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Write a
will require several functions:
bool isLeapYear (int year) ;
This function should return true if year is a leap year and false if it is not. Here is pseudocode to determine a leap year:
leap_year = ((year divisible by 400) or (year divisible by 4 and year not divisible by 100))
int getCenturyValue (int year) ;
This function should take the first two digits of the year (i.e., the century), divide by 4, and save the remainder. Subtract the remainder from 3 and return this value multiplied by 2. For example, the year 2008 becomes
int getYearValue (int year) ;
This function computes a value based on the years since the beginning of the century. First, extract the last two digits of the year. For example, 08 is extracted for 2008. Next, factor in leap years. Divide the value from the previous step by 4 and discard the remainder. Add the two results together and return this value. For example, from 2008 we extract 08. Then
int getMonthValue (int month, int year) ;
This function should return a value based on the following table and will require invoking the isLeapYear function:
Finally, to compute the day of the week, compute the sum of the date’s day plus the values returned by getMonthVaIue, getYearVaIue, and getCenturyVaIue. Divide the sum by 7 and compute the remainder. A remainder of 0 corresponds to Sunday, 1 corresponds to Monday, etc.—up to 6—which corresponds to Saturday. For example, the date July 4, 2008 should be computed as
Your program should allow the user to enter any date and output the corresponding day of the week in English.
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C++ Programming: From Problem Analysis to Program...Computer ScienceISBN:9781337102087Author:D. S. MalikPublisher:Cengage Learning