程序：运用函数计算1000到2000的闰年

#include<stdio.h>
int is_leap_year(int n)
{
  if ((n % 4 == 0 && n % 100 != 0) || (n % 400 == 0))
  {
    return 1;
  }
  else
  { 
    return 0;
  }
  
}
int main()
{
  int year = 1000;
  int count = 0;
  for (year = 1000; year <= 2000; year++)
  {
    int m =is_leap_year(year);
    if (m == 1)
    {
      printf("%d ", year);
      count++;
    }
  }
  printf("\n 有%d个闰年", count);
  return 0;
}