C语言：输入年月日计算该日为本年第几天，含闰年判断代码实现

以下是使用 C 语言解决此问题的代码： ```c #include <stdio.h> struct Date { int year; int month; int day; }; int isLeapYear(int year) { return (year % 4 == 0 && year % 100!= 0) || year % 400 == 0; } int daysInMonth(int month, int year) { int days; switch (month) { case 1: case 3: case 5: case 7: case 8: case 10: case 12: days = 31; break; case 4: case 6: case 9: case 11: days = 30; break; case 2: if (isLeapYear(year)) { days = 29; } else { days = 28; } break; } return days; } int main() { struct Date date; printf("请输入年、月、日，格式为：年 月 日\n"); scanf("%d%d%d", &date.year, &date.month, &date.day); int dayCount = 0; for (int i = 1; i < date.month; i++) { dayCount += daysInMonth(i, date.year); } dayCount += date.day; printf("%d\n", dayCount); return 0; } ``` ######[AI写代码神器 | 385点数解答 | 2024-10-30 16:50:05]