if (year %04==0) if (year %100=0)leap=1; else if (year%0400==0)leap=1; f if(leap) printf("%od is a leap year. In"year) else printf("%/od 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 Rint year printf("Please input the year: ) scan nf("od", &year) if((year%4=0&&year%100=0)(year%400=0) printf("od is a leap year. In"year) 上 else printf("%d is not a leap year. n" year) 下=顶返回本章首页 下一页 上一页 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); }