how to count the number divisible by N in given range (A to B) for multiple cases (C) in python-CodePudding

So basically, there are multiple cases that need to be solve by counting the number that divisible by N in given range A to B.

for example, there are 2 cases.

case 1 has A=1, B=10, N=3

case 2 has A=8, B=20, N=4

but on one condition that the input must be like this:

2        #<-- number of cases
1        #<-- A 
10       #<-- B 
3        #<-- N
8        #<-- A
20       #<-- B
4        #<-- N

the output will be like:

Case 1: 3 #because between 1 until 10, only 3,6,9 that divisible by 3
Case 2: 4 #because between 8 until 20, only 8,12,16,20 that divisible by 4

I have this function for reference:

def CountDiv(A,B,N):
    count = 0
    
    for i in range(A, B 1):
        if i % N == 0:
            count = count   1
    
    return count

My explanation is bad but I just don't know how to make it more clearer. I hope someone will understand and enlightened me about this problem. Thank you

CodePudding user response：

You don't need to loop over all values. You can only generate the multiple from the smallest starting point ((A N-1)//N*N):

def ndiv(A, B, N):
    return len(list((A N-1)//N*N, B 1, N)))

Even better, calculate directly the number of values using:

def ndiv(A, B, N):
    if B<A:
        return 0
    return (B N-(A N-1)//N*N)//N

example:

>>> ndiv(8,20,4)
4

>>> ndiv(1,10,3)
3

>>> ndiv(1,1,3)
0

CodePudding user response：

I'm assuming that you are asking how to properly slice your input list. Assuming you have your input sequence in a list, this should work

seq = [ 2,        #<-- number of cases
        1,        #<-- A 
        10,       #<-- B 
        3,        #<-- N
        8,        #<-- A
        20,       #<-- B
        4,        #<-- N
]


def slices(lst, n):

    number_of_cases = lst.pop(0)
    
    for i in range(0, n * number_of_cases, n):
        yield lst[i:i   n]

def CountDiv(A,B,N):
    
    count = 0
    for i in range(A, B 1):
        if i % N == 0:
            count = count   1
    
    return count

print([CountDiv(*sub) for sub in [*slices(seq, n=3)]])
# [3, 4]

If you want the exact output you described, you can do this

for idx, sub in enumerate([*slices(seq, n=3)]):
    print(f"Case {idx}: {CountDiv(*sub)}")

# Case 0: 3
# Case 1: 4

You should also combine @mozway's and my answer, like so.

def slices(lst, n):
   
    number_of_cases = lst.pop(0)
     
    for i in range(0, n * number_of_cases, n):
        yield lst[i:i   n]

def ndiv(A, B, N):
    return (B N-(A N-1)//N*N)//N if B>=A else 0

for idx, sub in enumerate([*slices(seq, n=3)]):
    print(f"Case {idx}: {ndiv(*sub)}")