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вернуться в форумA little hint Amount of rectangles is equal to the number of divisiors of tiles count wich are no more than Square_Root(tiles count). So the part of code solving this problem may be the following:     For L:=1 To 10000-K Do Begin     DivisiorsCount:=1;     For J:=2 To Trunc(Sqrt(L)) Do If L Mod J=0 Then       Inc(DivisiorsCount);       If DivisiorsCount=M Then Begin       DivisiorsCount2:=1;       For J:=2 To Trunc(Sqrt(L+K)) Do If (L+K) Mod J=0 Then         Inc(DivisiorsCount2);         If DivisiorsCount2=N Then Begin         SoulutionExists:=True;         Break;       End;     End;   End;   You should output L+K if SolutionExists, 0 - otherwise.  |  
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