Private Sub Command1_Click() s = p(1) + p(2) + p(3) + p(4) P
来源:学生作业帮 编辑:大师作文网作业帮 分类:综合作业 时间:2024/11/18 21:15:19
Private Sub Command1_Click() s = p(1) + p(2) + p(3) + p(4) Print s;
Private Sub Command1_Click()
s = p(1) + p(2) + p(3) + p(4)
Print s;
End Sub
Public Function p(n As Integer)
Static sum
For I = 1 To n
sum = sum + I
Next I
p = sum
End Function
Private Sub Command1_Click()
s = p(1) + p(2) + p(3) + p(4)
Print s;
End Sub
Public Function p(n As Integer)
Static sum
For I = 1 To n
sum = sum + I
Next I
p = sum
End Function
static sum 是静态的变量,他不会自动清空.p(1)=0+1=1;p(2)=sum+1+1=3;p(3)=sum+1+1+1=6;t同理p(4)=10;s=1+3+6+10=20.我把vb放下很久了.试着做一下.
答案是:20.
答案是:20.
公用签1p 2p 3p 4p 5p
*p
p
因式分解 2p^3-5p^2+4p-1
private sub command1_click( ) dim a a=array(1,2,3,4) j=1:s=0
断路器1P 2P 3P 4P 有什么区别.求详解..
1P、2P、3P、4P断路器是什么意思?使用有什么要求?
断路器里面的1P,2P,3P,4P各是什么意思?
配电箱中的断路器的1P,2P,3P,4P如何选用?
【p^2+2p-1】/【p^3+p^2+p+1】dp=-1/x dx 如何积分,
Private Sub Command1_Click() s = 0 Do s = (s + 1) * (s + 2)
开关上的1P.2P.3P.