PROGRAMMING:Judge leap year and day of week
Enter the value of month, year and day (all integer numbers), output whether the year is a leap year, and output the day of the week.
Among them, the legal value range of year is [18202020], the legal value range of month is [1,12], and the legal value range of date is [1,31];
The algorithm to judge the day of the week is as follows: assume that January 1, A.D. 0001 is Monday, so as to calculate the number of days between the current input date and January 1, 0001, and then divide the number of days by 7 to find the remainder. When the remainder is 0, it is Sunday; when the remainder is 1, it is Monday; and so on, when the remainder is 6, it is Saturday.
Requirement: the main class must contain the following methods with the signature as follows:
```
public static void main(String[] args)；// Main method;
public static boolean isLeapYear(int year) ;// Judge whether year is leap year, return Boolean type;
public static int numOfDays(int year,int month ,int day) ;// Calculate the distance days from year-month-day to 0001-1-1, and return the integer number;
public static String getWhatDay(int days) ; // Returns the day of the week according to the number of days, in which the parameter days is the number of days, integer number, and returns the English word of the day of the week.
```
**Note: Java date related classes and methods are not allowed**
###Input format:
Enter the values of month, year and day in one line. They are all integer numbers. You can separate them with one or more spaces or carriage returns.
###Output format:
1. When the input data is illegal and the input date does not exist, output "wrong format";
1. When the input date is legal, output two lines of data in the following format (note that there are two at the end of the two lines.)
*The first line: year (value) is a leap year
*The second line: Year Month Day is the day of the week
###Input sample 1:
Here is a set of inputs. For example:
```in
2020 3 9
```
###Output sample 1:
The corresponding output is given here. For example:
```out
2020 is a leap year.
2020-3-9 is Monday.
```
###Input sample 2:
Here is a set of inputs. For example:
```in
1835 12 31
```
###Output sample 2:
The corresponding output is given here. For example:
```out
1835 is not a leap year.
1835-12-31 is Thursday.
```
###Input sample 3:
Here is a set of inputs. For example:
```in
1999 9 31
```
###Output sample 3:
The corresponding output is given here. For example:
```out
Wrong Format
```
answer：If there is no answer, please comment
1. Pay attention to the situation of leap year;
1. Pay attention to the maximum number of days in each month;
1. Pay attention to the valid range of input data.
Among them, the legal value range of year is [18202020], the legal value range of month is [1,12], and the legal value range of date is [1,31];
The algorithm to judge the day of the week is as follows: assume that January 1, A.D. 0001 is Monday, so as to calculate the number of days between the current input date and January 1, 0001, and then divide the number of days by 7 to find the remainder. When the remainder is 0, it is Sunday; when the remainder is 1, it is Monday; and so on, when the remainder is 6, it is Saturday.
Requirement: the main class must contain the following methods with the signature as follows:
```
public static void main(String[] args)；// Main method;
public static boolean isLeapYear(int year) ;// Judge whether year is leap year, return Boolean type;
public static int numOfDays(int year,int month ,int day) ;// Calculate the distance days from year-month-day to 0001-1-1, and return the integer number;
public static String getWhatDay(int days) ; // Returns the day of the week according to the number of days, in which the parameter days is the number of days, integer number, and returns the English word of the day of the week.
```
**Note: Java date related classes and methods are not allowed**
###Input format:
Enter the values of month, year and day in one line. They are all integer numbers. You can separate them with one or more spaces or carriage returns.
###Output format:
1. When the input data is illegal and the input date does not exist, output "wrong format";
1. When the input date is legal, output two lines of data in the following format (note that there are two at the end of the two lines.)
*The first line: year (value) is a leap year
*The second line: Year Month Day is the day of the week
###Input sample 1:
Here is a set of inputs. For example:
```in
2020 3 9
```
###Output sample 1:
The corresponding output is given here. For example:
```out
2020 is a leap year.
2020-3-9 is Monday.
```
###Input sample 2:
Here is a set of inputs. For example:
```in
1835 12 31
```
###Output sample 2:
The corresponding output is given here. For example:
```out
1835 is not a leap year.
1835-12-31 is Thursday.
```
###Input sample 3:
Here is a set of inputs. For example:
```in
1999 9 31
```
###Output sample 3:
The corresponding output is given here. For example:
```out
Wrong Format
```
answer：If there is no answer, please comment
1. Pay attention to the situation of leap year;
1. Pay attention to the maximum number of days in each month;
1. Pay attention to the valid range of input data.