Code Review Asked on January 2, 2022
Problem Statement
In a School, Chocolate bars have to be distributed to children waiting in a queue. Each Chocolate bar is rectangular in shape. Consider its side lengths are integer values.
The distribution procedure is as follows:-
If bar is not square in shape, then the largest possible square piece of Chocolate is broken and given to the first child in queue.
If bar is square in shape, then complete bar is given to the first child in queue.
Once a child receives his share of Chocolate, he leaves the queue. The remaining portion of the Chocolate bar is dealt in same fashion and the whole or a portion of it is given to the next child in the queue.School has got a carton of Chocolate bars to be distributed among the children all over the School. The Chocolate bars in the carton are of different sizes. A bar of length
iand breadthjis considered to be different from a bar of length j and breadthi. For everyisuch that $M le i le N$ and everyjsuch that $Ple j le Q$ (where M, N, P and Q are integers). Each Chocolate bar in carton is unique in length (i) and breath(j).Given the values of M, N, P and Q (where M, N values are the ranges for length of Chocolate and P, Q values are the ranges for breadth of the chocolate). Find the number of children who will receive Chocolate from the carton.
Input Specification:
M, N, P, Q are of integer type (M, N values are the ranges for length of chocolate bar. P, Q values are the ranges for breadth of chocolate bar).
Output Specification:
Number of children who will receive Cadbury bar from the carton.
M = 5, N = 6, P = 3, Q=4 Here,
ican be from 5 to 6 andjcan be from 3 to 4. So the four bars will be in carton of sizes 5×3, 5×4, 6×3, 6×4.First we choose a Cadbury bar of size 5×3 → first child would receive 3×3 portion ( remaining 2×3 portion ) → next child would receive 2×2 portion ( remaining 2×1 portion ) → now the remaining portion are 2 square pieces of (1×1), which can be given to 2 more children
So the Cadbury bar with the size of 5×3 can be distributed to 4 children.
Similarly we can find out number of children for rest of the combinations (i.e. 5×4, 6×3, 6×4) in the given range as follows
Can anyone suggest improvements?
m, n, p, q = raw_input().split(":")
ll = int
bb = int
count = 0
def cadbury(m, n, p, q):
leng = [m, n]
brd = [p, q]
for l in leng:
for b in brd:
call(l, b)
def call(l, b):
#print l,b
global count
area = l * b
if l > b:
bb = l
rem = bb - b
#print rem
rem_part1 = rem
#print rem_part1
rem_part2 = b
#print rem_part2
l = rem_part1
#print l
b = rem_part2
#print b
if l != b:
if (l==1) or (b==1):
count += 1
count += (l*b)
#print count
return
else:
count += 1
#print count
call(l, b)
if l==b:
count+=2
#print count
return
elif b > l:
ll = b
#print ll
rem = ll - l
#print rem
rem_part1 = rem
#print rem_part1
rem_part2 = l
#print rem_part2
l = rem_part1
#print l
b = rem_part2
#print b
if l != b:
if (l==1) or (b==1):
count += 1
count += (l*b)
#print count
return
else:
count += 1
#print count
call(l, b)
if l==b:
count+=2
#print count
return
cadbury(int(m), int(n), int(p), int(q))
print count
Make your code more readable so that you can more easily reason about its logic.
minimum_length instead of m)for length in [minimum_length, maximum_length] instead of for l in leng)Answered by Andrew Hoos on January 2, 2022
I made it a little shorter: The logic is also little more simple.
def TotalCount(M,N,P,Q):
count=0
for l in range(M,N+1):
for b in range(P,Q+1):
count +=CountPerChocolateBar(l,b)
return count
Your cadbury function stayed almost the same. But your version has a bug: if the difference between M and N, or P and Q is not one (in your example it was one), your code fails:
a = 7
b = 10
for element in [a,b]:
print (element)
result is 7, 10 . NOT the expected 7,8,9,10
def CountPerChocolateBar(l,b):
count = 0
while True:
longerr=max(l,b)
shorterr=min(l,b)
count+=1
diff=longer-shorter
if diff==0:
return count
else :
l=min(l,b)
b=diff
Your call function can be simplified a little: you can handle l>b ,b>l without branches, because it doesn't matter which one is bigger. You count the difference, increase the count variable, and based on the difference you return the count number or calculate the new b and l values. When the difference is 0, the while loop ends with the return statement
while True:
numbers=raw_input("Number: ")
M=int(numbers.split()[0])
N=int(numbers.split()[1])
P=int(numbers.split()[2])
Q=int(numbers.split()[3])
tc=TotalCount(M,N,P,Q)
print (tc)
You can test my solution with this little loop. :)
Answered by user3598726 on January 2, 2022
ll = b
rem_part1 = rem
rem_part2 = l
l = rem_part1
Why do you continually assign x = y? Why don't you just keep each value in a single variable?
ll = int
bb = int
Also to be underlined ll that is LETTER L LETTER L looks really similar to 11 in some fonts that is NUMBER ONE NUMBER ONE especially at small font-sizes making it even more confusing to read the code.
rem = bb - b
This name is misleading rem stands for remainder, that is the result of division: from Wikipedia
In arithmetic, the remainder is the integer "left over" after dividing one integer by another to produce an integer quotient (integer division).
count global?The count variable should be defined inside the function and returned out. While it does not really matter in such a small program, avoiding global variables is a good habit.
Answered by Caridorc on January 2, 2022
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