33. 查询成绩比该课程平均成绩低的同学的成绩表. select * from score a where a.degree between 0 and( select avg(degree) from score b where a.cno=b.cno ) 34. 查询所有任课教师的Tname和Depart. select tname,depart from teacher where tno in( select tno from course ) 35 . 查询所有未讲课的教师的Tna
SELECT SUM(CASE WHEN C# ='001' THEN score ELSE 0 END)/SUM(CASE C# WHEN '001' THEN 1 ELSE 0 END) AS 企业管理平均分 ,100 * SUM(CASE WHEN C# = '001' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = '001' THEN 1 ELSE 0 END) AS 企业管理及格百分数 ,SUM(CASE WHEN C# =
29.查询选修编号为"3-105"课程且成绩至少高于选修编号为"3-245"的同学的Cno.Sno和Degree,并按Degree从高到低次序排序. select tname,prof from teacher where depart = '计算机系' and prof not in ( select prof from teacher where depart = '电子工程系') 30.查询选修编号为"3-105"且成绩高于选修编号为&q
SELECT S# as 学生ID ,(SELECT score FROM SC WHERE SC.S#=t.S# AND C#='004') AS 数据库 ,(SELECT score FROM SC WHERE SC.S#=t.S# AND C#='001') AS 企业管理 ,(SELECT score FROM SC WHERE SC.S#=t.S# AND C#='006') AS 英语 ,COUNT(*) AS 有效课程数, AVG(t.score) AS 平均成绩 FROM SC
23.查询"张旭"教师任课的学生成绩. select * from score s where cno in ( select cno from course where tno in( select tno from teacher where tname = '张旭')) 24.查询选修某课程的同学人数多于5人的教师姓名. select tname from teacher where tno in ( select tno from course where cno in ( s
求平均成绩 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 58415 Accepted Submission(s): 13985 Problem Description 假设一个班有n(n<=50)个学生,每人考m(m<=5)门课,求每个学生的平均成绩和每门课的平均成绩,并输出各科成绩均大于等于平均成绩的学生数量. Inpu
求平均成绩 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 74055 Accepted Submission(s): 17809 Problem Description 假设一个班有n(n<=50)个学生,每人考m(m<=5)门课,求每个学生的平均成绩和每门课的平均成绩,并输出各科成绩均大于等于平均成绩的学生数量. Inpu
求平均成绩 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 61842 Accepted Submission(s): 14812 Problem Description 假设一个班有n(n<=50)个学生,每人考m(m<=5)门课,求每个学生的平均成绩和每门课的平均成绩,并输出各科成绩均大于等于平均成绩的学生数量. Inp
C语言课程设计(成绩管理系统) 翻到了大学写的C语言课程设计,缅怀一下 内容: 增加学生成绩 查询学生成绩 删除 按照学生成绩进行排序 等 #include <stdio.h> #include <string.h> #include <stdlib.h> #define N 20 struct student { int num; ]; int chinese; int math; int english; int sum; }; struct student stu