Fairshare between two and more: Difference between revisions
Fairshare between two and more en FreeBASIC
(Added Quackery.) |
(Fairshare between two and more en FreeBASIC) |
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Line 602:
11 -> { 0 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 0 2 3 4 }
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=={{header|FreeBASIC}}==
{{trans|Visual Basic .NET}}
<lang freebasic>
Function Turn(mibase As Integer, n As Integer) As Integer
Dim As Integer sum = 0
While n <> 0
Dim As Integer re = n Mod mibase
n \= mibase
sum += re
Wend
Return sum Mod mibase
End Function
Sub Fairshare(mibase As Integer, count As Integer)
Print Using "mibase ##:"; mibase;
For i As Integer = 1 To count
Dim As Integer t = Turn(mibase, i - 1)
Print Using " ##"; t;
Next i
Print
End Sub
Sub TurnCount(mibase As Integer, count As Integer)
Dim As Integer cnt(mibase), i
For i = 1 To mibase
cnt(i - 1) = 0
Next i
For i = 1 To count
Dim As Integer t = Turn(mibase, i - 1)
cnt(t) += 1
Next i
Dim As Integer minTurn = 4294967295 'MaxValue of uLong
Dim As Integer maxTurn = 0 'MinValue of uLong
Dim As Integer portion = 0
For i As Integer = 1 To mibase
Dim As Integer num = cnt(i - 1)
If num > 0 Then portion += 1
If num < minTurn Then minTurn = num
If num > maxTurn Then maxTurn = num
Next i
Print Using " With ##### people: "; mibase;
If 0 = minTurn Then
Print Using "Only & have a turn"; portion
Elseif minTurn = maxTurn Then
Print minTurn
Else
Print Using "& or &"; minTurn; maxTurn
End If
End Sub
Fairshare(2, 25)
Fairshare(3, 25)
Fairshare(5, 25)
Fairshare(11, 25)
Print "How many times does each get a turn in 50000 iterations?"
TurnCount(191, 50000)
TurnCount(1377, 50000)
TurnCount(49999, 50000)
TurnCount(50000, 50000)
TurnCount(50001, 50000)
Sleep
</lang>
{{out}}
<pre>
Igual que la entrada de Visual Basic .NET.
</pre>
=={{header|Go}}==
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