(Ungraded) C Language - Write a program that takes in a year and determines whether that year is a leap year. ---### Leap Year Determination in the Gregorian CalendarA year in the modern Gregorian Calendar consists of 365 days. In reality, the earth takes longer to rotate around the sun. To account for the difference in time, every 4 years, a leap year takes place. A leap year is when a year has 366 days: An extra day, February 29th. The requirements for a given year to be a leap year are:1. The year must be divisible by 42. If the year is a century year (1700, 1800, etc.), the year must be evenly divisible by 400; therefore, both 1700 and 1800 are not leap years.Some example leap years are 1600, 1712, and 2016.#### Example Inputs and Outputs- If the input is 1712, the output is: ``` 1712 - leap year ```- If the input is 1913, the output is: ``` 1913 - not a leap year ```#### Program ExampleBelow is an example program written in C, which takes in a year and determines whether that year is a leap year.```c#include #include int main(void) { int inputYear; bool isLeapYear = false; scanf("%d", &inputYear); /* Type your code here. */ return 0;}```In this program:- `#include ` and `#include ` are directives to include the standard input-output and boolean libraries respectively.- `int main(void)` is the main function where the program starts execution.- `int inputYear;` declares an integer variable `inputYear` to hold the year input.- `bool isLeapYear = false;` initializes a boolean variable `isLeapYear` to false.- `scanf("%d", &inputYear);` reads the input year from the user.The actual logic to determine whether `inputYear` is a leap year should be typed where it says `/* Type your code here. */`.---This transcription includes all necessary details for understanding the leap year concept and a basic example of how to implement a leap year check programmatically.

Here's the algorithm for the provided code:1. Declare the necessary variables:inputYear: to store…

A year in the modern Gregorian Calendar consists of 365 days. In reality, the earth takes longer to rotate around the sun. To account for the difference in time, every 4 years, a leap year takes place. A leap year is when a year has 366 days: An extra day, February 29th. The requirements for a given year to be a leap year are: 1) The year must be divisible by 4 2) If the year is a century year (1700, 1800, etc.), the year must be evenly divisible by 400; therefore, both 1700 and 1800 are not leap years Some example leap years are 1600, 1712, and 2016. Write a program that takes in a year and determines whether that year is a leap year. Ex: If the input is 1712, the output is: 1712 leap year Ex: If the input is 1913, the output is: 1913 not a leap year 1 #include 2 #include 3 4 int main(void) { 5 6 7 8 9 10 11 12} int inputYear; bool isLeapYear = false; scanf("%d", &inputYear); /* Type your code here. */ return 0;

A year in the modern Gregorian Calendar consists of 365 days. In reality, the earth takes longer to rotate around the sun. To account for the difference in time, every 4 years, a leap year takes place. A leap year is when a year has 366 days: An extra day, February 29th. The requirements for a given year to be a leap year are: 1) The year must be divisible by 4 2) If the year is a century year (1700, 1800, etc.), the year must be evenly divisible by 400; therefore, both 1700 and 1800 are not leap years Some example leap years are 1600, 1712, and 2016. Write a program that takes in a year and determines whether that year is a leap year. Ex: If the input is 1712, the output is: 1712 leap year Ex: If the input is 1913, the output is: 1913 not a leap year 1 #include 2 #include 3 4 int main(void) { 5 6 7 8 9 10 11 12} int inputYear; bool isLeapYear = false; scanf("%d", &inputYear); /* Type your code here. */ return 0;

C++ Programming: From Problem Analysis to Program Design

8th Edition

ISBN:9781337102087

Author:D. S. Malik

Publisher:D. S. Malik

Chapter8: Arrays And Strings

Section: Chapter Questions

Problem 10PE

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When you borrow money to buy a house, a car, or for some other purpose, you repay the loan by making periodic payments over a certain period of time. Of course, the lending company will charge interest on the loan. Every periodic payment consists of the interest on the loan and the payment toward the principal amount. To be specific, suppose that you borrow $1,000 at an interest rate of 7.2% per year and the payments are monthly. Suppose that your monthly payment is $25. Now, the interest is 7.2% per year and the payments are monthly, so the interest rate per month is 7.2/12 = 0.6%. The first months interest on $1,000 is 1000 0.006 = 6. Because the payment is $25 and the interest for the first month is $6, the payment toward the principal amount is 25 6 = 19. This means after making the first payment, the loan amount is 1,000 19 = 981. For the second payment, the interest is calculated on $981. So the interest for the second month is 981 0.006 = 5.886, that is, approximately $5.89. This implies that the payment toward the principal is 25 5.89 = 19.11 and the remaining balance after the second payment is 981 19.11 = 961.89. This process is repeated until the loan is paid. Write a program that accepts as input the loan amount, the interest rate per year, and the monthly payment. (Enter the interest rate as a percentage. For example, if the interest rate is 7.2% per year, then enter 7.2.) The program then outputs the number of months it would take to repay the loan. (Note that if the monthly payment is less than the first months interest, then after each payment, the loan amount will increase. In this case, the program must warn the borrower that the monthly payment is too low, and with this monthly payment, the loan amount could not be repaid.)
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