C语言打印某年月日是本年第几天，以及某年第几天是本年几月几日

最新推荐文章于 2021-12-05 21:42:14 发布

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最新推荐文章于 2021-12-05 21:42:14 发布

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本文链接：https://blog.youkuaiyun.com/zjx102938/article/details/42706597

本文深入探讨了日期转换算法，包括如何从年、月、日转换为天数，以及从天数转换回年、月、日。详细介绍了非闰年和闰年的计算方法，并通过C语言实现示例代码。

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题目：

Consider the problem of date conversion, from day of month to day of year and vice versa.For example, March 1 is the 60th day of a non-leap year, and the 61st day of a leap year.

#include <stdio.h>

int day_of_year(int year,int month,int day);
void month_day(int year,int yearday,int *mm,int *dd);


void main(){

	int y = 2015;
	int m = 2;
	int d = 1;

	int yd = 32;
	int mm = 0;
	int dd = 0;

	
	printf("%d",day_of_year(y,m,d));

	month_day(y,yd,&mm,&dd);

	printf("%d%d",mm,dd);

}


int non_leap_year[] = {0,31,28,31,30,31,30,31,31,30,31,30,31};

int leap_year[] = {0,31,29,31,30,31,30,31,31,30,31,30,31};

static int *day_of_month[]={non_leap_year,leap_year};

int day_of_year(int year,int month,int day){

    int leap = year%4 == 0&&year%100 != 0||year%400 == 0;

	for(int i = 1;i < month;i++){
        day += day_of_month[leap][i];
	}

    return day;
}


void month_day(int year,int yearday,int *mm,int *dd){

	 int i;

	 int leap = year%4 == 0&&year%100 != 0||year%400 == 0;

	 for(i = 1;yearday > day_of_month[leap][i];i++){	
		yearday -= day_of_month[leap][i];
	 }

	 *mm = i;
	 *dd = yearday;
}