A friend of mine runs a pharmacy.   He has a specialized function in his Dispensary application which receives a decimal value of currency and replaces 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:

>=  0.00  <  0.06  :=  0.10
>=  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  :=  0.50
>=  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  :=  0.70
>=  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  :=  0.90
>=  0.86  <  0.91  :=  0.94
>=  0.91  <  0.96  :=  0.98
>=  0.96  <  1.01  :=  1.00



 
#lang racket
 
(define table
  '([0 #f]
    [0.06 0.10] [0.11 0.18] [0.16 0.26] [0.21 0.32] [0.26 0.38] [0.31 0.44]
    [0.36 0.50] [0.41 0.54] [0.46 0.58] [0.51 0.62] [0.56 0.66] [0.61 0.70]
    [0.66 0.74] [0.71 0.78] [0.76 0.82] [0.81 0.86] [0.86 0.90] [0.91 0.94]
    [0.96 0.98] [1.01 1.00])
 
  ;; returns #f for negatives or values >= 1.01
(define (convert x) (for/or ([c table]) (and (< x (car c)) (cadr c))))
 