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第一节:单项填空(共15小题;每小题1分,满分15分)从A、B、C、D四个选项中,选出可以填入空白处的最佳选项。例:Stand over there ______ you’ll be able to see it better.A. or B. and C. but D. while答案是B。1. I’m trying to break the _______ of getting up too late.A. tradition B. convenience C. habit D. leisure正确答案:C
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请教:北京华建集团SQL面试题第1大题第1小题如何解答?【题目描述】1.已知关系模式:S (SNO,SNAME) 学生关系。SNO 为学号,SNAME 为姓名C (CNO,CNAME,CTEACHER) 课程关系。CNO 为课程号,CNAME 为课程名,CTEACHER 为任课教师SC(SNO,CNO,SCGRADE) 选课关系。SCGRADE 为成绩要求实现如下5个处理: 1. 找出没有选修过“李明”老师讲授课程的所有学生姓名 2. 列出有二门以上(含两门)不及格课程的学生姓名及其平均成绩 3. 列出既学过“1”号课程,又学过“2”号课程的所有学生姓名 4. 列出“1”号课成绩比“2”号同学该门课成绩高的所有学生的学号 5. 列出“1”号课成绩比“2”号课成绩高的所有学生的学号及其“1”号课和“2”号课的成绩 【参考答案分析】: 1.找出没有选修过“李明”老师讲授课程的所有学生姓名 --实现代码:Select Sname As 学生姓名 From s Where Not Exists (Select * From c, Sc Where c.Cno = Sc.Cno And Cteacher = 李明 And Sc.Sno = s.Sno);Select Sname As 学生姓名 From s Where Sno Not In (Select Sno From c, Sc Where c.Cno = Sc.Cno And Cteacher = 李明);2. 列出有二门以上(含两门)不及格课程的学生姓名及其平均成绩 --实现代码:Select s.Sno As 学生学号, s.Sname As 学生姓名, Avg(Sc.Scgrade) As 平均成绩 From s, Sc Where Sc.Sno = s.Sno And Sc.Sno In (Select Sc.Sno From Sc Where Sc.Scgrade < 60 Group By Sc.Sno Having Count(*) > 2) Group By s.Sno, s.Sname;3. 列出既学过“01”号课程,又学过“02”号课程的所有学生姓名 --实现代码:select s.sno as 学生学号,s.sname as 学生姓名 from s where sno in(select sc.sno as 学生学号 from c,sc where c.cno=sc.cno and c.cno in(01,02) group by sno having count(distinct sc.cno)=2);4. 列出“01”号课成绩比“02”号同学该门课成绩高的所有学生的学号 --实现代码:select sc1.sno as 学生学号 from sc as sc1,c as c1,sc as sc2,c as c2where sc1.cno=c1.cno and c1.cno=01 and sc2.cno=c2.cno and c2.cno=02and sc1.scgrade>sc2.scgrade group by sc1.sno;5. 列出“01”号课成绩比“02”号课成绩高的所有学生的学号及其“01”号课和“02”号课的成绩 --实现代码:select sc1.sno as 学生学号, sc1.scgrade as no1grade ,sc2.scgrade as no2gradefrom sc as sc1,c as c1,sc as sc2,c as c2where sc1.cno=c1.cno and c1.cno=01 and sc2.cno=c2.cno and c2.cno=02and sc1.scgrade>sc2.scgrade group by sc1.sno;答案是:1.找出没有选修过“李明”老师讲授课程的所有学生姓名--实现代码:Select Sname As 学生姓名 From sWhere Not Exists (Select * From c, Sc Where c.Cno = Sc.Cno And Cteacher = 李明 And Sc.Sno = s.Sno);Select Sname As 学生姓名 From sWhere Sno Not In (Select Sno From c, Sc Where c.Cno = Sc.Cno And Cteacher = 李明);2. 列出有二门以上(含两门)不及格课程的学生姓名及其平均成绩 --实现代码:Select s.Sno As 学生学号, s.Sname As 学生姓名, Avg(Sc.Scgrade) As 平均成绩 From s, ScWhere Sc.Sno = s.Sno And Sc.Sno In (Select Sc.Sno From Sc Where Sc.Scgrade < 60 Group By Sc.Sno Having Count(*) > 2)Group By s.Sno, s.Sname;3. 列出既学过“01”号课程,又学过“02”号课程的所有学生姓名 --实现代码:select s.sno as 学生学号,s.sname as 学生姓名 from s where sno in(select sc.sno as 学生学号 from c,sc where c.cno=sc.cno and c.cno in(01,02) group by sno having count(distinct sc.cno)=2);4. 列出“01”号课成绩比“02”号同学该门课成绩高的所有学生的学号 --实现代码:select sc1.sno as 学生学号 from sc as sc1,c as c1,sc as sc2,c as c2where sc1.cno=c1.cno and c1.cno=01 and sc2.cno=c2.cno and c2.cno=02and sc1.scgrade>sc2.scgrade group by sc1.sno;5. 列出“01”号课成绩比“02”号课成绩高的所有学生的学号及其“01”号课和“02”号课的成绩 --实现代码:select sc1.sno as 学生学号, sc1.scgrade as no1grade ,sc2.scgrade as no2gradefrom sc as sc1,c as c1,sc as sc2,c as c2where sc1.cno=c1.cno and c1.cno=01 and sc2.cno=c2.cno and c2.cno=02and sc1.scgrade>sc2.scgrade group by sc1.sno;
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单选题设f是由群到群的同态映射,则Ker(f)是( )。A
G`的子群B
G的子群C
包含G`D
包含G正确答案:C解析:同态的核定义为Ker(f)={g∈G:f(g)=1`}。因为:①f(1)=1`,所以1∈Ker(f);②若g1,g2∈Ker(f),那么f(g1×g2)=f(g1)*f(g2)=1`*1`=1`,则g1×g2∈Ker(f);③若g∈Ker(f),即f(g)=1`,那么1`=f(1)=f(g-1×g)=f(g-1)*f(g)=f(g-1)*1`,所以f(g-1)=1`,即g-1∈Ker(f)。上面三点说明ker(f)是G的子群,故本题选B。
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请教:2009年7月自考概率论与数理统计(经管类)真题第2大题第1小题如何解答?【题目描述】二、填空题(本大题共15小题,每小题2分,共30分)请在每小题的空格中填上正确答案。错填、不填均无分。11.将三个不同的球随机地放入三个不同的盒中,则出现两个空盒的概率为______. 【参考答案分析】:11.1/9
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请教:2011年上半年信息处理技术员上午试卷(参考答案版)第1大题第27小题如何解答?【题目描述】●下面记录的是某班36人期末考试的数学成绩:97 100 95 96 100 8796 100 89 100 93 6999 89 100 81 88 91100 87 98 75 85 8984 92 86 100 100 6187 51 81 92 95 97这些成绩的中位数为 (27) 。(27)A.90 B.91 C.92 D.93 正确答案:C
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单选题6阶有限群的任何子群一定不是( )阶。A
2B
3C
4D
6正确答案:B解析:有限群子群的阶数一定能整除该有限群的阶数,所以6阶子群的有限群不可能是4阶的,故本题选C。
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请教:2010年6月英语四级考试真题及答案(A卷word版)第1大题第1小题如何解答?【题目描述】第 1 题1. 如今不少学生在英语学习中不重视拼写2. 出现这种情况的原因3. 为了改变这种状况,我认为…Due Attention Should Be Given To Spelling 【参考答案分析】:Due Attention Should Be Given To SpellingCorrect spelling is a basic skill in English study. However, nowadays many students do not pay much attention to it.They have their own reasons for misspelling. First of all, they like an easy way of studying, which causes some omissions and changes in spelling. Second, the teachers might not be very strict in students’ spelling. In China, teachers seem to be more concerned with grammar and vocabulary but not spelling.To change this situation, in my opinion, the teachers and the students should work together. On one and, the teachers should give more attention to students’ spelling, asking the students to be conscious of the importance of correct spelling from the very beginning of their English study. On the other hand, the students themselves are supposed to be aware that correct spelling is a must in English study.To sum up, correct spelling is so important that both students and the teachers should spare no efforts to achieve correct spelling.
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单选题筛径确定的一只筛可以将泥沙群分开成小于(通过)和大于(通不过)筛径的两组子群,N只筛径不同的筛,可将泥沙群分开成()组子群。A
10B
N-1C
ND
N+1正确答案:D解析:暂无解析
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请教:2016年计算机二级考试C++模拟试题简答题4答案已知学生的记录由学号和学习成绩构成,n名学生的数据已存入a结构体数组中。请编写函数fun,该函数的功能是:找出成绩最低的学生记录,通过形参返回主函数(规定只有一个最低分)。void fun(stu a[],stu *s) {int i,min; min=a[0].s; for(i=0;i if(a.s {min=a.s; *s=a;}}
考题
单选题设f是由群到群的同态映射,则Ker(f)是( )。A
G`的子群B
G的子群C
包含G`D
包含G正确答案:C解析:同态的核定义为Ker(f)={g∈G:f(g)=1`}。因为:①f(1)=1`,所以1∈Ker(f);②若g1,g2∈Ker(f),那么f(g1×g2)=f(g1)*f(g2)=1`*1`=1`,则g1×g2∈Ker(f);③若g∈Ker(f),即f(g)=1`,那么1`=f(1)=f(g-1×g)=f(g-1)*f(g)=f(g-1)*1`,所以f(g-1)=1`,即g-1∈Ker(f)。上面三点说明ker(f)是G的子群,故本题选B。