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用vb证明一个数组可以经过换行和换列的方法等于另一个数组 已有1人参与
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用vb证明一个数组可以经过换行和换列的方法等于另一个数组，如下文，vb运行就卡住了，是运算方法太复杂了，还是我写的程序有错？求你一个能解决问题的方法 Dim zj(11, 12) As Integer For i = 0 To 11 For j = 0 To 12 zj(i, j) = 0 Next Next zj(1, 1) = 1 zj(1, 2) = 1 zj(1, 3) = 1 zj(1, 4) = 1 zj(1, 5) = 1 zj(1, 6) = 1 zj(1, 7) = 2 zj(1, 8) = 2 zj(1, 9) = 2 zj(1, 10) = 2 zj(1, 11) = 2 zj(1, 12) = 2 zj(2, 1) = 1 zj(2, 2) = 1 zj(2, 3) = 1 zj(2, 4) = 2 zj(2, 5) = 2 zj(2, 6) = 2 zj(2, 7) = 1 zj(2, 8) = 1 zj(2, 9) = 1 zj(2, 10) = 2 zj(2, 11) = 2 zj(2, 12) = 2 zj(3, 1) = 1 zj(3, 2) = 1 zj(3, 3) = 2 zj(3, 4) = 1 zj(3, 5) = 2 zj(3, 6) = 2 zj(3, 7) = 2 zj(3, 8) = 2 zj(3, 9) = 1 zj(3, 10) = 2 zj(3, 11) = 1 zj(3, 12) = 1 zj(4, 1) = 1 zj(4, 2) = 1 zj(4, 3) = 2 zj(4, 4) = 2 zj(4, 5) = 1 zj(4, 6) = 2 zj(4, 7) = 2 zj(4, 8) = 1 zj(4, 9) = 2 zj(4, 10) = 1 zj(4, 11) = 2 zj(4, 12) = 1 zj(5, 1) = 1 zj(5, 2) = 1 zj(5, 3) = 2 zj(5, 4) = 2 zj(5, 5) = 2 zj(5, 6) = 1 zj(5, 7) = 1 zj(5, 8) = 2 zj(5, 9) = 2 zj(5, 10) = 1 zj(5, 11) = 1 zj(5, 12) = 2 zj(6, 1) = 1 zj(6, 2) = 2 zj(6, 3) = 1 zj(6, 4) = 1 zj(6, 5) = 2 zj(6, 6) = 2 zj(6, 7) = 1 zj(6, 8) = 2 zj(6, 9) = 2 zj(6, 10) = 1 zj(6, 11) = 2 zj(6, 12) = 1 zj(7, 1) = 1 zj(7, 2) = 2 zj(7, 3) = 1 zj(7, 4) = 2 zj(7, 5) = 1 zj(7, 6) = 2 zj(7, 7) = 2 zj(7, 8) = 2 zj(7, 9) = 1 zj(7, 10) = 1 zj(7, 11) = 1 zj(7, 12) = 2 zj(8, 1) = 1 zj(8, 2) = 2 zj(8, 3) = 1 zj(8, 4) = 2 zj(8, 5) = 2 zj(8, 6) = 1 zj(8, 7) = 2 zj(8, 8) = 1 zj(8, 9) = 2 zj(8, 10) = 2 zj(8, 11) = 1 zj(8, 12) = 1 zj(9, 1) = 1 zj(9, 2) = 2 zj(9, 3) = 2 zj(9, 4) = 1 zj(9, 5) = 1 zj(9, 6) = 2 zj(9, 7) = 1 zj(9, 8) = 1 zj(9, 9) = 2 zj(9, 10) = 2 zj(9, 11) = 1 zj(9, 12) = 2 zj(10, 1) = 1 zj(10, 2) = 2 zj(10, 3) = 2 zj(10, 4) = 1 zj(10, 5) = 2 zj(10, 6) = 1 zj(10, 7) = 2 zj(10, 8) = 1 zj(10, 9) = 1 zj(10, 10) = 1 zj(10, 11) = 2 zj(10, 12) = 2 zj(11, 1) = 1 zj(11, 2) = 2 zj(11, 3) = 2 zj(11, 4) = 2 zj(11, 5) = 1 zj(11, 6) = 1 zj(11, 7) = 1 zj(11, 8) = 2 zj(11, 9) = 1 zj(11, 10) = 2 zj(11, 11) = 2 zj(11, 12) = 1 Dim pb(11, 12) As Integer For i = 0 To 11 For j = 0 To 12 pb(i, j) = 0 Next Next pb(1, 1) = 2 pb(1, 2) = 2 pb(1, 3) = 1 pb(1, 4) = 2 pb(1, 5) = 2 pb(1, 6) = 2 pb(1, 7) = 1 pb(1, 8) = 1 pb(1, 9) = 1 pb(1, 10) = 2 pb(1, 11) = 1 pb(1, 12) = 1 pb(2, 1) = 1 pb(2, 2) = 2 pb(2, 3) = 2 pb(2, 4) = 1 pb(2, 5) = 2 pb(2, 6) = 2 pb(2, 7) = 2 pb(2, 8) = 1 pb(2, 9) = 1 pb(2, 10) = 1 pb(2, 11) = 2 pb(2, 12) = 1 pb(3, 1) = 2 pb(3, 2) = 1 pb(3, 3) = 2 pb(3, 4) = 2 pb(3, 5) = 1 pb(3, 6) = 2 pb(3, 7) = 2 pb(3, 8) = 2 pb(3, 9) = 1 pb(3, 10) = 1 pb(3, 11) = 1 pb(3, 12) = 1 pb(4, 1) = 1 pb(4, 2) = 2 pb(4, 3) = 1 pb(4, 4) = 2 pb(4, 5) = 2 pb(4, 6) = 1 pb(4, 7) = 2 pb(4, 8) = 2 pb(4, 9) = 2 pb(4, 10) = 1 pb(4, 11) = 1 pb(4, 12) = 1 pb(5, 1) = 1 pb(5, 2) = 1 pb(5, 3) = 2 pb(5, 4) = 1 pb(5, 5) = 2 pb(5, 6) = 2 pb(5, 7) = 1 pb(5, 8) = 2 pb(5, 9) = 2 pb(5, 10) = 2 pb(5, 11) = 1 pb(5, 12) = 1 pb(6, 1) = 1 pb(6, 2) = 1 pb(6, 3) = 1 pb(6, 4) = 2 pb(6, 5) = 1 pb(6, 6) = 2 pb(6, 7) = 2 pb(6, 8) = 1 pb(6, 9) = 2 pb(6, 10) = 2 pb(6, 11) = 2 pb(6, 12) = 1 pb(7, 1) = 2 pb(7, 2) = 1 pb(7, 3) = 1 pb(7, 4) = 1 pb(7, 5) = 2 pb(7, 6) = 1 pb(7, 7) = 2 pb(7, 8) = 2 pb(7, 9) = 1 pb(7, 10) = 2 pb(7, 11) = 2 pb(7, 12) = 1 pb(8, 1) = 2 pb(8, 2) = 2 pb(8, 3) = 1 pb(8, 4) = 1 pb(8, 5) = 1 pb(8, 6) = 2 pb(8, 7) = 1 pb(8, 8) = 2 pb(8, 9) = 2 pb(8, 10) = 1 pb(8, 11) = 2 pb(8, 12) = 1 pb(9, 1) = 2 pb(9, 2) = 2 pb(9, 3) = 2 pb(9, 4) = 1 pb(9, 5) = 1 pb(9, 6) = 1 pb(9, 7) = 2 pb(9, 8) = 1 pb(9, 9) = 2 pb(9, 10) = 2 pb(9, 11) = 1 pb(9, 12) = 1 pb(10, 1) = 1 pb(10, 2) = 2 pb(10, 3) = 2 pb(10, 4) = 2 pb(10, 5) = 1 pb(10, 6) = 1 pb(10, 7) = 1 pb(10, 8) = 2 pb(10, 9) = 1 pb(10, 10) = 2 pb(10, 11) = 2 pb(10, 12) = 1 pb(11, 1) = 2 pb(11, 2) = 1 pb(11, 3) = 2 pb(11, 4) = 2 pb(11, 5) = 2 pb(11, 6) = 1 pb(11, 7) = 1 pb(11, 8) = 1 pb(11, 9) = 2 pb(11, 10) = 1 pb(11, 11) = 2 pb(11, 12) = 1 Dim hl(11, 12), hh(11, 12) '定义过程数组：换行和换列，分别采用循环穷举所有可能的排列，在每一个换列的数组穷举换行的所有结果 For i = 0 To 11 For j = 0 To 12 hl(i, j) = 0 hh(i, j) = 0 Next Next For i1 = 1 To 11 For i2 = 1 To 11 For i3 = 1 To 11 For i4 = 1 To 11 For i5 = 1 To 11 For i6 = 1 To 11 For i7 = 1 To 11 For i8 = 1 To 11 For i9 = 1 To 11 For i10 = 1 To 11 For i11 = 1 To 11 For j1 = 1 To 11 For j2 = 1 To 12 If j1 = 1 Then hl(j1, j2) = pb(i1, j2) If j1 = 2 Then hl(j1, j2) = pb(i2, j2) If j1 = 3 Then hl(j1, j2) = pb(i3, j2) If j1 = 4 Then hl(j1, j2) = pb(i4, j2) If j1 = 5 Then hl(j1, j2) = pb(i5, j2) If j1 = 6 Then hl(j1, j2) = pb(i6, j2) If j1 = 7 Then hl(j1, j2) = pb(i7, j2) If j1 = 8 Then hl(j1, j2) = pb(i8, j2) If j1 = 9 Then hl(j1, j2) = pb(i9, j2) If j1 = 10 Then hl(j1, j2) = pb(i10, j2) If j1 = 11 Then hl(j1, j2) = pb(i11, j2) Next Next For k1 = 1 To 12 For k2 = 1 To 12 For k3 = 1 To 12 For k4 = 1 To 12 For k5 = 1 To 12 For k6 = 1 To 12 For k7 = 1 To 12 For k8 = 1 To 12 For k9 = 1 To 12 For k10 = 1 To 12 For k11 = 1 To 12 For k12 = 1 To 12 For l2 = 1 To 12 For l1 = 1 To 11 If l2 = 1 Then hh(l1, l2) = hl(l1, k1) If l2 = 2 Then hh(l1, l2) = hl(l1, k2) If l2 = 3 Then hh(l1, l2) = hl(l1, k3) If l2 = 4 Then hh(l1, l2) = hl(l1, k4) If l2 = 5 Then hh(l1, l2) = hl(l1, k5) If l2 = 6 Then hh(l1, l2) = hl(l1, k6) If l2 = 7 Then hh(l1, l2) = hl(l1, k7) If l2 = 8 Then hh(l1, l2) = hl(l1, k8) If l2 = 9 Then hh(l1, l2) = hl(l1, k9) If l2 = 10 Then hh(l1, l2) = hl(l1, k10) If l2 = 11 Then hh(l1, l2) = hl(l1, k11) If l2 = 12 Then hh(l1, l2) = hl(l1, k12) Next Next Dim test2 = 0 For x = 1 To 11 For y = 1 To 12 If hl(x, y) = zj(x, y) Then test2 = test2 + 1 End If Next Next If test2 = 11 * 12 Then TextBox1.Text = "成功"+ TextBox1.Text End If Next Next Next Next Next Next Next Next Next Next Next Next Next Next Next Next Next Next Next Next Next Next Next |
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deephill
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Micah1011: 金币+200, ★★★★★最佳答案, 感谢回答，请问我用visual stidio2013能让他多线程吗？需要如何配置 2015-03-21 22:17:27
感谢参与，应助指数 +1
Micah1011: 金币+200, ★★★★★最佳答案, 感谢回答，请问我用visual stidio2013能让他多线程吗？需要如何配置 2015-03-21 22:17:27
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你这个程序一定会卡住的，循环套循环，层数多了，就卡住。 1 要是你怕程序不正确，需要调试。在程序的关键位置 加上stop ，暂停时就会输出结果，可以检查结果，这样不用等到程序运行完就强制结束程序。 2 vb的程序老版本不能开多线程，运算速度不会快。只能等，让程序算，看似停止响应，但程序在计算着了，时间到了，自然出结果。我估计你这个程序可能要等20分钟。 3 有人为了不让程序假死，可以加上sleep(1) 语句。 |
2楼2015-03-21 13:16:23
3楼2015-03-21 22:18:06
deephill
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4楼2015-03-21 23:43:50
