C-language judgment leap year between 0~3000

Cooked words "Four years a run, a century not run, 400 years again run." So let's use programming to find leap years!

#include <stdio.h>int main () { int year,leap=1; printf ("Leap year between \t\t\t judgment 0~3000 \ n");  printf ("Please enter the year between 0~3000 \ n"), flag: scanf ("%d", &year),  //enter the year  if (year>0&&year<=     //four years a run, a century not run, 400 years re-run    {         if (year%4!=0)           leap=0;        else if (year%100==0)             Leap=0;     if (year%400==0)            leap=1;                          if (Leap)//leap is 1, a leap year               printf ("%d is", year);            else             printf ("%d is  Not ", year);              printf ("  a  leap year.\n ");           goto flag;     }   else      printf ("Input error, please enter the year between 0~3000 \ n");          goto flag; return 0;}

NOTE: Flag will continue to operate with the label Goto Flag shifted to the label position.

If you are looking for a leap year after 3,000 years, please revise the 3000 data yourself.

You are welcome to criticize and correct me.

This article is from the "Na-dimensional Snow" blog, please be sure to keep this source http://1536262434.blog.51cto.com/10731069/1697314

C-language judgment leap year between 0~3000