 的边。所有点的 $T - 1$ 时刻向汇点连一条容量为 $1$ 费用为该方格最大幸福值的边。 还有每个格子同时刻不能有多条蛇呆在上面，再把每个点每个时刻拆成两个点，容量为 $1$ 费用为 $0$。跑最大费用最大流即可。 #include <bits/stdc++.h> using namespace std; const int N = 300; const int NN = 1e5 + 7, M = 5e5 + 7; const int INF = 0x3f3f3f3f; template<class T> inline void checkmax(T &a, T b) { if (a < b) a = b; } template<class T> inline void checkmin(T &a, T b) { if (a > b) a = b; } struct E { int v, ne, f, c; } e[M]; int head[NN], cnt, tol; int id[N][N]

leetcode http://leetcode.com/ careerup http://www.careercup.com/ http://hawstein.com/posts/ctci-solutions-contents.html glassdoor http://www.glassdoor.com/index.htm topcoder http://www.topcoder.com/ zoj http://acm.zju.edu.cn/onlinejudge/ 七月在线 https://www.julyedu.com/question 牛客网 https://www.nowcoder.com/ mitbbs http://www.mitbbs.com/bbsdoc/JobHunting.html poj http://poj.org/ acm http://acm.hdu.edu.cn/ codeforces http://codeforces.com/ 复杂数据结构的讲解及实现GeeksForGeeks： http://www.geeksforgeeks.org/ 一个比较牛逼的coding challenge网站： https://www.hackerrank.com/ 他总结出的《程序员编程艺术》： https://github.com/julycoding/The

所谓海量数据处理，无非就是基于海量数据上的存储、处理、操作。何谓海量，就是数据量太大，所以导致要么是 无法在较短时间内迅速解决 ，要么是数据太大，导致 无法一次性装入内存 。 处理海量数据的基本方法： 分而治之/hash映射 + hash统计 + 堆/快速/归并排序； 双层桶划分 Bloom filter/Bitmap； Trie树/数据库/倒排索引； 外排序； 分布式处理之Hadoop/Mapreduce。 参考内容 教你如何迅速秒杀掉：99%的海量数据处理面试题. https://blog.csdn.net/v_july_v/article/details/7382693 十道海量数据处理面试题与十个方法大总结. https://blog.csdn.net/v_JULY_v/article/details/6279498 来源： https://www.cnblogs.com/lvjincheng/p/11337461.html