传送门:http://www.eeworld.com.cn/Test_and_measurement/2014/0610/article_9152.html i:对于2端口无损耗网络,可以根据S11的值求出S21的值,但实际上么有那么多无损耗网络,但至少  说明S11与S21是相关的 ii;当S11高达-10dB时,30%的信号被反射,S21的传输损耗为-0.5dB?原文有公式 iii;如果S11小于-13db,那么我们不会看到反射对传输信号的影响;但如果S11大于-13db,对S21的影响将会

多层pcb,背地共面波导微带线: 建模时如下结构:Top+diel1+diel2+diel3+diel4+diel5,参考层都应该在diel内部,否则会出现如下报错: [warning] Port refinement, process hf3d: All absorbing boundary conditions have been overridden, this will cause problems in linked projects. (1:00:44 下午 2月 02, 2023)

题目描述 Given a string S, find the longest palindromic substring in S. You may assume that the maximum length of S is 1000, and there exists one unique longest palindromic substring. 即给定一个字符串,返回该字符串最长的回文子串 如给出"acabcddcbadike",返回"abcddcba"

题目描述 Given a string S, find the longest palindromic substring in S. You may assume that the maximum length of S is 1000, and there exists one unique longest palindromic substring. 即给定一个字符串,求它的最长回文子串的长度(或者最长回文子串). 解法一 对于一个问题,一定可以找到一个傻的可爱的暴力解法,本题的暴力解法即

MySQL5.6中引入了MRR,专门来优化:二级索引的范围扫描并且需要回表的情况.它的原理是,将多个需要回表的二级索引根据主键进行排序,然后一起回表,将原来的回表时进行的随机IO,转变成顺序IO.文档地址:http://dev.mysql.com/doc/refman/5.6/en/mrr-optimization.html Reading rows using a range scan on a secondary index can result in many random disk ac

昨天因为误操作把一个写了一上午的代码给删了,找到的这个,以前竟然还没发现有这个功能- -! 具体操作: 1.建立同路径同名的文件 2.文件上右键 --> Compare With --> Local History 3.根据历史记录回滚文件历史

在上一篇Log4net(日志文件篇)中,我们使用"log4net.Appender.FileAppender"将日志信息输出到一个单一的文件中,随着应用程序的持续使用,该日志文件会越来越庞大,进而影响系统的性能.因此,有必要对日志文件按某种条件进行切分,要切分日志文件,我们可以使用"log4net.Appender.RollingFileAppender"输出源,使用该输出源我们可以按照文件大小或者日期对日志文件进行切分,下面我们分别描述之. 一.按文件大小切分日志

出栈的方法:getActivity().getSupportFragmentManager().popBackStackImmediate(CreateOneFragment.class.getName(), FragmentManager.POP_BACK_STACK_INCLUSIVE); 即弹出Tag为CreateOneFragment之上的所有(包括自身)的Fragment. popBackStackImmediate(name,flag); 第二个参数:只能是 0 或者 1(POP_B

前面的文章已经实现相关的布局,本文接着进行相关的功能实现 本文地址:http://www.cnblogs.com/wuyudong/p/5951794.html,转载请注明出处. 读取系统联系人 当点击“选择联系人”按钮后,弹出联系人列表,读取系统联系人分如下几个步骤: 系统联系人提供了一个内容提供者,通过内容解析器,匹配Url地址 1,内容解析器 2,Url地址,查看系统联系人数据库,内容提供者源码 先看api文档的清单文件,后看java类(联系人数据库有多张表) contents://com

回表 简单来说就是数据库根据索引找到了指定的记录所在行后,还需要根据rowid再次到数据块里取数据的操作. "回表"一般就是指执行计划里显示的"TABLE ACCESS BY INDEX ROWID". 例如select的字段里有索引不包含的列 根据tom的oracle编程艺术,建表big_table,300W数据. 建索引: create index idx_big_table_created on big_table(created); 下面语句不会回表,因为只

1.遇到的问题 当我们一个方法里面有多个数据库保存操作的时候,中间的数据库操作发生的错误.伪代码如下: public method() { Dao1.save(Person1); Dao1.save(Person2); Dao1.save(Person2);//假如这句发生了错误,前面的两个对象会被保存到数据库中 Dao1.save(Person2); } 期待的情况:发生错误之前的所有数据库保存操作都回滚,即不保存 正常情况:前面的数据库操作会被执行,而发生数据库操作错误开始及之后的所有的数据

Given a string which consists of lowercase or uppercase letters, find the length of the longest palindromes that can be built with those letters. This is case sensitive, for example "Aa" is not considered a palindrome here. Note: Assume the leng

Given a list of unique words. Find all pairs of distinct indices (i, j) in the given list, so that the concatenation of the two words, i.e. words[i] + words[j] is a palindrome. Example 1:Given words = ["bat", "tab", "cat"]Ret

Given a string s, return all the palindromic permutations (without duplicates) of it. Return an empty list if no palindromic permutation could be form. For example: Given s = "aabb", return ["abba", "baab"]. Given s = "a

Given a string, determine if a permutation of the string could form a palindrome. For example,"code" -> False, "aab" -> True, "carerac" -> True. Hint: Consider the palindromes of odd vs even length. What difference d

Given a singly linked list, determine if it is a palindrome. Follow up: Could you do it in O(n) time and O(1) space? 这道题让我们判断一个链表是否为回文链表,LeetCode中关于回文串的题共有六道,除了这道,其他的五道为Palindrome Number 验证回文数字,Validate Palindrome 验证回文字符串,Palindrome Partitioning 拆分回文

Given a string S, you are allowed to convert it to a palindrome by adding characters in front of it. Find and return the shortest palindrome you can find by performing this transformation. For example: Given "aacecaaa", return "aaacecaaa&qu

Given a string s, partition s such that every substring of the partition is a palindrome. Return the minimum cuts needed for a palindrome partitioning of s. For example, given s = "aab", Return 1 since the palindrome partitioning ["aa"

Given a string s, partition s such that every substring of the partition is a palindrome. Return all possible palindrome partitioning of s. For example, given s = "aab",Return [ ["aa","b"], ["a","a","

Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases. For example, "A man, a plan, a canal: Panama" is a palindrome. "race a car" is not a palindrome. Note: Have you consider that

Determine whether an integer is a palindrome. Do this without extra space. click to show spoilers. Some hints: Could negative integers be palindromes? (ie, -1) If you are thinking of converting the integer to string, note the restriction of using ext