sat

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: In guidance with Material Design, how much darker should the statusbar be than the actionbar? I have a color set for the actionbar at runtime and have no way of knowing this colour at programming time, so how can I get the correct statusbar colour? I know i can darken a colour using this this.getSupportActionBar().setBackgroundDrawable(new ColorDrawable(colorPrimary)); float[] hsv = new float[3]; Color.colorToHSV(colorPrimary, hsv); hsv[2] *= 0.8f; int colorPrimaryDark = Color.HSVToColor(hsv); if(Build.VERSION.SDK_INT>=21) Chat.this

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: [Sat Apr 09 18:27:29.953008 2016] [:error] [pid 3230:tid 140635784853248] [client 103.14.196.22:53950] mod_wsgi (pid=3230): Target WSGI script '/var/www/FlaskApps/FlaskApps.wsgi' cannot be loaded as Python module. [Sat Apr 09 18:27:29.953045 2016] [:error] [pid 3230:tid 140635784853248] [client 103.14.196.22:53950] mod_wsgi (pid=3230): Exception occurred processing WSGI script '/var/www/FlaskApps/FlaskApps.wsgi'. [Sat Apr 09 18:27:29.953065 2016] [:error] [pid 3230:tid 140635784853248] [client 103.14.196.22:53950] Traceback (most recent call

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 am trying to solve a SAT problem with 12000+ boolean variables using Z3. I expect that most of the variables will evaluate to false in the solution. Is there a way to guide or hint Z3 as SAT solver to try "polarity false" first? I've tried it with cryptominisat 2 and got good results. 回答1: Z3 is a collection of solvers and preprocessors. We can provide hints for some of the solvers. When the command (check-sat) is used, Z3 will select the solver automatically for us. We should (check-sat-using ) if we want to select the solver ourselves.

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'm using an Arduino Nano, previously an UNO, to collect data from sensors and then send the data back over a USB connection through a powered hub to a Raspberry Pi model B running Jessie 4.1.19+ Mar, 15 2016 and Python 2.7. The python code uses pyserial readline to get the data. In general it works fine but occasionally (typically a couple of times a day) it fails with the above message. It seemed the only way to fix it was to unplug the USB connection, reconnect it and restart the program. I altered the code to catch the exception but it

T13_REF_COUNTRY_SAT_TEST COUNTRY_CD NAME_CN NAME_EN T13_REF_CITY_SAT_TEST CITY_CD NAME_CN NAME_EN T13_REF_AIRPORT_SAT_TEST IATA_CD NAME_CN NAME_EN 0 接口中不存在 1 记录数量只有1条(新增) 2 ... ---1 T13_REF_COUNTRY_SAT_TEST 总`327` SELECT COUNT(COUNTRY_CD) FROM T13_REF_COUNTRY_SAT_TEST 去重`259` SELECT COUNT(DISTINCT COUNTRY_CD) FROM T13_REF_COUNTRY_SAT_TEST 总`82` SELECT COUNT(COUNTRY_CD) FROM T13_REF_COUNTRY_SAT_TEST WHERE IS_EXISTS='0' SELECT COUNTRY_CD,NAME_CN,NAME_EN FROM T13_REF_COUNTRY_SAT_TEST WHERE IS_EXISTS='0' 总`177` SELECT COUNT(COUNTRY_CD) FROM T13_REF_COUNTRY_SAT_TEST WHERE IS_EXISTS='1' 新增`129`

#!/bin/bash a="The cat sat on the mat" test="cat" awk -v a="$a" -v b="$test" 'BEGIN{print index(a,b)}' 输出 bogon:Desktop macname$ ./test.sh 5 参考： 来源： https://www.cnblogs.com/sea-stream/p/11403014.html

I have some theoretical/practical problem and I don't have clue for now on how to manage, Here it is: I create a SAT solver able to find a model when one is existing and to prove the contradiction when it's not the case on CNF problems in C using genetics algorithms. A SAT-problem looks basically like this kind of problem : My goal is to use this solver to find solutions in a lot of different NP-completes problems. Basically, I translate different problems into SAT, solve SAT with my solver and then transform the solution into a solution acceptable for the original problem. I already succeed

I am using the Python bindings for the Z3 theorem prover (Z3Py). I have N boolean variables, x1,..,xN. I want to express the constraint that exactly K out of N of them should be true. How can I do that, in Z3Py? Is there any built-in support for that? I checked the online documentation, but the Z3Py docs don't have any mention of any API for that. For one-out-of-N constraints, I know I can separately express that at least one is true (assert Or(x1,..,xN)) and that at most one is true (assert Not(And(xi,xj)) for all i,j). I also know of other ways to manually express the 1-out-of-N and K-out-of

2-sat 一个变量两种状态 符合条件的状态建边 找强连通，两两成立 1 - n 为第一状态 （n + 1） - （n + n） 为第二状态 例题模板 链接一 POJ 3207 Ikki's Story IV - Panda's Trick（2- sat） 链接二 Katu Puzzle POJ - 3678 （2 - sat） 链接三 HDU3622 Bomb Game(二分+2-SAT) 来源： https://www.cnblogs.com/DWVictor/p/11355038.html

题面 https://www.luogu.org/problem/P3825 题解 我发现这些年我学的$2-sat$竟然一直是假的，醉了醉了。 我们看一下$2-sat$的形式： 如果$p$，那么$q$。 数学上学过，一个命题成立，那么它的逆否命题成立。在这里，他的逆否命题也是可以完全确定的。 所以，如果非$q$，那么非$p$。 形象理解一下，如果非$q$，那么$p$变量的值肯定不是随便取的，如果它成立，而$q$又不成立，那就假了。所以也要非$p$ 所以在连边的时候，也要连它的“逆否边”。 我认为，$2-sat$是可解的，和“它的逆否命题唯一确定”的性质是很有关系的，如果只连原命题的边，解集会形成一个闭合子图的形式。。。。。。。。 #include<stack> #include<vector> #include<cstdio> #include<cstring> #include<iostream> #define ri register int #define N 100500 using namespace std; int n,d,m,d1[N],t1[N],d2[N],t2[N],sani,c[N],sat[N]; int dfn[N],low[N],cnt,clo,bel[N]; bool ins[N],vis[N]; char s[N]; stack<int> sta;