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Mit ír ki a program

a(1)= 5, a(2)= 1, a(3)= 10, a(4)= 3. i, m : egész a(1..4) : egész. i = 1. hamis. kiír( a(4), m). i<= 3. v ége. igaz. igaz. a(i) > a(i+1). m= a(i) a(i)= a(i+1) a(i+1)= m. hamis. kiír( a(i), m) i = i+1. Mit ír ki a program. a(1)= 5, a(2)= 1, a(3)= 10, a(4)= 3.

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Mit ír ki a program

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  1. a(1)= 5, a(2)= 1, a(3)= 10, a(4)= 3 i, m : egész a(1..4) : egész i = 1 hamis kiír( a(4), m) i<= 3 vége igaz igaz a(i) > a(i+1) m= a(i) a(i)= a(i+1) a(i+1)= m hamis kiír( a(i), m) i = i+1 Mit ír ki a program

  2. a(1)= 5, a(2)= 1, a(3)= 10, a(4)= 3 i = 1 a(1)= 5, a(2)= 1, a(3)= 10, a(4)= 3 i, m : egész a(1..4) : egész igaz a(i) > a(i+1) m= a(i) a(i)= a(i+1) a(i+1)= m i = 1 hamis hamis kiír( a(4), m) i<= 3 vége igaz igaz kiír( a(i), m) i = i+1 a(i) > a(i+1) m= a(i) a(i)= a(i+1) a(i+1)= m hamis igaz kiír( a(4), m) i> 3 hamis kiír( a(i), m) i = i+1 vége A blokk-diagram átalakítása

  3. a(1)= 5, a(2)= 1, a(3)= 10, a(4)= 3 i = 1 igaz a(i) > a(i+1) m= a(i) a(i)= a(i+1) a(i+1)= m hamis kiír( a(i), m) i = i+1 igaz kiír( a(4), m) i> 3 hamis vége Program megírása Sub mit_ir_ki() End Sub Dim a%(5), i%, m% a(1)= 5: a(2) = 1 a(3) = 10: a(4) = 3 i = 1 Do Loop Until a > 3 If a(i) > a(i + 1) Then Else End if m = a(i) a(i) = a(i + 1) a(i + 1) = m Cells( i , 1) = a(i): Cells( i , 2) = m i = i + 1 Cells(i,1) = a(i): Cells(i,2) = m

  4. Tömb feltöltés (eljárás nélkül) Sub feltolt() Dim A%(50, 100), m%, n% m = InputBox("sor száma = m = ? ") n = InputBox("oszlop száma = n = ?") For i = 1 To m Forj = 1 To n A(i, j) = i - j Cells(i, j) = A(i, j) Next j Nexti End Sub pl. m = 5 n = 7 esetén

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