RosettaCodeData/Task/Chinese-remainder-theorem/PureBasic/chinese-remainder-theorem.basic
2023-07-01 13:44:08 -04:00

77 lines
1.6 KiB
Text

EnableExplicit
DisableDebugger
DataSection
LBL_n1:
Data.i 3,5,7
LBL_a1:
Data.i 2,3,2
LBL_n2:
Data.i 11,12,13
LBL_a2:
Data.i 10,4,12
LBL_n3:
Data.i 10,4,9
LBL_a3:
Data.i 11,22,19
EndDataSection
Procedure ErrorHdl()
Print(ErrorMessage())
Input()
EndProcedure
Macro PrintData(n,a)
Define Idx.i=0
Print("[")
While n+SizeOf(Integer)*Idx<a
Print("( ")
Print(Str(PeekI(a+SizeOf(Integer)*Idx)))
Print(" . ")
Print(Str(PeekI(n+SizeOf(Integer)*Idx)))
Print(" )")
Idx+1
Wend
Print(~"]\nx = ")
EndMacro
Procedure.i Produkt_n(n_Adr.i,a_Adr.i)
Define p.i=1
While n_Adr<a_Adr
p*PeekI(n_Adr)
n_Adr+SizeOf(Integer)
Wend
ProcedureReturn p
EndProcedure
Procedure.i Eval_x1(a.i,b.i)
Define b0.i=b, x0.i=0, x1.i=1, q.i, t.i
If b=1 : ProcedureReturn x1 : EndIf
While a>1
q=Int(a/b)
t=b : b=a%b : a=t
t=x0 : x0=x1-q*x0 : x1=t
Wend
If x1<0 : ProcedureReturn x1+b0 : EndIf
ProcedureReturn x1
EndProcedure
Procedure.i ChineseRem(n_Adr.i,a_Adr.i)
Define prod.i=Produkt_n(n_Adr,a_Adr), a.i, b.i, p.i, Idx.i=0, sum.i
While n_Adr+SizeOf(Integer)*Idx<a_Adr
b=PeekI(n_Adr+SizeOf(Integer)*Idx)
p=Int(prod/b) : a=p
sum+PeekI(a_Adr+SizeOf(Integer)*Idx)*Eval_x1(a,b)*p
Idx+1
Wend
ProcedureReturn sum%prod
EndProcedure
OnErrorCall(@ErrorHdl())
OpenConsole("Chinese remainder theorem")
PrintData(?LBL_n1,?LBL_a1)
PrintN(Str(ChineseRem(?LBL_n1,?LBL_a1)))
PrintData(?LBL_n2,?LBL_a2)
PrintN(Str(ChineseRem(?LBL_n2,?LBL_a2)))
PrintData(?LBL_n3,?LBL_a3)
PrintN(Str(ChineseRem(?LBL_n3,?LBL_a3)))
Input()