Price fraction
You are encouraged to solve this task according to the task description, using any language you may know.
A friend of mine runs a Pharmacy. He has a specialised rounding function in his Dispensary application which receives a decimal value of currency and forces it to a standard value, This value is regulated by a government department.
Task: Given a floating point value between 0.00 and 1.00, rescale according to the following table:
>= 0 < 0.06 := 0.1
>= 0.06 < 0.11 := 0.18
>= 0.11 < 0.16 := 0.26
>= 0.16 < 0.21 := 0.32
>= 0.21 < 0.26 := 0.38
>= 0.26 < 0.31 := 0.44
>= 0.31 < 0.36 := 0.5
>= 0.36 < 0.41 := 0.54
>= 0.41 < 0.46 := 0.58
>= 0.46 < 0.51 := 0.62
>= 0.51 < 0.56 := 0.66
>= 0.56 < 0.61 := 0.7
>= 0.61 < 0.66 := 0.74
>= 0.66 < 0.71 := 0.78
>= 0.71 < 0.76 := 0.82
>= 0.76 < 0.81 := 0.86
>= 0.81 < 0.86 := 0.9
>= 0.86 < 0.91 := 0.94
>= 0.91 < 0.96 := 0.98
>= 0.96 < 1.01 := 1.00
Clipper
<lang dbase>FUNCTION PriceFraction( npQuantDispensed )
LOCAL aPriceFraction := { {0,.06,.1},;
{.06,.11,.18}, ;
{.11,.16,.26}, ;
{.16,.21,.32}, ;
{.21,.26,.38}, ;
{.26,.31,.44}, ;
{.31,.36,.5}, ;
{.36,.41,.54}, ;
{.41,.46,.58}, ;
{.46,.51,.62}, ;
{.51,.56,.66}, ;
{.56,.61,.7}, ;
{.61,.66,.74}, ;
{.66,.71,.78}, ;
{.71,.76,.82}, ;
{.76,.81,.86}, ;
{.81,.86,.9}, ;
{.86,.91,.94}, ;
{.91,.96,.98} }
LOCAL nResult
LOCAL nScan
IF npQuantDispensed = 0
nResult = 0
ELSEIF npQuantDispensed >= .96
nResult = 1
ELSE
nScan := ASCAN( aPriceFraction, ;
{ |aItem| npQuantDispensed >= aItem[ 1 ] .AND.;
npQuantDispensed < aItem[ 2 ] } )
nResult := aPriceFraction[ nScan ][ 3 ]
END IF
RETURN nResult</lang>