a = 218500000000
s = 6
f = 2
k = 49
d = k + f + s
r = a
i = 0
while (r >= d):
r = r - d
#print ( r = ,r)
i = i+1
#print ( i = ,i)
print (i)
I think it does what I expect it to, but its way too slow to calculate such a large number, I waited 5 mins for i to print (while python used 100% cpu to calculate..), but it didn t. Is there a more efficient way of rewriting this piece of code so I can see how many iterations (i) it takes to complete?
Many thanks
r = (a % d)
i = (a / d)
Use the modulo and division operators.
There is also a divmod function to calculate both together:
i, r = divmod(a,d)
You can use i = a/d. :D
Isn t a division what you re looking for?
try 3833333333.3333333333333333333333. AKA r / d.
Looks like you are doing truncating division to me. That is, you want to find out how many times d goes into a without knowing the remainder.
a = 218500000000
s = 6
f = 2
k = 49
d = k + f + s
i = a // d
print (i)
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