Question

Suppose that today is April 1st of the year 2019. By entering your birthday, print what is a day of the week you were born and how many days passed since you were born. Remember that January 1st of the year 2019 is Tuesday See the example below Read 1/1/2019 - Tue 90 Read 4/1/2019 - Mon 0 Read 4/1/2018 - Sun 365 Read 4/1/2017 -> Sat 730 Read 4/1/2016 Fri 1095 Read 4/1/2015-> Wed 1461 Bye... Wikipedia description tells that every year that is exactly divisible by four is a leap year, except for years that are exactly divisible by 100, but these centurial years are leap years if they are exactly divisible by 400 For example, the years 1700, 1800, and 1900 are not leap years, but the years 1600 and 2000 are o create a test case, you have to type in the following input (for the above case) 1 1 2019 4 1 2019 4 1 2018 4 1 2017 4 1 2016 4 1 2015

C++ question

Answer 1

#include <iostream>

using namespace std;

int daysInMonths[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };
string strDays[] = { "Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday" };

bool isLeapYear(int year) {
        return ((year % 4 == 0) && (year % 100 != 0)) || (year % 400 == 0);
}



int aprilOneDay() {
        int y = 2019;
        int d = 1;
        int m = 1;
        int day = 2;  // start 1 Jan is on tuesday
        while(m != 4 || d != 1) {
                d++;
                // 2019 is not leap, so no need to worry.
                if(d > daysInMonths[m]) {
                        d = 1;
                        m++;
                }
                day = (day + 1) % 7;
        }
        return day;
}

int getDaysInMonth(int m, int y) {
        if(isLeapYear(y) && (m == 2)) {
                return 29;
        }
        return daysInMonths[m];
}

int main() {
        int m, d, y;
        int day = 0;
        cout << "Enter brith day(M D Y): ";
        cin >> m >> d >> y;
        int currentDay = aprilOneDay();
        int count = 0;

        while(m != 4 || d != 1 || y != 2019) {
                d++;
                // 2019 is not leap, so no need to worry.
                if(d > getDaysInMonth(m, y)) {
                        d = 1;
                        m++;
                        if(m > 12) {
                                m = 1;
                                y++;
                        }
                }
                day = (day + 1) % 7;
                count++;
        }
        day = (7 - day + currentDay) % 7;
        cout << strDays[day] << " " << count << endl;
}

please upvote.

main.cpp saved https://UpsetFrighteningTriggers.anjaligupta000.re 36 37 38 int main() 39 40 41 42 43 clang version 7.0.0-3-ubuntuo. 18.04.1 (tags/ Enter brith day(M D Y): 4 1 2016 Friday 1095 int m, d, y; int day-0; cout << "Enter brith day (MDY): "; cin >> m >> d >> y; int currentDayaprilOneDay (); int count 0; 45 46 47 48 49 50 51 52 53 54 while(m != 4 II d != 1 11 y != 2019) { d+ // 2019 is not leap, so no need to worry if(d> getDaysInMonth(m, y)) ( if(m 12) [ 56 57 58 59 60 61 62 day (day + 1) % 7; count+; day- (7 - day + currentDay) % 7; cout << strDays [day] << "" << count << endl;