Kindly solve this complex code for me in python please mentioned in the discription experts .

Question

5. A year is a leap year if it is divisible by 4, except that years divisible by 100 are not leap
years unless they are also divisible by 400. Ask the user to enter a year, and, using the l/
operator, determine how many leap years there have been between 1600 and that year.

Answer 1

we don't have to reinvent the wheel everytime. Use calendar.isleap()

import calendar

year = int(input())

# counter
n=0

# starting year
start = 1600

#if the input year is less than the start year, reverse the values
if start > year:
start, year = year, start

for y in range(start, year+1):
if calendar.isleap(y):
n+=1

print(n)

https://code.sololearn.com/cgwjQJ4IzcFD/?ref=app

Answer 2

Brother couldn't tell in the output if the year is leap year or not.

Answer 3

Example:
import calendar

year=1600
print(calendar.isleap(year))

>>True if it is a leap year
>>False if not.

Your original question was only about counting the number of leap years. calendar.isleap() takes care of determining if the year is a leap year or not.

It is trivial to add that functionality.
<edit>
Ok, I have added the print function for declaring if the input is a leap year or not.

Answer 4

Thanks brother really appreciate your help

Answer 5

Does this is what you want ?
year = int(input())
leaps = 0
for i in range(min(1600, year), max(1600, year)) :
leaps += (1 if(not(i % 4) and (i % 100 or not(i % 400))) else 0)
print(leaps)

Answer 6

Nope

Answer 7

Why ?

Answer 8

Let me show you how I made

Answer 9

year = int(input("Enter year: "))

if year % 400 == 0 :
print(year, "is a Leap Year")
elif year % 100 == 0 :
print(year, "is not a Leap Year")
elif year % 4 == 0 :
print(year, "is a Leap Year")
else :
print(year, "is not a Leap Year")
print("Leap year between 1600 and that Year is :")
print("================================================")
for year in range(1600):

if year % 400 == 0 :
print(year, "==> is a Leap Year")
elif year % 100 == 0 :
print(year, "==> is not a Leap Year")
elif year % 4 == 0 :
print(year, "==> is a Leap Year")
else :
print(year, "==> is not a Leap Year")

Answer 10

This code has a problem

Answer 11

Working fine for me

Answer 12

It's running

Answer 13

Did you try with 2021 ?

Answer 14

Yes working 2021 is not a leap year

Answer 15

Of course it's running but try with 1610, the result should be 2

Answer 16

It's working paste it on your compiler.

Answer 17

Brother I have to determine whether the year is leap year or not that's the question the range is given 1600

Answer 18

Can't go above 1600 in this.

Answer 19

the first part of your code works fine,
but you also must determine how many leap years there have been between 1600 and that year