How does C language judge leap year and leap year judgment conditions?

1, first open the C-Free 5 compiler in the computer, int year = 0；; Define the year variable as an integer and initialize the value to 0 to store the year.

2. then Cout

3、? CIN & gt； & gt year; The input data is saved in a year integer variable.

4. If (year < = 0) judges whether the year variable is less than or equal to 0, the year has no negative number. Illegal input exits the program.

5. Then enter if ((year% 4 = = 0&; & amp Year% 100! = 0)| | year % 400 = = 0)cout & lt； & lt Year & lt& lt "Year is a leap year" < & ltendlelsecout & lt& lt "Year is an average year" < & ltendl How to judge whether a year is a leap year or not?

6. Return 0; Return the value to main () and end the program, as shown in the figure below.