Kindly solve this complex code for me in python please mentioned in the discription experts . | Sololearn: Learn to code for FREE!

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

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.

12/2/2021 8:59:54 PM

SAfiullaH KhoKhar

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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

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

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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.

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Thanks brother really appreciate your help

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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)

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Why ?

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Let me show you how I made

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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")

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This code has a problem

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Working fine for me

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It's running

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Did you try with 2021 ?

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Yes working 2021 is not a leap year

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

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It's working paste it on your compiler.

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

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Can't go above 1600 in this.

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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