好吧，表示太菜了，新手再来求大神指点~
Private Sub Form_click()Dim a As String
Dim i As Integer
Dim kk As Integer
Dim j As Double
Dim he!
Dim v1!, v2!, v3!
Dim z!(), v!() '水库水位和库容曲线
Dim zx!(), qx!() '下游水位和流量关系
Dim ql!(), qxx!() '上游来水和下游需水
Dim capacity!, bzcl!, kn! '装机容量，保证出力，出力系数
Dim minlever!, diaoduchu!, diaodumo! '死水位，调度初水位，调度末水位
Dim sum1!
Dim ht#() '调度线
Dim s!, s1!
Dim n!(), n1!(), n2!(), n3!() '中间过程发电量
Dim qq1 As Double
Dim qq2 As Double
i = 0
Open App.Path & "\z-v.txt" For Input As #1
'Input #1, a
Do While Not EOF(1)
i = i + 1
ReDim Preserve z(i), v(i)
Input #1, z(i), v(i)
'Print z(i), v(i)
Loop
Close 1
i = 0
Open App.Path & "\z-q.txt" For Input As #2
Do While Not EOF(2)
i = i + 1
ReDim Preserve zx(i), qx(i)
Input #2, zx(i), qx(i)
Print zx(i), qx(i)
Loop
Close 2
i = 0
Open App.Path & "\上游来水.txt" For Input As #3
Do While Not EOF(3)
i = i + 1
ReDim Preserve ql(i)
Input #3, ql(i)
'Print ql(i)
Loop
Close 3
i = 0
Open App.Path & "\下游需水.txt" For Input As #4
Do While Not EOF(4)
i = i + 1
ReDim Preserve qxx(i)
Input #4, qxx(i)
'Print qxx(i)
Loop
Close 4
capacity = 32
bzcl = 12.8
kn = 0.8
minlever = 135
buchang = 0.1
diaoduchu =160
diaodumo = 160
sum1 = UBound(ql)
'Print sum1
ReDim ht(sum1 + 1), n(sum1 + 1), n3(sum1 + 1), qx(sum1), qfd(sum1), qql(sum1)
kk = (160 - minlever) / buchang
For i = 1 To sum1 + 1
ht(i) = diaoduchu
n(i) = 0
'Print n(i)
Next i
s = 0
s1 = 0
Do
For i = 2 To sum1 - 1 Step 1
For j = 160 To minlever Step -buchang
kk = (160 - j) / buchang
v1 = cha(z(), v(), ht(i - 1))
v2 = cha(z(), v(), j)
v3 = cha(z(), v(), ht(i + 1)) '第一次调用cha函数
qq1 = (v1 - v2) * 38.58 + ql(i - 1) '38.58=(10^8/30*24*3600)
qq2 = (v2 - v3) * 38.58 + ql(i)
h1 = (ht(i - 1) + j) / 2 - cha(qx(), zx(), qq1)
h2 = (ht(i + 1) + j) / 2 - cha(qx(), zx(), qq2) '第二次调用cha函数
n1(kk + 1) = kn * qq1 * h1 / 10000
n2(kk + 1) = kn * qq2 * h2 / 10000
If n1(kk + 1) > capacity Then
n1(kk + 1) = capatity
End If
If qq1 < qxx(i) Or n1(kk + 1) < bzcl Then
n1(kk + 1) = -10000000
End If
If n2(kk + 1) > capacity Then
n2(kk + 1) = capatity
End If
If qq2 < qxx(i) Or n2(kk + 1) < bzcl Then
n2(kk + 1) = -10000000
End If
no = n1(kk + 1) + n2(kk + 1)
If no > n(i - 1) + n(i) Then
n(i - 1) = n1(kk + 1)
n(i) = n2(kk + 1)
ht(i) = j
x = n1(kk + 1)
y = n2(kk + 1)
End If
Next j
s = s + x
Next i
s = s + y
he = Abs(s - s1) / s
s1 = s
Loop While he > 0.01
For i = 1 To sum1 - 1 Step 1
Print ht(i)
Next
End Sub
Public Function cha!(x!(), y!(), a#)
Dim zhi As Single
n = UBound(x)
For i = 1 To n
Debug.Print x(2)
Next
If x(1) >= a Then
zhi = y(1)
ElseIf x(n) < a Then
zhi = y(n)
Else
For i = 1 To n
If x(i) > a Or x(i) = a Then
ml = i
Exit For
End If
Next i
zhi = y(ml - 1) + (y(ml) - y(ml - 1)) * (a - x(ml - 1)) / (x(ml) - x(ml - 1))
End If
cha = zhi
End Function
'如上面程序所示我在调用第二次cha函数时，显示x!()的ubound及n为12,可是我的txt格式里的数据只有7个，为什么n会等于12呢，求大神指点。。。
[ 本帖最后由 liu141231893 于 2014-10-29 17:00 编辑 ]
