Python program to check Leap Year; Program Explanation; How to determine a leap year? So, write codes to check leap year program in python language. We saw two methods to write a leap year program in python. 2018 is not a Leap Year 2006 is not a Leap Year 2020 is a Leap Year 2000 is a Leap Year Python Code We will use nested if…else to solve this problem. AND (&&) Year should not be evenly divisible by 100 (year % 100 != 0) 3. In this Python Program we have to find the year is a leap year or not.In a year, there are only 365 Days but after every four year there are 366 days in a year. Leap Year Program in Python - Using if...else statement check year is divisible by 4 or not. Output: Enter a year: 2024 2024 is a leap year Enter a year: 1900 1900 is not a leap year Program to Check Leap Year. Source. Some leap years examples are - 1600, 1988, 1992, 1996, and 2000. Python Source Code: Leap Year Check (Calendar Library) In this method, we first import calendar library using import calendar. 1. Dear Reader, In this post I am going to write a core Python program to print Leap years!. It mostly occurs every 4 years but every 100 years we skip a leap year unless it is divisible by 400. I will be discussing 3 different ways. year = int(input("Enter a year: ")) if (year % 4) == 0: if (year % 100) == 0: if (year % 400) == 0: print("{0} is a leap year".format(year)) else: print("{0} is not a leap year".format(year)) else: print("{0} is a leap year".format(year)) else: print("{0} is not a leap year".format(year)) Output: Here is the complete python program to find leap year between two given years: This is the easy and beginner level python program to learn python using if else statement. Sitesbay - Easy to Learn . Next, use nested if statement to check whether the user entered year is Leap year or not. In this article you will get to know about leap year program. It is not a leap year. The name print_leap_years is, IMO, more descriptive than loop_year. In this Program, we are going to learn how to make sure the given year whether leap year or not using different methods in Python language. I am using the range function to make the code smaller and avoid the x = 0 and x += 1 statements. If it is not divisible by 4. This is standard library in python. Python Leap Year … If a Leap Year is a century year (years ending with 00) then it is a leap year only whne it is completely divisible by 400. After that we call isleap method by passing year given by user i.e. Using (1) multiple if-else statements … #A python program to check whether a given year is a leap year or not def check_leap_year(): #allow user input the year year = int(input('Entire the year you wish to check: ')) #define the conditions if year % 400 == 0: return(f"Yes, {year} is a leap year") if year % 100 == 0: return(f"No, {year} is not a leap year") if year % 4 == 0: return(f"Yes, {year} is a leap year") return(f"No, {year} is not a leap … #include int main() { int year; printf("Enter a year: "); scanf("%d", &year); // leap year if perfectly visible by 400 if (year % 400 == 0) { printf("%d is a leap year. Given a year, determine whether it is a leap year. Python program to check Leap Year|Python language. Program to Check Leap Year. ", year); } // not a leap year if visible by 100 // but not divisible by 400 else if (year % 100 == 0) { printf("%d is not a leap year. For example, 1997 is not a leap year. This means that in the Gregorian calendar, the years 2000 and 2400 are leap years, while 1800, 1900, 2100, 2200, 2300 and 2500 are NOT leap years. A normal year contains 366 days while a leap year contains 365 years. "); num = input ("Enter year: "); if num == 'x': exit (); try: year = int (num); except ValueError: print ("Please, enter year...exiting..."); else: if ( (year%4 == 0) and (year%100 != 0)): print (year, "is a Leap … See this example: year = int(input("Enter a year: ")) if (year % 4) == 0: if (year % 100) == 0: if (year % 400) == 0: print("{0} is a leap year".format(year)) else: print("{0} is not a leap year".format(year)) else: print("{0} is a leap year".format(year)) else: print("{0} is not a leap year".format(year)) : # python program to check whether a given year is a function provided in calendar module for simple calendars!, watch python script to check whether a given year is a leap except. 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