F-statistic vs. constant model: 5.36, p-value = 0.0161什么意思啊？

1个回答

csdn_xinyi 谢谢你
6 个月之前 回复

java.text.ParseException: Unparseable date
library(tseries) library(forecast) library(urca) library(vars) set.seed(12345) u<-rnorm(500) x<-cumsum(u) y<-x+u ##如果用adf.test adf.test(y) ##结果是： Augmented Dickey-Fuller Test data: y Dickey-Fuller = -2.5536, Lag order = 7, p-value = 0.344 alternative hypothesis: stationary ##请问是平稳序列吗？？看哪些数值？ ##对同一个序列，如果用ur.df summary(ur.df(y)) ##结果是 ############################################### # Augmented Dickey-Fuller Test Unit Root Test # ############################################### Test regression none Call: lm(formula = z.diff ~ z.lag.1 - 1 + z.diff.lag) Residuals: Min 1Q Median 3Q Max -5.5235 -1.3095 0.0289 1.4407 5.4889 Coefficients: Estimate Std. Error t value Pr(>|t|) z.lag.1 0.0007778 0.0028307 0.275 0.784 z.diff.lag -0.3831155 0.0415501 -9.221 <2e-16 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 2.051 on 496 degrees of freedom Multiple R-squared: 0.1464, Adjusted R-squared: 0.143 F-statistic: 42.54 on 2 and 496 DF, p-value: < 2.2e-16 Value of test-statistic is: 0.2748 Critical values for test statistics: 1pct 5pct 10pct tau1 -2.58 -1.95 -1.62 ##请问是平稳序列吗？看哪些数值？ ##如果两个结果不一样，为什么？

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Lost Cows C语言来实现
Problem Description N (2 <= N <= 8,000) cows have unique brands in the range 1..N. In a spectacular display of poor judgment, they visited the neighborhood 'watering hole' and drank a few too many beers before dinner. When it was time to line up for their evening meal, they did not line up in the required ascending numerical order of their brands. Regrettably, FJ does not have a way to sort them. Furthermore, he's not very good at observing problems. Instead of writing down each cow's brand, he determined a rather silly statistic: For each cow in line, he knows the number of cows that precede that cow in line that do, in fact, have smaller brands than that cow. Given this data, tell FJ the exact ordering of the cows. Input * Line 1: A single integer, N * Lines 2..N: These N-1 lines describe the number of cows that precede a given cow in line and have brands smaller than that cow. Of course, no cows precede the first cow in line, so she is not listed. Line 2 of the input describes the number of preceding cows whose brands are smaller than the cow in slot #2; line 3 describes the number of preceding cows whose brands are smaller than the cow in slot #3; and so on. Output * Lines 1..N: Each of the N lines of output tells the brand of a cow in line. Line #1 of the output tells the brand of the first cow in line; line 2 tells the brand of the second cow; and so on. Sample Input 5 1 2 1 0 Sample Output 2 4 5 3 1

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spring-servlet.xml ``` <beans:bean id="myDataSource" class="org.apache.commons.dbcp.BasicDataSource"> <beans:property name="driverClassName" value="org.postgresql.Driver"/> <beans:property name="url" value="jdbc:postgresql://192.168.20.61:5432/atomdb"/> <beans:property name="username" value="postgre"/> <beans:property name="password" value="postgre"/> </beans:bean> <beans:bean id="sqlSessionFactory" class="org.mybatis.spring.SqlSessionFactoryBean"> <beans:property name="dataSource" ref="myDataSource"/> <beans:property name="configLocation" value="/WEB-INF/db/SqlMapConfig.xml"/> </beans:bean> <beans:bean class="org.mybatis.spring.mapper.MapperScannerConfigurer"> <beans:property name="basePackage" value="Mapper"/> </beans:bean> <beans:bean id="transactionManager" class="org.springframework.jdbc.datasource.DataSourceTransactionManager"> <beans:property name="dataSource" ref="myDataSource"/> </beans:bean> ``` SqlMapConfig.xml ``` <?xml version="1.0" encoding="UTF-8"?> <!DOCTYPE configuration PUBLIC "-//mybatis.org//DTD Config 3.0//EN" "http://mybatis.org/dtd/mybatis-3-config.dtd"> <configuration> <mappers> <mapper resource="ApiServiceRepoMapper.xml" /> </mappers> </configuration> ``` ApiServiceRepoMapper.xml ``` <?xml version="1.0" encoding="UTF-8" ?> <!DOCTYPE mapper PUBLIC "-//mybatis.org//DTD Mapper 3.0//EN" "http://mybatis.org/dtd/mybatis-3-mapper.dtd"> <mapper namespace="com.shangpin.EChars.repo.ComplexQueryRepo"> <select id="count" resultType="Integer"> SELECT "count"(*) FROM "public"."search_keyword_statistic_day" </select> </mapper> ``` Controller.java ``` @Controller @RequestMapping("/search") public class Controller { @Autowired private SearchService searchService ; @RequestMapping("/get") public void ss(){ System.out.println(searchService.count()); } } ``` SearchService.java ``` @Service public class SearchService { @Autowired ComplexQueryRepo complexQueryRepo; @Autowired public int count(){ System.out.println("222"); int a = complexQueryRepo.count(); return a; } } ``` ComplexQueryRepo.java ``` public interface ComplexQueryRepo{ public int count(); } ``` 项目能启动，发送请求/search/get后会输出222然后 ``` int a = complexQueryRepo.count(); ``` 这一行报空指针
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Lost Cows
Problem Description N (2 <= N <= 8,000) cows have unique brands in the range 1..N. In a spectacular display of poor judgment, they visited the neighborhood 'watering hole' and drank a few too many beers before dinner. When it was time to line up for their evening meal, they did not line up in the required ascending numerical order of their brands. Regrettably, FJ does not have a way to sort them. Furthermore, he's not very good at observing problems. Instead of writing down each cow's brand, he determined a rather silly statistic: For each cow in line, he knows the number of cows that precede that cow in line that do, in fact, have smaller brands than that cow. Given this data, tell FJ the exact ordering of the cows. Input * Line 1: A single integer, N * Lines 2..N: These N-1 lines describe the number of cows that precede a given cow in line and have brands smaller than that cow. Of course, no cows precede the first cow in line, so she is not listed. Line 2 of the input describes the number of preceding cows whose brands are smaller than the cow in slot #2; line 3 describes the number of preceding cows whose brands are smaller than the cow in slot #3; and so on. Output * Lines 1..N: Each of the N lines of output tells the brand of a cow in line. Line #1 of the output tells the brand of the first cow in line; line 2 tells the brand of the second cow; and so on. Sample Input 5 1 2 1 0 Sample Output 2 4 5 3 1
Free DIY Tour 编写程序来实现
Problem Description Weiwei is a software engineer of ShiningSoft. He has just excellently fulfilled a software project with his fellow workers. His boss is so satisfied with their job that he decide to provide them a free tour around the world. It's a good chance to relax themselves. To most of them, it's the first time to go abroad so they decide to make a collective tour. The tour company shows them a new kind of tour circuit - DIY circuit. Each circuit contains some cities which can be selected by tourists themselves. According to the company's statistic, each city has its own interesting point. For instance, Paris has its interesting point of 90, New York has its interesting point of 70, ect. Not any two cities in the world have straight flight so the tour company provide a map to tell its tourists whether they can got a straight flight between any two cities on the map. In order to fly back, the company has made it impossible to make a circle-flight on the half way, using the cities on the map. That is, they marked each city on the map with one number, a city with higher number has no straight flight to a city with lower number. Note: Weiwei always starts from Hangzhou(in this problem, we assume Hangzhou is always the first city and also the last city, so we mark Hangzhou both 1 and N+1), and its interesting point is always 0. Now as the leader of the team, Weiwei wants to make a tour as interesting as possible. If you were Weiwei, how did you DIY it? Input The input will contain several cases. The first line is an integer T which suggests the number of cases. Then T cases follows. Each case will begin with an integer N(2 ≤ N ≤ 100) which is the number of cities on the map. Then N integers follows, representing the interesting point list of the cities. And then it is an integer M followed by M pairs of integers [Ai, Bi] (1 ≤ i ≤ M). Each pair of [Ai, Bi] indicates that a straight flight is available from City Ai to City Bi. Output For each case, your task is to output the maximal summation of interesting points Weiwei and his fellow workers can get through optimal DIYing and the optimal circuit. The format is as the sample. You may assume that there is only one optimal circuit. Output a blank line between two cases. Sample Input 2 3 0 70 90 4 1 2 1 3 2 4 3 4 3 0 90 70 4 1 2 1 3 2 4 3 4 Sample Output CASE 1# points : 90 circuit : 1->3->1 CASE 2# points : 90 circuit : 1->2->1
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Stats

``` HttpHelper hh = new HttpHelper(); HttpResult hr = hh.GetHtml(new HttpItem() { Method = "POST", URL = "http://www.liepin.com/user/ajaxlogin/", PostDataType = PostDataType.String, Postdata = "isMd5=1&layer_from=wwwindex_rightbox_new&user_login=13345453073%40qq.com&user_pwd=a7c53b22f15684ec246921116a2f8faf&chk_remember_pwd=on", ContentType = "application/x-www-form-urlencoded", }); Literal1.Text = hr.Html; HttpResult hr2 = hh.GetHtml(new HttpItem() { Method = "GET", URL = "http://statistic.liepin.com/statisticPlatform/tLog?page_id=&url=http%3A%2F%2Fwww.liepin.com%2F&resolution=1920X1080&uuid=1444489384702.90&sessionId=1446362465522.16&if_mscid=00000000&il_mscid=00000000&ef_mscid=00000000&el_mscid=00000000&v_stay_time=83507&type=v&user_id=20817384&user_kind=0&session_seq=3&uv_seq=3&t=1446362572121", }); Literal1.Text += "<hr />" + hr2.Html; HttpResult hr3 = hh.GetHtml(new HttpItem() { Method = "GET", URL = "http://statistic.liepin.com/statVisit.do?site=1&userId=20817384&userKind=0&url=http%3A%2F%2Fwww.liepin.com%2F&resolution=1920x1080&h=15&m=22&s=52&cookie=1&ref=&puuid=14463624886176532258288&stay_time=84000&rand=0.057066346751526", }); Literal1.Text += "<hr />" + hr3.Html; HttpResult hr4 = hh.GetHtml(new HttpItem() { Method = "GET", URL = "http://c.liepin.com/?time=1446362571809", Cookie = "__uuid=1444489384702.90; _uuid=5246D851FFA8420C798600EBFAC2AB4A; gr_user_id=88684461-82ec-4a67-96dc-b9e120daaecc; JSESSIONID=4DD6C93314B2978337B9767477DC099C; _ltu=%7B%22level%22%3A1%2C%22goldcard%22%3Afalse%2C%22id%22%3A%2220817384%22%2C%22v%22%3A%221.1%22%7D; rand=89c3e5a340a877f518d751403671bee5; __tlog=1446362465522.16%7C00000000%7C00000000%7C00000000%7C00000000; __session_seq=3; __uv_seq=3; Hm_lvt_a2647413544f5a04f00da7eee0d5e200=1445584941,1446198379,1446207568,1446362466; Hm_lpvt_a2647413544f5a04f00da7eee0d5e200=1446362489; _mscid=00000000; _lpcdn=s%2C1%7Ccore.pc%2C1%7Ca.pc%2C1%7Cwww.pc%2C1%7Cjob.pc%2C1%7Ccompany.pc%2C1%7Cit.pc%2C1%7Cc.pc%2C1%7Clpt.pc%2C1%7Ch.pc%2C1%7Carticle.pc%2C1%7Cats.pc%2C1%7Ccampus.pc%2C1%7Ccity.pc%2C1%7Cclt.pc%2C1; gr_session_id=317eeb56-bd19-49d1-b83f-6e9701cf3a1b; lt_auth=uugIMyQCnFn6tnLb32Nb4qdJ2oiqVGrA9nwMhB9RgYC0DaW04PziSwqArbkDxBMhxEwkcMULNbT4%0D%0ANu78zHRL7kAT%0D%0A; autologin=true; user_login=164423073%40qq.com; user_name=%E6%9D%A8%E6%B6%9B; user_kind=0; user_id=20817384; user_photo=55557f3b28ee44a8919620ce01a.gif%3F0.9157210058157577; user_vip=0", }); Literal1.Text += "<hr />" + hr4.Html; // Response.Redirect("http://c.liepin.com");这个跳转到c.liepin.com后还是要提示登录 Response.Write(hr4.Html); //这个可以输出登录成功后的页面HTML代码 Response.End(); ``` 以上是相关的ASP.NET代码，如果是单纯获取到登录成功后的HTML页面的源代码是可以了。。但我想要的效果是用户名和密码在我的网站上输入，点击登录后跳转到猎聘网的用户中心，即网址从http://localhost:67465跳到http://c.liepin.com 后还是登录的状态的。。。请问这个可以搞吗?

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