The prompt is:
Implement a function that reads in a string containing a textual description of a calendar date and that prints out the corresponding day of the week (Monday–Sunday). The two valid input formats for this function are:
mm/dd/yyyy
Example: 03/04/2014 Output: Tuesday
Month dd, yyyy
Example: March 04, 2014 Output: Tuesday
where dd is the numeric day, mm is the numeric month, yyyy is the year and Month is the name of the month. All days and months are specified using two digits (i.e. for March, use 03 instead of 3). In the second valid format, there is a single space between Month and dd and between dd, and yyyy.
In order to receive full credit on this task, your program should print out the correct day of the week for any input in a correct format.
The code I have so far is able to give me the number of the day of the year that I input but from there I don't know what to do so that it will give me the day of the week because each year starts from a different weekday to begin with.
#include<stdio.h>
int main() {
int month, day, year, dm, dn, leap;
printf("enter the month:");
scanf("%d",&month);
printf("enter the day:");
scanf("%d",&day);
printf("enter the year:");
scanf("%d",&year);
if((year%100 == 0 && year%400 == 0) || (year%4==0)) {
if(month==1)
dm=0;
if(month==2)
dm=31;
if(month==3)
dm=60;
if(month==4)
dm=91;
if(month==5)
dm=121;
if(month==6)
dm=152;
if(month==7)
dm=182;
if(month==8)
dm=213;
if(month==9)
dm=244;
if(month==10)
dm=274;
if(month==11)
dm=305;
if(month==12)
dm=335;
}
else {
if(month==1)
dm=0;
if(month==2)
dm=31;
if(month==3)
dm=59;
if(month==4)
dm=90;
if(month==5)
dm=120;
if(month==6)
dm=151;
if(month==7)
dm=181;
if(month==8)
dm=212;
if(month==9)
dm=243;
if(month==10)
dm=273;
if(month==11)
dm=303;
if(month==12)
dm=334;
}
dn = dm+day;
printf("the day number is :%d",dn);
return 0;
}
Related
I have two dates
1.statdate : 12-11-2022
2.enddate. : 02-20-2023
if the start date is 1-15th day then the result should be the current month ex: mm-dd-yyyy[12-11-2022] Then December 1st
if the start date is 16th-31st day then the result should be the current month EX: mm-dd-yyyy[12-22-2022] Then January 1st.
if the EndDate is 1-15th day then the result should be the current month+1 EX: mm-dd-yyyy[02-11-2022] Then January 31st
if the EndDate is 16-31sh day then the result should be the current month+1 EX: mm-dd-yyyy[02-20-2022] Then Feb 28th
input start date result: December 1st
input end date result: Feb 28th
Result[3] which is three months from the start date to the end date.
Can we do this in the formula field? I am able to do it in apex it worked but I was unable to do it in the formula field any help would be appreciated.
ROUND(((IF(DAY(Return_To_Work__c) <= 15, DATE( YEAR(Return_To_Work__c) ,
MONTH(Return_To_Work__c) -1,(DAY(Return_To_Work__c)-
DAY(Return_To_Work__c)+
28 + MOD(((MONTH(Return_To_Work__c) -1) +
FLOOR((MONTH(Return_To_Work__c) -1)/8)), 2) + MOD(2,
(MONTH(Return_To_Work__c) -1)) + 2 * FLOOR(1/(MONTH(Return_To_Work__c)
-1))))
,IF(DAY(Return_To_Work__c) >= 16,DATE( YEAR(Return_To_Work__c)
,MONTH(Return_To_Work__c),(DAY(Return_To_Work__c)-
DAY(Return_To_Work__c)+28 + MOD(((MONTH(Return_To_Work__c)) +
FLOOR((MONTH(Return_To_Work__c))/8)), 2) + MOD(2,
(MONTH(Return_To_Work__c))) + 2 *
FLOOR(1/(MONTH(Return_To_Work__c))))),NULL)) -
IF(DAY(First_Day_Of_Leave__c) <= 15, DATE( YEAR(First_Day_Of_Leave__c)
, MONTH(First_Day_Of_Leave__c) ,(DAY(First_Day_Of_Leave__c)-
DAY(First_Day_Of_Leave__c)+1)),IF(DAY(First_Day_Of_Leave__c) >=
16,DATE( YEAR(First_Day_Of_Leave__c)
,MONTH(First_Day_Of_Leave__c)+1,(DAY(First_Day_Of_Leave__c)-
DAY(First_Day_Of_Leave__c)+1)),NULL)))/30),0)
As a little background: in October 1582 the Gregorian calendar was introduced to correct problems with the Julian calendar, which was in use until then. The Gregorian calendar specifies the length of a year more precisely, adds a new leap year calculation and removes 10 days from the Julian calendar on transition (moving from 4.Oct.1582 to 15.Oct.1582)
// Julian leap year calculation
((year % 4) == 0)
// Gregorian leap year calculation
((year % 4) == 0 && (year % 100) != 0) || (year % 400) == 0)
To sum up all calendar days from 1 AD to 2001 AD I use:
#include <stdio.h>
int main( void)
{
int year;
int daysInYear;
int total;
total = 0;
for( year = 1; year <= 2001; year++)
{
//
daysInYear = 365;
if( year < 1582)
daysInYear += ((year % 4) == 0);
else
if( year > 1582)
daysInYear += ((year % 4) == 0 && (year % 100) != 0) || ((year % 400) == 0);
else
daysInYear -= 10; // 1582 exactly
total += daysInYear;
printf( "%d: +%d = %d\n", year, daysInYear, total);
}
return( 0);
}
Now running this exhaustively over all years from 1 to 2000 gives this pastebin: https://pastebin.com/bR7hwNr1
The most interesting bits show that the leap year and day-skip calculation is correct and also produces the result:
1: +365 = 365
2: +365 = 730
3: +365 = 1095
4: +366 = 1461
...
100: +366 = 36525
...
400: +366 = 146100
...
1581: +365 = 577460
1582: +355 = 577815
1583: +365 = 578180
...
1900: +365 = 693962
...
1996: +366 = 729026
1997: +365 = 729391
1998: +365 = 729756
1999: +365 = 730121
2000: +366 = 730487
Now the question is this: the 1.1.2001 is at 730487 elapsed calendar days. Given the length of the year as defined by the Gregorian calendar as 365.2425, I would expect to see 2000*365.2425 = 730485 days. If one uses tropical days or solar days it would be 730484. But 740487 is clearly out of range.
Assuming that no-one miscounted any days in the Gregorian Calendar range, there are some extra days in the Julian Calendar, that don't match with the tropical days. But the Gregorian was created to correct the Julian calendar and I don't think they would have made that large an error in 1582.
This could be more a historical question than a programming question I would guess.
The goal of the Gregorian reform of the calendar was to reset the date of the northern hemisphere vernal equinox to March 21, and keep it there. This date was chosen because that was the approximate average date of the equinox at the time of the Council of Nicaea in AD 325. At the time of the council, the date of the equinox had slipped from about March 25 in AD 1.
For all the details read the book Gregorian Reform of the Calendar which is the proceedings of a conference held by the Vatican Observatory and published in 1983.
To clarify, when evaluating the Gregorian calendar compared to the Julian, using a mix of Julian and Gregorian dates incorporates the one-time correction of 10 days which was intended to correct for accumulated excess leap years from about AD 325 to AD 1582. If one wishes to use some start date other than AD 325, the entire calculation should use ONLY Gregorian dates or ONLY Julian dates.
Basically I'm attempting a question on how to compute fee. However, my code is extremely long. I ended up using a heck tons of 'if' and 'else' for my code. I was wondering if there is any better way on approaching this question.
I initially attempted to use loops to keep stacking the fee but I face a problem when my time cross the critical juncture. I added time to TimeIn along with my fees, but as it crosses the 7am mark, I need to reset it back to 7am so that my fees will be charge properly. And that was when I gave up when my code became very long too.
Eg. loop adds from 0630 to 0730, fees are properly increased, the first 30 minutes of 0700 will be skipped and fees are wrongly charged.
The question I attempted:
Mr. Wu has been going to work every day by taxi for many years. However, the taxi fare has been increasing rather quickly in recent years. Therefore, he is considering driving to work instead.
One of the costs for driving is the parking fee. The parking rates of the car park at Mr. Wu’s workplace are as shown in the table below.
Weekday Saturday Sunday
4am ~ 7am $2.00 per hour $2.50 per hour $5
7am ~ 6pm $1.20 per 30 minute $1.50 per 30 minutes per
6pm ~ midnight $5.00 per entry $7.00 per entry entry
Special note:
1. The car park opens at 4am and closes at midnight. All vehicles must leave
by midnight.
2. There is a grace period of 10 minutes on any day (i.e., it is completely
free to park for 10 minutes or less regardless of day and time.)
3. There is a 10% surcharge for parking more than 10 hours on a weekday and
20% for Saturday. There is no surcharge for Sunday.
4. There is an additional $3.00 fee for exiting after 10pm on any day.
(Surcharge is not applicable on this fee.)
Your program should read in one integer, which is an integer between 1 and 7
representing the day of the week (1 being Monday and 7 being Sunday). It should also
read in two numbers representing the time-in and time-out in 24-hour format. It
should then calculate and display the parking fee (with two decimal places).
You may assume that the inputs are valid (i.e., the day is within the specified range,
both time-in and time-out are between 4am and midnight in 24-hour format, and timeout
is no earlier than time-in).
My extremely long code:
#include <stdio.h>
#include <math.h>
double computeFee(int, int, int);
int main(void){
int day, timeIn, timeOut;
scanf("%d %d %d", &day, &timeIn, &timeOut);
printf("Enter day: %d\n", day);
printf("Enter time-in: %d\n", timeIn);
printf("Enter time-out: %d\n", timeOut);
printf("Parking fee is $%.2lf\n", computeFee(day, timeIn, timeOut));
return 0;
}
double computeFee(int day, int timeIn, int timeOut){
double fee = 0;
double TimeIn, TimeOut;
TimeIn = 60*(floor(timeIn/100)) + (timeIn%100);
TimeOut = (60*(floor(timeOut/100)) + (timeOut%100));
if((day>=1)&&(day<=5)){
if(TimeIn<420){
if(TimeOut<420){
fee += 2*ceil((TimeOut - TimeIn)/60);
}
else{
fee += 6;
if(TimeOut>=1080){
fee += 31.4;
}
else{
fee += 1.2*ceil((TimeOut - 420)/30);
}
}
}
if(TimeIn>=420){
if(TimeIn>=1080){
fee = 5;
}
else{
if(TimeOut>=1080){
fee += (1.2*ceil((1080 - TimeIn)/30) + 5);
}
else{
fee += (1.2*ceil((TimeOut - TimeIn)/30));
}
}
}
if((TimeOut-TimeIn)>=600){
fee *= 1.1;
}
}
if(day == 6){
if(TimeIn<420){
if(TimeOut<420){
fee += (2.5*ceil((TimeOut - TimeIn)/60));
}
else{
fee += 7.5;
if(TimeOut>=1080){
fee += 40;
}
else{
fee += (1.5*ceil((TimeOut - 420)/30));
}
}
}
if(TimeIn>=420){
if(TimeIn>=1080){
fee = 7;
}
else{
if(TimeOut>=1080){
fee += (1.5*ceil((1080 - TimeIn)/30) + 7);
}
else{
fee += (1.5*ceil((TimeOut - TimeIn)/30));
}
}
}
if((TimeOut-TimeIn)>=600){
fee *= 1.2;
}
}
if(day == 7){
fee = 5;
}
if((timeOut/100)>=22){
fee += 3;
}
if((timeOut - timeIn)<=10){
fee = 0;
}
return fee;
}
Examples on how fees are calculated:
Example 1: Tuesday, 4:29am to 7:50am.
• 4:29am to 7am is charged as 3 1-hour slots: $2.00 * 3 = $6.00
• 7am to 7:50am is charged as 2 30-minute slots: $1.20 * 2 = $2.40
• Total fee = $6.00 + $2.40 = $8.40
Example 2: Saturday, 7:01am to 7:49pm.
• 7:01am to 6pm is charged as 22 30-minute slots: $1.50 * 22 = $33.00
• 6pm to 7:49pm is charged as one entry: $7.00
• 20% Surcharge for parking more than 10 hours: ($33.00 + $7.00) * 20% =
$8.00
• Total fee = $33.00 + $7.00 + $8.00 = $48.00
Example 3: Sunday, 3pm to 10:01pm.
• 3pm to 10:01pm is charged as one entry: $5.00
• Additional fee for exiting after 10pm: $3.00
• Total fee = $5.00 + $3.00 = $8.00
Example 4: Thursday, 11:49pm to 11:59pm.
• Grace period
• Total fee = $0.00
Example 5: Monday, 12pm to 10:01pm.
• 12pm to 6pm is charged as 12 30-minute slots: $1.20 * 12 = $14.40
• 6pm to 10:01pm is charged as one entry: $5.00
• 10% Surcharge for parking more than 10 hours: ($14.40 + $5.00) * 10% =
$1.94
• Additional fee for exiting after 10pm: $3.00
• Total fee = $14.40 + $5.00 + $1.94 + $3.00 = $24.34
Thanks for reading my long question. And thanks in advance for the help.
note: I havent learn arrays and anything beyond that. Only learn loops and selection statement so far(reading K&R programming tutorial, until chapter 17, using Online GeekforGeek as compiler). However, I will still greatly appreciate solutions using other methods.
I am very new to angular JS but since morning struggling with this.
I have a datepicker with "MM/yyyy" format, the date value returned here is first day of month.
i.e. February 1, 2017 but i want the date as February 28, 2017 i.e last day of month.
Just to update i am using moment function.
Please suggest some work around for the same!
I infer from your question that you are using momentjs.
This lib provides you with a built in function endof
const date = new Date(2017, 1) // 1st Feb
moment(date).endOf('month');
This should handle most cases directly including leap years
If you have a JavaScript Date instance d, you can simply use
const d = new Date(2017, 1) // 1st Feb
d.setMonth(d.getMonth() + 1)
d.setDate(d.getDate() - 1)
console.info(d.toLocaleString())
Now d will be the last day of the month.
Note: this easily handles year boundaries without any extra code. For example
const d = new Date(2017, 11) // 1st Dec
console.info('Before', d.toLocaleString())
d.setMonth(d.getMonth() + 1)
d.setDate(d.getDate() - 1)
console.info('After', d.toLocaleString())
You just need add 1 month to the date, and then substract 1 day. Here's an example:
// Let's suppose the date selected from the picker was February 1st, 2017
// Remember months in JS are zero-index based, so 0=Jan, 1=Feb, etc.
var selectedDate = new Date(2017, 1, 1);
var month = selectedDate.getMonth();
var year = selectedDate.getFullYear();
month ++;
if(month > 11){ // Last month number is 11 (December)
month = 0; // January
year ++;
}
var oneMonthAheadDate = new Date(year, month, 1);
var lastDayOfSelectedMonthDate = new Date(oneMonthAheadDate.getTime() - (1000 * 3600 * 24)); // We substract 1 day (1000 ms x 3600 secs in an hour x 24 hours in a day)
Your needed date will be in lastDayOfSelectedMonthDate
The prompt is:
Implement a function that reads in a string containing a textual description of a cal- endar date and that prints out the corresponding day of the week (Monday–Sunday). The two valid input formats for this function are:
mm/dd/yyyy
Example: 03/04/2014
Output: Tuesday
Month dd, yyyy
Example: March 04, 2014
Output: Tuesday
where dd is the numeric day, mm is the numeric month, yyyy is the year and Month is the name of the month. All days and months are specified using two digits (i.e. for March, use 03 instead of 3). In the second valid format, there is a single space between Month and dd and between dd, and yyyy.
In order to receive full credit on this task, your program should print out the correct day of the week for any input in a correct format.
So as of right now i am able to get the correct days for every single day except in the years 2005 2009 2013 2017 etc etc... they are always a day behind, i notice that its going by a trend of every 4 years the days end up 1 day behind. Im not sure whats wrong. is it cause my method of using 365.25 as each year is wrong?
My code:
#include<stdio.h>
int main()
{
int month,day1,day2,totdays,year,dm,dn,leap,rmd;
printf(" ");
scanf("%d/%d/%d",&month,&day1,&year);
if(((year%4==0) && (year%100!=0)) || (year%400==0))
{
if(month==1)
dm=0;
if(month==2)
dm=31;
if(month==3)
dm=60;
if(month==4)
dm=91;
if(month==5)
dm=121;
if(month==6)
dm=152;
if(month==7)
dm=182;
if(month==8)
dm=213;
if(month==9)
dm=244;
if(month==10)
dm=274;
if(month==11)
dm=305;
if(month==12)
dm=335;
}
else
{
if(month==1)
dm=0;
if(month==2)
dm=31;
if(month==3)
dm=59;
if(month==4)
dm=90;
if(month==5)
dm=120;
if(month==6)
dm=151;
if(month==7)
dm=181;
if(month==8)
dm=212;
if(month==9)
dm=243;
if(month==10)
dm=273;
if(month==11)
dm=304;
if(month==12)
dm=334;
}
day2=(year-1970)*(365.25);
dn=dm+day1;
totdays=day2+dn;
rmd=totdays%7;
if(rmd==5)
{
printf("Monday \n");
}
if(rmd==6)
{
printf("Tuesday \n");
}
if(rmd==0)
{
printf("Wednesday \n");
}
if(rmd==1)
{
printf("Thursday \n");
}
if(rmd==2)
{
printf("Friday \n");
}
if(rmd==3)
{
printf("Saturday \n");
}
if(rmd==4)
{
printf("Sunday \n");
}
return 0;
}
1969 wasn't a leap year, 1972 was. When you do
day2=(year-1970)*(365.25);
to discover how many days off January 1st of year year is, you'll count
0 days for '70
365.25 days for '71
730.5 days for '72
1095.75 days for '73
1461 days for '74
The fractional portion of the floating point calculation is truncated, so day2 isn't going to count the extra day from 02/29/1972 until 01/01/1974, instead of 01/01/1973 as it should.
Put another way, you are making the assumption that 1970 was the first year after a leap year, so a leap day won't be counted until four years later.
The day2 calculation won't work. There are 1461 days in every four year period. First you need to compute how many 4 year periods have passed. Then figure out how many days there were to the beginning of the specified year, similar to what you did for the months.
The year%100 and year%400 exceptions add a little complexity, but fortunately the year 2000 was a leap year, so the first time you have to deal with that little wrinkle is the year 2100.