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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: How do I expand the hexadecimal number 0x1234 to 0x11223344 in a high-performance way? unsigned int c = 0x1234, b; b = (c & 0xff) << 4 | c & 0xf | (c & 0xff0) << 8 | (c & 0xff00) << 12 | (c & 0xf000) << 16; printf("%p -> %p\n", c, b); Output: 0x1234 -> 0x11223344 I need this for color conversion. Users provide their data in the form 0xARGB, and I need to convert it to 0xAARRGGBB . And yes, there could be millions, because each could be a pixel. 1000x1000 pixels equals to one million. The actual case is even more complicated, because a single

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: so my own valid dates only are from 1520 to 3999 and im new to this site and im new to java please help i want to know what did i do wrong or what to do in this code. int x; Scanner in = new Scanner (System.in); do { System.out.print ("Enter a date "); x = in.nextInt(); } while (x<1520&&x>3999); { System.out.println ("Invalid Gregorian Calendar date."); System.out.print ("Please Input a valid Gregorian Calendar date: "); } }} 回答1: In the while loop, get another integer whenever the input was invalid. Also, you need to change && to || in your

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: The installation problems pointed out in an earlier question are still present. I have tried to install Z3 4.3.0 and 4.1 under Windows XP SP3 32-bit and under Windows 7 64-bit. None of the combinations work! I am able to do the " from z3 import * ", but the init() of the Z3 dll fails. My Python version is 2.7.3. Z3 stand-alone and Python stand-alone do work, but they don't work together without lots of complaints. It would help to get an up-to-date installation recipe which answers the following questions: Which Z3 download (source version,

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: As from Chrome version 37, pre-flighted, cross-domain requests are failing (again) if the server has authentication enabled, even though all CORS headers are set correctly. This is on localhost (my dev PC). Some of you may be aware of the history of Chrome/CORS/auth bugs, especially when HTTPS was involved. My problem does not involve HTTPS: I have an AngularJS application served from localhost:8383 talking to a Java (Jetty) server on localhost:8081 that has HTTP BASIC auth activated. GETs work fine, but POSTs fail with a 401: XMLHttpRequest

Ignatius has just come back school from the 30th ACM/ICPC. Now he has a lot of homework to do. Every teacher gives him a deadline of handing in the homework. If Ignatius hands in the homework after the deadline, the teacher will reduce his score of the final test. And now we assume that doing everyone homework always takes one day. So Ignatius wants you to help him to arrange the order of doing homework to minimize the reduced score. Input The input contains several test cases. The first line of the input is a single integer T that is the number of test cases. T test cases follow. Each test

Rimi learned a new thing about integers, which is - any positive integer greater than 1 can be divided by its divisors. So, he is now playing with this property. He selects a number N. And he calls this D. In each turn he randomly chooses a divisor of D (1 to D). Then he divides D by the number to obtain new D. He repeats this procedure until D becomes 1. What is the expected number of moves required for N to become 1. Input Input starts with an integer T (≤ 10000), denoting the number of test cases. Each case begins with an integer N (1 ≤ N ≤ 105). Output For each case of input you have to

HITCON 2019 Lost Modular again writeup 算是基础题，有很多之前题的影子，做不出来纯属菜。 题目 加密脚本 from Crypto.Util.number import * class Key: def __init__(self, bits): assert bits >= 512 self.p = getPrime(bits) self.q = getPrime(bits) self.n = self.p * self.q self.e = 0x100007 self.d = inverse(self.e, (self.p-1)*(self.q-1)) self.dmp1 = self.d%(self.p-1) self.dmq1 = self.d%(self.q-1) self.iqmp = inverse(self.q, self.p) self.ipmq = inverse(self.p, self.q) def encrypt(self, data): num = bytes_to_long(data) result = pow(num, self.e, self.n) return long_to_bytes(result) def decrypt(self, data): num = bytes_to_long(data) v1

An inorder binary tree traversal can be implemented in a non-recursive way with a stack. For example, suppose that when a 6-node binary tree (with the keys numbered from 1 to 6) is traversed, the stack operations are: push(1); push(2); push(3); pop(); pop(); push(4); pop(); pop(); push(5); push(6); pop(); pop(). Then a unique binary tree (shown in Figure 1) can be generated from this sequence of operations. Your task is to give the postorder traversal sequence of this tree. Figure 1 Input Specification: Each input file contains one test case. For each case, the first line contains a positive

CF306C White, Black and White Again 洛谷评测传送门 题目描述 Polycarpus is sure that his life fits the description: "first there is a white stripe, then a black one, then a white one again". So, Polycarpus is sure that this rule is going to fulfill during the next n n days. Polycarpus knows that he is in for w w good events and b b not-so-good events. At least one event is going to take place during each day. As each day is unequivocally characterizes as a part of a white or a black stripe, then each day is going to have events of the same type only (ether good or not-so-good). What is the number of

Tongue Twister - Jessica 白洁 http://www.unsv.com/material/Food-Fashion-Fun/2016/06/22/ # 用于课后练习 Schwa Sound 英音 口语 ETJ English https://www.bilibili.com/video/av24311199/ Cup o f tea 轻读“f” a pice o f bread 规则一： R 结尾的字母上 <-- schwa Bette r Teache r What are you doing fo r Christmas? 规则二： 非重读音节上 P o lice = Puhlice Sof a = Sofuh A bout = Uhbout A gain = Uhgain Probl e m = Problum Pres i dent = Presuhdent Bott o m = Bottum Alb u m = Albuhm S y ringe = Suhringe 重读 /i:/ eg. Sk y yy < 非重读 /i/ eg. bab y an d 放中间不动读“d” # 正课内容 来源： https://www.cnblogs.com/Cong0ks/p/11570282.html