my homework is in c. program:check if user's input is not leap year then find the nearest leap year to user's input. variable:y 1300<y<1400

I think the best way is to write it with while loop like.

#include <stdio.h>
int main()
{
   int year;
   scanf("%d", &year);
   while(1300<year<1400)
   {
     if (year % 100 == 0)
     {
        //i don't know:(
     }
   }
   return 0;
}

CodePudding user response：

How I would solve this :

#include <stdio.h>
 int main()
   {
       int year;
       int moreyear;
       int lessyear;

       scanf("%d", &year);
       moreyear= year;   // Different var to avoid the cases where the incremented year goes after 1400 while there are leap years before to search
       lessyear = year; // Here to check if a year before is leap before a year after
       while (1300 < year && year < 1400)
       {
         if (moreyear % 4 == 0 && moreyear % 100 != 0 && moreyear < 1400)
         {
            printf("%d is leap !", moreyear);
            return 0;               // stops the program as soon as the year is found
         }
         else if (lessyear % 4 == 0 && lessyear % 100 != 0 && lessyear > 1300) // I don't know if the leap year must be after 1300 but then, you can add another condition with && lessyear > 1300
         {
            printf("%d is leap !", lessyear);
            return 0;               // stops the program as soon as the year is found
         }
         moreyear   ;
         lessyear--;
       }
       printf("No leap year found =(");  // This case would be surprising if user entered a year between the interval
       return 0;
    }

CodePudding user response：

First of all, the task is invalid, since the Gregorian calendar wasn't in use between 1300 and 1400 BCE. But let's ignore that...

So, you need to remind yourself of a proper rule for determining if a year is a leap year or not:

In the Gregorian calendar, three criteria must be taken into account to identify leap years:

• The year must be evenly divisible by 4.
• If the year can also be evenly divided by 100, it is not a leap year;
• Unless the year is also evenly divisible by 400. Then it is a leap year.

(Taken from here)

Now, I suggest you write a function which, given a year, returns true if it's a leap year and false otherwise. It's usually a good idea, once you've defined a piece of your computation with distinct inputs and outputs, to separate it out into a function, rather than write everything within a single large main program.

Finally, use the function in the loop in your main program. Note that, as @stark points out, you can't write the loop the way that you have, i.e. C doesn't support a lower_bound < variable < upper_bound notation like we're used to from continued math equations; you have to separate that into two expressions with a logical "and": (lower_bound < variable) && (variable < upper_bound).

Also, since you're just iterating through all year, you might want to consider a for loop:

for(int year = 1301; year < 1400; year  ) {
    // do things with year 
}