how to write a code that prints the next 20 leap years using a for loop
can someone help me Write a program that prints the next 20 leap years. ( You must use a for loop in your code )
Last edited on
Start at the year 2016 and keep incrementing by one.
What is a leap year?
Years divisible by 4 are leap years, except for those that are also divisible by 100. Years divisible by 400 are leap years regardless.
So you will need a series of if-statements in your loop to decide whether you will print the current year or not. Have a counter variable that you increment every time you print a leap year. Once that variable reaches 20, you can end the loop.
Write the pseudo code first based on the rules above, figure out how you can do this using a for-loop, and then translate to C++ code. Post here if you get stuck.
What is a leap year?
Years divisible by 4 are leap years, except for those that are also divisible by 100. Years divisible by 400 are leap years regardless.
So you will need a series of if-statements in your loop to decide whether you will print the current year or not. Have a counter variable that you increment every time you print a leap year. Once that variable reaches 20, you can end the loop.
Write the pseudo code first based on the rules above, figure out how you can do this using a for-loop, and then translate to C++ code. Post here if you get stuck.
Last edited on
#include <iostream>
using namespace std;
int main()
{
int yr=2016;
for(int i=0;i<20;i++)
{
cout<<yr<<endl;
yr=yr+4;
}
return 0;
}
using namespace std;
int main()
{
int yr=2016;
for(int i=0;i<20;i++)
{
cout<<yr<<endl;
yr=yr+4;
}
return 0;
}
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