判断一年的第几天

参考代码:

#include<stdio.h>

struct date {

    int year;

    int month;

    int day;

};

int main() {

    struct date today;

    scanf("%d %d %d", &today.year, &today.month, &today.day);

    int sum = 0; // 初始化sum为0，用于累加天数

    int i;

    int a[12] = {31, 0, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; // 月份天数数组，注意2月为0，稍后会根据年份调整

    // 判断闰年并调整2月的天数

    if (today.year % 4 == 0 && (today.year % 100 != 0 || today.year % 400 == 0)) {

        a[1] = 29;

    } else {

        a[1] = 28;

    }

    // 累加前(today.month-1)个月的天数

    for (i = 0; i < today.month - 1; i++) {

        sum += a[i];

    }

    // 加上当前月份的天数

    sum += today.day;

    // 输出结果

    printf("%d\n", sum);

    return 0;

}