if (year %04==0) if (year %100!=0)leap=1; y else if (year%400=0)leap=1; i if (leap)printf(" %d is a leap year. n" year); else printf(" %d is not a leap year. n"year) 程序演示 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: maino Sint year, printf("Please input the year: scanf(%od",&year) if((year%4=0&year%100!-=0)|(year%400=0) printf("%/od is a leap year. In"year) else printf("od is not a leap year. n",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); } [程序演示] 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); }