C语言之经典习题

/*功能：判断闰年*/

main()

{int year,leap=0; /* leap=0：预置为非闰年*/

printf(Please input the year:);

scanf(%d,&year);

if (year % 4==0) {if (year % 100 != 0) leap=1;}
else {if (year%400==0) leap=1; }
if (leap) printf(%d is a leap year.\n,year);
else printf(%d is not a leap year.\n,year);
}

利用逻辑运算能描述复杂条件的特点，可将上述程序优化如下：

main()
{int year;
printf(Please input the year:);
scanf(%d,&year);
if ((year%4==0 && year%100!=0)||(year%400==0))
printf(%d is a leap year.\n,year);
else
printf(%d is not a leap year.\n,year);
}