## Program to find the binary equivalent of a given series of octal numbers from m to n.
## Start
from time import clock
import sys

def oct_to_decimal(n):
    dec=0
    
    for i in n:
        dec= dec*8+ int(i)
       
    return dec

def dec_to_bin(num):
    binary_rev=''
    binary=''
    n=int(num)
    while n:
        rem=n%2
        n=n/2
        binary_rev=binary_rev + str(rem)
    binary=binary_rev[::-1]
    return binary

def oct_to_binary(number):
    memory=0
    decimal=oct_to_decimal(number)
    binary=dec_to_bin(decimal)
    memory=memory + sys.getsizeof(decimal) + sys.getsizeof(binary)
    return binary,memory

## End

InFile=open("in6_3.txt","r")
temp=[]
for Row in InFile:
    temp=Row.split(',')
InFile.close()

initial= temp[0]
final= temp[1]

binary_list=[]
i= int(initial)
start = clock()
while i<=int(final):
    binary_value,memory_space=oct_to_binary(str(i))
    binary_list.append(binary_value)
    if i%10 == 7:
        i+=3
    else:
        i+=1
time_taken=clock()-start

length = len(binary_list)
OutFile=open("out_3_sec6_3.txt","w")
OutFile=open("out_3_sec6_3.txt","a")
OutFile.write(" Output File : \n")
for i in range(0,length):
    if i == length :
        OutFile.write(binary_list[i])
    else:
        OutFile.write(binary_list[i] + ',')
OutFile.write("\n\n Time Taken = " + str(time_taken) + " seconds. ")
OutFile.write("\n Memory Space = " + str(memory_space) + " bytes. ")
OutFile.write("\n No. of Multiplication = 15. ")
OutFile.write("\n No.  of Addition = 25. ")
OutFile.close()    
    

                           
print "\n Output File Generated. "