equal

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): 由 翻译 强力驱动 问题: I have a time.Time created using time.Date() . I then calculate the number of nanoseconds between 1970/1/1 00:00:00.000000000 and that time. I then take the nanoseconds and turn them back into a time.Time using time.Unix() . However, if I compare the reconstituted time with the original using == , it returns false. If I subtract these 2 times, the resulting duration is 0. If I compare these 2 times using time.Equal() , it returns true. If I create another time using time.Date() with the same values as the first time, using == to

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: My app have a basic login screen, an outer box and in it, some textfield and buttons. I set the box to fill the screen. However, on some devices this box will to big. I want to set a maximum width and height to it. How do I set maximum width and height to a view? 回答1: You can use auto-layout constraints so that the box adapts to the screen size, but does not exceed a given width and height. To do this, set a "less than or equal to" constraint on the width and height. Add top, bottom, leading, trailing, width and height constraints to the box

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: Im creating my first sql trigger, CREATE OR REPLACE TRIGGER totalsalary AFTER INSERT ON Employee FOR EACH ROW WHEN ( NEW.Dno IS NOT NULL ) BEGIN UPDATE Department SET totalSalary totalSalary + NEW.salary WHERE Dno = NEW.Dno; END ; but i got this error message and i dont know how to fix it Error at line 3: PL/SQL: ORA-00927: missing equal sign 1. CREATE OR REPLACE TRIGGER SueldoTotal 2. AFTER INSERT ON EMPLEADO 3. FOR EACH ROW 4. WHEN ( NEW.Dno IS NOT NULL ) 5. BEGIN 回答1: SET totalSalary totalSalary + NEW.salary You have a missing equal sign

F. Equalizing Two Strings You are given two strings s and t both of length n and both consisting of lowercase Latin letters. In one move, you can choose any length len from 1 to n and perform the following operation: Choose any contiguous substring of the string s of length len and reverse it; at the same time choose any contiguous substring of the string t of length len and reverse it as well. Note that during one move you reverse exactly one substring of the string s and exactly one substring of the string t. Also note that borders of substrings you reverse in s and in t can be different,

The only difference between easy and hard versions is the number of elements in the array. You are given an array a a consisting of n n integers. In one move you can choose any a i ai and divide it by 2 2 rounding down (in other words, in one move you can set a i := a i 2 You can perform such an operation any (possibly, zero) number of times with any a i ai. Your task is to calculate the minimum possible number of operations required to obtain at least k k equal numbers in the array. Don't forget that it is possible to have a i = 0 ai=0 after some operations, thus the answer always exists.

题目如下： matrix Return the maximum number of rows that have all values equal after some number of flips. Example 1: Input: [[0,1],[1,1]] Output: 1 Explanation: After flipping no values, 1 row has all values equal. Example 2: Input: [[0,1],[1,0]] Output: 2 Explanation: After flipping values in the first column, both rows have equal values. Example 3: Input: [[0,0,0],[0,0,1],[1,1,0]] Output: 2 Explanation: After flipping values in the first two columns, the last two rows have equal values. Note: 1 <= matrix.length <= 300 1 <= matrix[i].length <= 300 matrix[i].length 's are equal matrix[i][j] 0 1

返回(x == y)元素的真值。 参数: 返回值: bool类型的张量。 注意: equal支持broadcasting。 https://tensorflow.google.cn/versions/r1.8/api_docs/python/tf/equal?hl=en 文章来源: https://blog.csdn.net/weixin_36670529/article/details/91822652

Created by Wang, Jerry, last modified on Oct 27, 2015 test ( "parse with fallback patterns" , function ( ) { var oLocaleEN = new sap . ui . core . Locale ( "en_US" ) , oLocaleDE = new sap . ui . core . Locale ( "de_DE" ) , oFormat , iCompare = new Date ( 1975 , 3 , 16 ) . getTime ( ) ; oFormat = sap . ui . core . format . DateFormat . getDateInstance ( { style : "long" } , oLocaleEN ) ; equal ( oFormat . parse ( "April 16, 1975" ) . getTime ( ) , iCompare , "Parse long style" ) ; equal ( oFormat . parse ( "Apr 16, 1975" ) . getTime ( ) , iCompare , "Parse fallback medium style" ) ; equal (

默认的equals只判断对象是不是同一个对象。 但是有时候即使不是同一个对象，我们也认为相等。例如我定义一个复数类，只要实部和虚部都相等，我就认为两个对象相等（即使它们不是同一个对象）。 最直接的例子是Collection。比如一个ArrayList，你判断其中有没有3+4i，你不需要自己写个循环，只要调用list.contains(new Complex(3,4))就可以知道list里有没有3+4i这个对象。这个contains方法就调用了equals来比较。如果你不重写equals()，这永远都返回false。 hashCode()需要重写的一个原因是Hash表。我们知道java里用HashTable和HashMap这些方便的数据结构可以直接用。我们往一个hash表里插入一个元素的时候，hash表通过hashCode来得到hash值，从而找这个元素来插入。同样例子，如果我们用Complex类做键，例如你想要3+4i对应一个值5，你以3+4i为主键每插入一次，应该都插入到同一位置。因为每个hashCode()函数对于3+4i都应该返回相同的值。可如果你只重写了equals()而不重写hashCode()，每个新的3+4i都返回新的hashCode()，你就会插入许多个3+4i. 所以你可以不重写equals（极其少的情况下），但只要重写了equals()

Monocarp has got two strings s s and t t having equal length. Both strings consist of lowercase Latin letters " a" and " b". Monocarp wants to make these two strings s s and t t equal to each other. He can do the following operation any number of times: choose an index p o s 1 pos1 in the string s s, choose an index p o s 2 pos2 in the string t t, and swap s p o s 1 spos1 with t p o s 2 tpos2. You have to determine the minimum number of operations Monocarp has to perform to make s s and t t equal, and print any optimal sequence of operations — or say that it is impossible to make these strings