September 2017 Update

Task/Chinese-remainder-theorem/PureBasic/chinese-remainder-theorem.purebasic

@@ -0,0 +1,77 @@
+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()