练习1:求数组中所有元素的和 var arr1 = [10, 20, 30, 40, 50]; var sum = 0; for (var i = 0; i < arr1.length; i++) { sum += arr1[i]; } console.log(sum); 练习2:求数组中所有元素的平均值 var arr2 = [1, 2, 3, 4, 5]; var sum2 = 0; for (var i = 0; i < arr2.length; i++) { sum2 += arr2[
SELECT L.C# As 课程ID,L.score AS 最高分,R.score AS 最低分 FROM SC L ,SC AS R WHERE L.C# = R.C# and L.score = (SELECT MAX(IL.score) FROM SC AS IL,Student AS IM WHERE L.C# = IL.C# and IM.S#=IL.S# GROUP BY IL.C#) AND R.Score = (SELECT MIN(IR.score) FROM SC AS I
两种不同方式获取最大值与最小值 代码1: #include <stdio.h> int main() { ], sum = , max, min; int i; printf("请输入5名童鞋的成绩:\n"); ; i < ; i = i + ) scanf_s("%f", &score[i]); max = min = score[]; ; i < ; i = i + ) { if (max <= score[i]) max
select * from ( selectclass 班级,subject,avg(grade) avg_gradefrom student_score group by class,subject) pivot (sum(avg_grade) for subject in ('语文', '数学','英语')) order by 班级 asc 求班级平均分select 学科,平均分 别名 from 表where class="121"就是group by 班级,平均分 1.121班的
Console.WriteLine("请输入学生人数:"); int n=int.Parse(Console.ReadLine()); ArrayList arr= new ArrayList(); ; i <n; i++) { Console.WriteLine(); arr.Add(int.Parse(Console.ReadLine())); } foreach(int a in arr) { Console.Write(a+"\t"); } Conso
