皮皮学,免费搜题
登录
logo - 刷刷题
搜题
【单选题】
Jim Lewis, software developer at Technology Systems wants to update the details, such as name, address, and telephone number, of all the employees in its Payroll Application. For this, Jim needs to generate the address list of all the employees working in the organization. He decides to implement a function that will accept the name, address, and the telephone number from each employee and then print a list containing the details. He has written the following steps to perform the above task: 1. Start the function 2. Declare “Counter” as a numeric variable 3. Declare “Name”, “Address”, and “TelNo” as character variables 4. Let Counter= 1 5. Check the condition Is (Counter <= 30). If the condition is True, then a)Accept the Name value from the employee b)Accept the Address value from the employee c)Accept the TelNo value from the employee d)Display the Name, Address, and TelNo values on the computer screen e) Counter = Counter - 1 f)Again check the condition, Is (Counter <= 30). If this condition is true, repeat the steps from a to d. 6. If the condition, Is (Counter <= 30) is False, then end the function. However, these steps does not produce the required results. What should Jim do to accomplish the task?
A.
Jim needs to replace the step e and f with following steps: e. Counter = Counter + 1 f. Check the condition, Is (Counter <= 30). If this condition is true, repeat the steps from a to f.
B.
Jim needs to replace the step e and f with following steps: e. Counter = Counter + 1 f. Check the condition, Is (Counter >= 30). If this condition is true, repeat the steps from a to f.
C.
Jim needs to replace the step e and f with following steps: e. Counter = Counter - 1 f. Check the condition, Is (Counter = 30). If this condition is False, repeat the steps from a to f.
D.
Jim needs to replace the step e and f with following steps: e. Counter = Counter + 2 f. Check the condition, Is (Counter <= 30). If this condition is true, repeat the steps from d to f.
手机使用
分享
复制链接
新浪微博
分享QQ
微信扫一扫
微信内点击右上角“…”即可分享
反馈
参考答案:
举一反三
【判断题】深度为k的二叉树至多有2k-1个结点
A.
正确
B.
错误
【单选题】深度为k的二叉树至多有______ 。
A.
2k个结点
B.
2k-1个结点
C.
2K-1结点
D.
2k-1-1个结点
【单选题】深度为k的二叉树至多有 ______。
A.
2k个结点
B.
2(k-1)结点
C.
2k-1结点
D.
2(k-1)-1个结点
【简答题】深度为k的二叉树至多有2^k-1个结点,最少有 ( )个结点。
【单选题】一带电体可作为点电荷处理的条件是
A.
电荷必须呈球形分布
B.
带电体的限度很小
C.
带电体的限度与其它有关长度相比可以忽略不计
D.
电荷量很小
【判断题】深度为 k 的二叉树至多有 2 k - 1 个结点, (k>=1) 。
A.
正确
B.
错误
【单选题】深度为k的二叉树至多有2k+1(k≥1)个结点。
A.
正确
B.
错误
【判断题】深度为K的二叉树至多有2K-1个结点。( )
A.
正确
B.
错误
【单选题】一带电体可作为点电荷处理的条件是
A.
电荷必须呈球形分布
B.
带电体的线度很小
C.
带电体的线度与其它有关长度相比可以忽略
D.
带电体的电量很小
【单选题】一带电体可作为点电荷处理的条件是
A.
( A )电荷必须呈球形分布
B.
( B )带电体的线度很小
C.
( C )带电体的线度与其它有关长度相比可忽略不计 (或者说:带电体的线度与它到空间点的距离比小很多 )
D.
( D )电量很小
相关题目:
参考解析:
知识点:
题目纠错 0
发布
创建自己的小题库 - 刷刷题