(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;

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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