計算機在化工中的應用實驗報告

1、計算機在化學化工中的應用實驗報告學院： 化學與化工學院 班級：12級碩勛勵志班 姓名: 徐凱杰 學號： 120702028實驗一 傳熱實驗中多變量的曲線的擬合一、實驗目的1） 熟悉VB編程平臺2） 掌握多變量曲線擬合的算法3） 編擬合所給的傳熱實驗模型的VB程序4） 通過實驗數(shù)據(jù)求出模型數(shù)據(jù)、并掌握解線性方程組的克拉默法則二、運行環(huán)境1） Microsoft Windows XP2） VB6.0三、實驗原理略四、vb代碼Private Sub Command1_Click()Dim m As Integer'm=inputbox(“實驗次數(shù)”)m = 7Dim x10, x20, y0

2、Dim i, j, k As IntegerDim a(1 To 10, 1 To 10), y(1 To 10), y1(1 To 10), a0, a1, a2Dim s, S1, S2, S3, b(1 To 10, 1 To 10), xxDim x1(1 To 10), x2(1 To 10), YY, sd'open"dem.dat"for input as#1'for i=1 to m' input#1,xx,YY' x1(i)=xx' x2(i0=xx2' y(i)=YY'next i'clos

3、e#1'7組努塞爾準數(shù)、雷諾數(shù)及普蘭德準數(shù)，數(shù)據(jù)最大時應采用直接從文件讀取方法x10 = Array(0, 100, 200, 300, 500, 100, 700, 800) '注意下標的起點處理（加0)x20 = Array(0, 2, 4, 1, 0.3, 5, 3, 4) '注意下標的起點處理（加0)y0 = Array(0, 1.127, 2.416, 2.205, 2.312, 1.484, 6.038, 7.325) '注意下標的起點處理（加0)For i = 1 To m x1(i) = Log(x10(i) x2(i) = Log(x20(i)

4、 y(i) = Log(y0(i)Next i'求解法方程系數(shù)矩陣a(1, 1) = ma(1, 2) = 0For i = 1 To m a(1, 2) = a(1, 2) + x1(i)Next ia(2, 1) = a(1, 2)a(1, 3) = 0For i = 1 To m a(1, 3) = a(1, 3) + x2(i)Next ia(3, 1) = a(1, 3)a(2, 2) = 0For i = 1 To m a(2, 2) = a(2, 2) + x1(i) * x1(i)Next ia(3, 3) = 0For i = 1 To m a(3, 3) = a(3

5、, 3) + x2(i) * x2(i)Next ia(2, 3) = 0For i = 1 To m a(2, 3) = a(2, 3) + x1(i) * x2(i)Next ia(3, 2) = a(2, 3)'求解法方程常數(shù)向量y1(1) = 0 For i = 1 To m y1(1) = y1(1) + y(i) Next iy1(2) = 0 For i = 1 To m y1(2) = y1(2) + x1(i) * y(i) Next iy1(3) = 0 For i = 1 To m y1(3) = y1(3) + x2(i) * y(i) Next i'(

6、利用克拉默法則解法方程/線性非常組)s = a(1, 1) * a(2, 2) * a(3, 3) + a(1, 2) * a(2, 3) * a(3, 1) + a(1, 3) * a(2, 1) * a(3, 2)s = s - a(1, 1) * a(2, 3) * a(3, 2) - a(1, 2) * a(2, 1) * a(3, 3) - a(1, 3) * a(2, 2) * a(3, 1)For j = 1 To 3 b(j, 1) = a(j, 1) a(j, 1) = y1(j)Next jS1 = a(1, 1) * a(2, 2) * a(3, 3) + a(1, 2)

7、 * a(2, 3) * a(3, 1) + a(1, 3) * a(2, 1) * a(3, 2)S1 = S1 - a(1, 1) * a(2, 3) * a(3, 2) - a(1, 2) * a(2, 1) * a(3, 3) - a(1, 3) * a(2, 2) * a(3, 1)For j = 1 To 3 a(j, 1) = b(j, 1)Next jFor j = 1 To 3 b(j, 2) = a(j, 2) a(j, 2) = y1(j) Next jS2 = a(1, 1) * a(2, 2) * a(3, 3) + a(1, 2) * a(2, 3) * a(3,

8、1) + a(1, 3) * a(2, 1) * a(3, 2)S2 = S2 - a(1, 1) * a(2, 3) * a(3, 2) - a(1, 2) * a(2, 1) * a(3, 3) - a(1, 3) * a(2, 2) * a(3, 1)For j = 1 To 3 a(j, 2) = b(j, 2) Next jFor j = 1 To 3 b(j, 3) = a(j, 3) a(j, 3) = y1(j)Next jS3 = a(1, 1) * a(2, 2) * a(3, 3) + a(1, 2) * a(2, 3) * a(3, 1) + a(1, 3) * a(2

9、, 1) * a(3, 2)S3 = S3 - a(1, 1) * a(2, 3) * a(3, 2) - a(1, 2) * a(2, 1) * a(3, 3) - a(1, 3) * a(2, 2) * a(3, 1)a0 = S1 / sa1 = S2 / sa2 = S3 / sText1.Text = Int(1000 * Exp(a0) + 0.5) / 1000 '四舍五入保留三位Text2.Text = Int(1000 * a1 + 0.5) / 1000Text3.Text = Int(1000 * a2 + 0.5) / 1000sd = 0For i = 1 T

10、o m sd = sd + Abs(a0 + a1 * x1(i) + a2 * x2(i) - y(i) '求 Nextsd = sd / mText4.Text = sd 'Int(1000 * sd + 0.5) / 1000Print Tab(50); "序號", "模型計算值", "實驗值"For i = 1 To mPrintPrint Tab(45); i; (Text1.Text) * (x10(i) (Text2.Text) * (x20(i) (Text3.Text); 0.023 * (x10(i

11、) 0.8) * (x20(i) 0.3)NextEnd Sub五、實驗結果截圖六、實驗后思考。 VB編程是一種簡單，并且效率高的可視化的、面向?qū)ο蠛筒捎檬录?qū)動方式的結構化高級程序設計語言。通過對本實驗的實際操作，我掌握了多變量曲線擬合的基本算法，了解了解線性方程組的克拉默法則。并且，同時在以后的工作中，可以通過這個實驗來解決大部分實驗數(shù)據(jù)及模型參數(shù)的擬合問題。實驗二 梯度法擬合蒸汽壓與溫度關系模型一 、實驗目的1) 掌握梯度法擬合的基本算法以及理解其普適性2) 編寫梯度法擬合蒸汽壓與溫度的關系的VB程序3) 通過實對程序進行驗證，并注意比較初值對運行速度和結果的影響二 、運行環(huán)境1） Mi

12、crosoft Windows XP2） VB6.0三 、實驗原理略四、 實驗VB程序代碼Private Sub Command1_Click(Index As Integer)Dim m, n As Integerm = 6Dim i, j, k As IntegerDim A, B, C, F, ee, P(1 To 10), T(1 To 10)Dim A1, B1, C1, TA, TB, TC, TT, f1, f2, f3Dim sd, W, S, EY, XX, YY'(由dem.dat輸入實驗數(shù)據(jù)XX = Array(-23.7, -10, 0, 10, 20, 30,

13、 40) '注意下標的起點處理（加0)YY = Array(0.101, 0.174, 0.254, 0.359, 0.495, 0.662, 0.88) '注意下標的起點處理（加0)Print "直接讀數(shù)據(jù)文件后計算"For i = 1 To m T(i) = XX(i) T(i) = 273.15 + T(i) P(i) = YY(i) * 7600 Print T(i), P(i)Next iClose i A = Val(InputBox("A") '指定初值 B = Val(InputBox("B")

14、 '指定初值 C = Val(InputBox("C") '指定初值1000 F = 0For i = 1 To m ee = FNP(A, B, C, T(i), P(i) ee = ee 2 F = F + eeNext if1 = 0A1 = A + 0.000001 * A'print"A,A1="A,A1For i = 1 To m ee = FNP(A1, B, C, T(i), P(i) ee = ee 2 f1 = f1 + eeNext iTA = (f1 - F) / (0.000001 * A)'pr

15、int f1,F,TA'A=val(inputbox("A")f2 = 0B1 = B + 0.00001 * BFor i = 1 To m ee = FNP(A, B1, C, T(i), P(i) ee = ee 2 f2 = f2 + eeNext iTB = (f2 - F) / (0.00001 * B)f3 = 0C1 = C + 0.00001 * CFor i = 1 To m ee = FNP(A, B, C1, T(i), P(i) ee = ee 2 f3 = f3 + eeNext iTC = (f3 - F) / (0.00001 * C

16、)TT = TA 2 + TB 2 + TC 2TT = Sqr(TT)If TT > 0.001 ThenA = A - 0.005 * TAB = B - 1.5 * TBC = C - 0.001 * TCGoTo 1000ElseEnd IfPrintsd = 0For i = 1 To m '/計算絕對平均相對誤差sd = sd + Abs(FNSD(A, B, C, T(i), P(i) / P(i)Print FNSD(A, B, C, T(i), P(i)Next isd = sd / mPrintPrint "A,B,C=" A, B, CP

17、rint "sd=" sd '/打印絕對平均相對誤差End SubPublic Function FNP(A, B, C, T, P)FNP = (A - B / (T + C) - Log(P)End FunctionPublic Function FNSD(A, B, C, T, P)FNSD = Exp(A - B / (T + C) - PEnd Function五 、實驗結果截圖六 、實驗后思考。本實驗是基于最小二乘原理，函數(shù)擬合的目標是使擬合函數(shù)和實際測量值之間的差的平方和最小。對于最小值的問題，梯度法是用負梯度方向作為優(yōu)化搜索方向。而梯度法是一個簡單的

18、迭代優(yōu)化計算方法。注意的是，負梯度的最速下降性是一個局部的性質(zhì)。在計算的前期使用此法，當接近極小點時，在改用其他的算法，如共軛梯度法。 實驗三 二分法求解化工中的非線性方程一、實驗目的1） 掌握二分法解非線性方程組的基本算法2） 編寫二分法邱玨非線性方程組的VB程序3） 通過實例的程序進行調(diào)試，并學習輸出數(shù)據(jù)格式化二、 運行環(huán)境1） Microsoft WindowsXP2） VB6.0三 、實驗原理略四 、實驗VB代碼Private Sub Command1_Click()Dim ax As SingleDim bx As SingleDim cx As SingleDim ay As Si

19、ngleDim by As SingleDim cy As SingleDim e As SingleDim num As Integer '累計次數(shù)變量Dim st As StringDim ch As StringDim sp As Stringch = Chr(13) + Chr(10)sp = Space(10)st = "二分法解方程" + chst = st + "求2,3-二甲基苯胺沸點（當 P=101325 時 解 lnP=59.7622-8013.69/T-5.081lnT)" + chax = 200bx = 500e = 0

20、.01st = st + "區(qū)間左端點初始值 ax=" + Str(ax) + chst = st + "區(qū)間右端點初始值 bx=" + Str(bx) + chst = st + "精度控制限 e=" + Str(e) + chst = st + "num" + sp + "ax" + Space(14) + "bx" + Space(14) + "|ax-bx|" + chay = F(ax)by = F(bx)num = 1Do While Abs(

21、ax - bx) > e cx = (ax + bx) / 2 cy = F(cx) If cy = 0 Then Exit Do '如果已得解，則退出循環(huán) If cy * ay > 0 Then ax = cx ay = cy Else bx = cx by = cy End If st = st + Format(num, "000") + sp + Format(ax, "000.00") + sp + Format(bx, "000.00") + sp + Format(Abs(ax - bx), &quo

22、t;0.0000") + ch num = num + 1Loopst = st + ch + "2,3-二甲基苯胺沸點:" + Format(cx, "#00.00") + "K" + ch + chst = st + "*時間:" + Str(Time) + Space(3) + "日期:" + Str(Date) + chText1.Text = ""Text1.Text = stEnd Sub'二分法求2,3-二甲基苯胺沸點所用函數(shù)Private F

23、unction F(ByVal u As Single)F = Log(101325) - 59.7622 + 8013.69 / u + 5.081 * Log(u) '注意對數(shù)運算End Function五、實驗結果截圖六 、實驗后思考。 通過應用微積分中的介值定理，是是用二分法的前提條件。如果我們所要求解的方程從物理意義上來講確實存在實根，但又不滿足f(a)f(b)<0,這時候，我們必須通過改變a和b的值來滿足二分法的應用條件。實驗四 主元最大高斯消元法解化工中的線性方程組一、實驗目的1） 掌握主元最大高斯消元法2） 編寫最大高斯消元法求解線性方程組的VB程序3） 通過實例

24、對程序進行調(diào)試，并比較一般的高斯消去法比較二 、運行環(huán)境1） Microsoft WIndowsXP_2） VB6.0三 、實驗原理略四 、實驗程序代碼Private Sub Command1_Click()Dim m, n As IntegerDim a(), z(), x(), w, aa(), s, t, k, ln = 4ReDim a(n + 2, 2 + n), z(n + 2, 2 + n), x(n + 1), aa(n + 2, 2 + n)Dim i, j, k1, k2, stDim ch As StringDim sp As Stringch = Chr(13) + C

25、hr(10)sp = Space(5)a(1, 1) = 6# / 123.1a(1, 2) = 6# / 93.13a(1, 3) = 3# / 73.1a(1, 4) = 2# / 43.07a(2, 1) = 5# / 123.1a(2, 2) = 7# / 93.13a(2, 3) = 7# / 73.1a(2, 4) = 6# / 43.07a(3, 1) = 1# / 123.1a(3, 2) = 1# / 93.13a(3, 3) = 1# / 73.1a(3, 4) = 0# / 43.07a(4, 1) = 2# / 123.1a(4, 2) = 0# / 93.13a(4,

26、 3) = 1# / 73.1a(4, 4) = 1# / 43.07a(1, 5) = 57.78 / 12.01a(2, 5) = 7.92 / 1.008a(3, 5) = 11.23 / 14.01a(4, 5) = 23.09 / 16st = st + "主元最大高斯消去法解線性方程組" + chst = st + "設有一混合物由硝基苯、苯胺、氨基丙酮、乙醇組成;" + chst = st + "對該混合物進行元素分析結果以百分數(shù)表示如下" + chst = st + "C%=57.78%;H%=7.92%;N

27、%=11.23%;O%=23.09%" + chst = st + "原子量：A(C)=12.01;A(H)=1.008;A(N)=14.01;A(O)=16.00" + chst = st + "分子量：硝基苯 123.1;苯胺 93.13;氨基丙酮 73.10;乙醇 43.07" + chst = st + "硝基苯分子C-6;H-5;N-1;O-2" + chst = st + "苯胺分子C-6;H-7;N-1;O-0" + chst = st + "氨基丙酮分子C-3;H-7;N-1;O

28、-1" + chst = st + "乙醇分子C-2;H-6;N-0;O-1" + chst = st + "確定上面四種化合物在混合物中所占的百分比" + ch + ch'尋找主元For i = 1 To n If i = n Then GoTo 200 For t = i + 1 To n If Abs(a(i, i) < Abs(a(t, i) Then For s = i To n + 1 aa(t, s) = a(i, s) a(i, s) = a(t, s) a(t, s) = aa(t, s) Next s Else

29、 End If Next t200'消去w = a(i, i) For j = 1 To n + 1 a(i, j) = a(i, j) / w Next jIf i = n Then GoTo 100For j = i + 1 To n For k = i + 1 To n + 1 z(i, k) = a(i, k) * a(j, i) a(j, k) = a(j, k) - z(i, k) Next kNext jNext i100'回代x(n + 1) = 0 For k = n To 1 Step -1 s = 0 For j = k + 1 To n s = s +

30、a(k, j) * x(j) Next j x(k) = a(k, n + 1) - s 'st=st+"x("+str(i)+")="+format(x(i),"00.00")+"%"+ch 'print"x("k;")="x(k) Next kFor i = 1 To n '輸出結果 st = st + "x(" + Str(i) + ")=" + Format(x(i), "00.00"

31、) + "%" + chNext ist = st + chst = st + "*時間:" + Str(Time) + Space(3) + "日期:" + Str(Date) + chText1.Text = ""Text1.Text = stEnd Sub五 、實驗結果截圖六、 實驗后思考高斯消去法不需要方程組的初值，也不需要重復迭代計算。只通過“消去”和“回代”2個過程就可以直接求出方程組的解。然后若是在消去的過程中，若碰到主元為0，則無法計算。所以，發(fā)展了“主元最大高斯消去法”。就是在主元所在的列中，尋找

32、到最大的元素，進行行與行之間的調(diào)換，并將該最大的元素作為主元，保證主元不為0。實驗五 松弛迭代法求解化工中的線性方程組一、實驗目的1） 掌握松弛迭代法的基本算法及和緊湊迭代的細微區(qū)別2） 編寫松弛迭代法求救線性方程組的VB代碼，注意學習從文件讀取數(shù)據(jù)3） 通過實例的程序進行驗證，并觀察松弛迭代因子對結果的影響二、運行環(huán)境1） Microsoft WIndowse XP2） VB6.0三、實驗原理略四、實驗程序代碼Private Sub Command1_Click()Dim n As IntegerDim i, j, ff, t, k, l, hDim st As StringDim a()

33、As SingleDim y() As SingleDim b() As SingleDim g() As SingleDim x1() As SingleDim x2() As SingleDim jk() As IntegerDim ch As StringDim sp As Stringch = Chr(13) + Chr(10)sp = Space(5)CommonDialog1.CancelError = True'on error goto errhandlerCommonDialog1.Filter = "數(shù)據(jù)文件（*.txt）|*.txt|拉圖文件（*.bmp

34、）|*.bmp|AllFiles(*.*)|*.*" '文件過濾CommonDialog1.FilterIndex = 0CommonDialog1.DialogTitle = "加載增廣矩陣數(shù)據(jù)文件"CommonDialog1.ShowOpen'*數(shù)據(jù)文件的行數(shù)就是方程的個數(shù)Open CommonDialog1.FileName For Input As #1Do While Not EOF(1) Line Input #1, st n = n + 1LoopClose #1'*數(shù)據(jù)文件的行數(shù)就是方程的個數(shù)ReDim a(1 To n,

35、1 To n) As SingleReDim b(1 To n, 1 To n) As SingleReDim x1(1 To n) As SingleReDim x2(1 To n) As SingleReDim g(1 To n) As SingleReDim y(1 To n) As Single'*讀數(shù)據(jù)Open CommonDialog1.FileName For Input As #1For i = 1 To n For j = 1 To n Input #1, a(i, j) '方程等號左端數(shù)據(jù) Next j Input #1, y(i) '方程等號右端數(shù)

36、據(jù)Next iClose #q'*讀數(shù)據(jù)st = "松弛迭代法解線性方程" + chst = st + Space(5) + "數(shù)據(jù)來源于" + CommonDialog1.FileName + ch + chst = st + "增廣矩陣如下（對二甲苯-間二甲苯-鄰二甲苯-乙苯-（混合物）：" + ch + chst = st + Space(5) + "第一行為12.5nm波長處摩爾吸收系數(shù)-混合物吸收" + chst = st + Space(5) + "第一行為13.0nm波長處摩爾吸收系

37、數(shù)-混合物吸收" + chst = st + Space(5) + "第一行為13.4nm波長處摩爾吸收系數(shù)-混合物吸收" + chst = st + Space(5) + "第一行為14.3nm波長處摩爾吸收系數(shù)-混合物吸收" + ch + ch'*輸出原始數(shù)據(jù) For i = 1 To n For j = 1 To n If a(i, j) >= 0 Then st = st + Space(5) + Format(a(i, j), "0.00000") Else st = st + Space(4) +

38、Format(a(i, j), "0.00000") End If Next j If y(i) >= 0 Then st = st + Space(5) + Format(y(i), "0.00000") + ch Else st = st + Space(4) + Format(y(i), "0.00000") + ch End If Next i'*輸出原始數(shù)據(jù)'-For i = 1 To n x1(i) = 0 x2(i) = 0Next i'for i = 1 to n' for j =

39、 1 to n' a(i,j) = InputBox("a("&i&","&j&")")' Print a(i,j),' next j'y(i) = InputBox("y("&i&")")'print" ",y(i)'Next i'產(chǎn)生迭代矩陣For i = 1 To n g(i) = y(i) / a(i, i) For j = 1 To n If j = i Then

40、 b(i, j) = 0 Else b(i, j) = -a(i, j) / a(i, i) End If Next j Next i e = InputBox("輸入松弛因子") '開始松弛迭代Do If k >= 1 Then For i = 1 To n x1(i) = x2(i) Next i End If For i = 1 To n s = g(i) For j = 1 To n s = s + b(i, j) * x2(j) Next j x2(i) = (1 - e) * x1(i) + e * s '注意 Next i eer = 0

41、 For i = 1 To n eer = cer + Abs(x1(i) - x2(i) '計算誤差 Next i k = k + 1 '累計次數(shù)Loop While (k < 100 And eer >= 0.001) Print kst = st + ch + "方程組的解為：" + ch + chFor i = 1 To n st = st + "x(" + Str(i) + ")=" + Format(x2(i), "0.00000") + ch Next ist = st +

42、 ch + "迭代次數(shù)為：" + Str(k) + ch 'format(k,"000")st = st + ch + "松弛因子為：" + Format(e, "0.0000") + chst = st + ch + "誤差為：" + Format(eer, "0.000000") + chst = st + chst = st + "*時間：" + Str(Time) + Space(3) + "日期：" + Str(Dat

43、e) + chText1.Text = ""Text1.Text = st End Sub五、實驗結果截圖六、 實驗后思考。松弛迭代法是數(shù)值計算中解線性代數(shù)方程組的一類迭代法。逐次超松弛迭代過程中，已知迭代方程及其系數(shù)矩陣，對任意的初始值，確定超松弛因子，用迭代矩陣來進行計算確定譜半徑，然后其絕對值小于一解出來超松弛因子。而緊湊迭代是當松弛因子為1的時候，叫做緊湊迭代。兩者的區(qū)別在于松弛因子的不同。實驗六 龍格庫塔法求解化工過程中的常微分方程一、實驗目的1） 掌握龍格庫塔法的基本原理2） 編寫龍格庫塔法解決常微分方程的VB程序3） 通過實例的程序進行調(diào)試和驗證，并觀察初值對

44、計算過程及結果的影響4） 掌握VB繪制二維曲線圖的方法和繪圖參數(shù)的設置二、 運行環(huán)境1） Microsoft WindowsXP2） VB6.0三 、實驗原理略四、 實驗程序截圖Private Sub Command1_Click()Const eps = 0.00001Dim t() As SingleDim x() As SingleDim y() As SingleDim z() As SingleDim J1, J2 As SingleDim K1, K2, K3, K4 As SingleDim Q1, Q2, Q3, Q4 As SingleDim S1, S2, S3, S4 A

45、s SingleDim h As SingleDim i As IntegerDim n As Integerh = 0.01J1 = 1J2 = 1.1n = Int(10 / h)ReDim t(n + 1), x(n + 1), y(n + 1), z(n + 1) As Singlet(0) = 0x(0) = 0y(0) = 0'z(0) = 0For i = 0 To n - 1K1 = -J1 * x(i)Q1 = J1 * x(i) - J2 * y(i)S1 = h * (J2 * y(i)K2 = -J1 * (x(i) + h * K1 / 2)Q2 = J1 *

46、 (x(i) + h * K1 / 2) - J2 * (y(i) + h * Q1 / 2)S2 = h * (J2 * (y(i) + Q1 / 2)K3 = -J1 * (x(i) + h * K2 / 2)Q3 = J1 * (x(i) + h * K2 / 2) - J2 * (y(i) + h * Q2 / 2)S3 = h * (J2 * (y(i) + Q2 / 2)K4 = -J1 * (x(i) + h * K3)Q4 = J1 * (x(i) + h * K3) - J2 * (y(i) + h * Q3)S4 = h * (J2 * (y(i) + Q3)x(i + 1

47、) = x(i) + h * (K1 + 2 * K2 + 2 * K3 + K4) / 6 '計算A物質(zhì)的濃度y(i + 1) = y(i) + h * (Q1 + 2 * Q2 + 2 * Q3 + Q4) / 6z(i + 1) = z(i) + (S1 + 2 * Q2 + 2 * Q3 + Q4) / 6z(i) = x(0) - x(i) - y(i)t(i + 1) = t(i) + h '計算反應時間tNext iDim axisname1 As String, axisname2 As Stringaxisname1 = "t"axisna

48、me2 = ""xy_axis picture1, t(), x(), axisname1, axisname2For i = 0 To n - 1 picture1.PSet (t(i), x(i)Next i'xy_axis picture1,t(),y(),axisname1,axisname2For i = 0 To n - 1 Picture.PSet (t(i), y(i) Next i 'xy_axis picture1,t(),z(),axisname1,axisname2 For i = 0 To n - 1 Picture.PSet (t

49、(i), z(i)Next iEnd Sub模塊代碼Sub xy_axis(pic As PictureBox, x1() As Single, y1() As Single, axisname1 As String, axisname2 As String)On Error GoTo problemx:Dim maxnumber As Single, minnumber As SingleDim leftx As Single, topy As SingleDim rightx As Single, bottomy As SingleDim n As Integern = UBound(x1)pic.Font