### Look what we did using python today.

During the class I was explaining the students why you need to keep on solving the programs, so that you can improve your speed. Speed and accuracy is what  managers expect from their developers in a professional setup.
I asked them can  we convert numbers to words i.e. spell the number. Cheque printing is the one area where we need this program. Even bill printing employs this logic.  Here is what we come up with. This program uses recursion and converts number to text Indian way. To learn python and other computer courses please visit nearest MICE center.

'''
Created on 23-Jun-2017
@author: Ravi
Created on Jun 23, 2017
@author: Manipal Institute of Computer education
'''

def numberToText(no):
ones = " ,one,two,three,four,five,six,seven,eight,nine,ten,eleven,tweleve,thirteen,fourteen,fifteen,sixteen,seventeen,eighteen,nineteen,twenty".split(',')
tens = "ten,twenty,thirty,fourty,fifty,sixty,seventy,eighty,ninety".split(',')
text = ""
if len(str(no))<=2:
if(no<20):
text = ones[no]
else:
text = tens[no//10-1] +" " + ones[(no %10)]
elif len(str(no))==3:
text = ones[no//100] +" hundred " + numberToText(no- ((no//100)* 100))
elif len(str(no))<=5:
text = numberToText(no//1000) +" thousand " + numberToText(no- ((no//1000)* 1000))
elif len(str(no))<=7:
text = numberToText(no//100000) +" lakh " + numberToText(no- ((no//100000)* 100000))
else:
text = numberToText(no//10000000) +" crores " + numberToText(no- ((no//10000000)* 10000000))
return text

def spellNumber(no):
# str(no) will result in  56.9 for 56.90 so we used the method which is given below.
strNo = "%.2f" %no
n = strNo.split(".")
rs = numberToText(int(n)).strip()
ps =""
if(len(n)>=2):
ps = numberToText(int(n)).strip()
rs = "" + ps+ " paise"  if(rs.strip()=="") else  (rs + " and " + ps+ " paise").strip()
return rs
print(spellNumber(0.67))
print(spellNumber(5858.099))
print(spellNumber(5083754857380.50))