用C语言写一个日期计算器

忘是亡心i 2022-05-23 04:21 228阅读 0赞

首先，我们应知道日期计算器包括哪些功能

1、明天的日期

2 、n天后的日期

3、两个日期之间的天数

我们先从第一个功能开始，首先创建一个日期的结构体，包括：年、月、日。

struct date
    {
    	int day;
    	int month;
    	int year;
    };

其次来看一下解决这个问题的思想：

![20180601224054376][]

然后我们要写一个判断闰年的函数：

int Leap(struct date d)
    {
    	int leap = false;
    	if ((d.year % 4 == 0 && d.year % 100 != 0) || d.year % 400 == 0)
    		leap = true;
    	return leap;
    }

上述代码中，我们需要先定义两个宏

#define true 1
    #define false 0

假如是闰年，便返回1，否则返回0；

接下来，我们要写一个判断今天是否为本月的最后一天的函数：

int lastdays(struct date d)
    {
    	int days;
    	int Month[12] = { 31,28,31,30,31,30,31,31,30,31,30,31 };
    	if (d.month == 2 && Leap(d))
    		days = 29;
    	else
    		days = Month[d.month - 1];
    	return days;
    }

上面我们先定义了一个数组，里面是每个月的天数，其中二月有28，或者29天，我们在数组中定义二月为28天，然后判断本月是否为2月且是闰年，如果是，便返回29天，否则，返回每月的天数即可；

下面便是主函数了：

int main()
    {
    	struct date today, tomorrow;
    	scanf("%d %d %d", &today.year, &today.month, &today.day);
    	if (today.day != lastdays(today))
    	{
    		tomorrow.day = today.day + 1;
    		tomorrow.month = today.month;
    		tomorrow.year = today.year;
    	}
    	else if (today.month == 12) {
    		tomorrow.day = 1;
    		tomorrow.month = 1;
    		tomorrow.year = today.year + 1;
    	}
    	else {
    		tomorrow.day = 1;
    		tomorrow.month = today.month + 1;
    		tomorrow.year = today.year;
    	}
    	printf("%d %d %d", tomorrow.year, tomorrow.month, tomorrow.day);
    	return 0;
    }

其实当我们知道如何算明天的日期，后面两个问题也就迎刃而解了。对于第二个问题，只需要在主函数里面加循环就可以了，假如n为50的话，为非就是50个明天。

int main()
    {
    	struct date today, tomorrow;
    	int n = 0;
    	scanf("%d %d %d", &today.year, &today.month, &today.day);
    	scanf("%d", &n);
    	while (n != 0) 
    	{
    		if (today.day != lastdays(today))
    		{
    			tomorrow.day = today.day + 1;
    			tomorrow.month = today.month;
    			tomorrow.year = today.year;
    		}
    		else if (today.month == 12) {
    			tomorrow.day = 1;
    			tomorrow.month = 1;
    			tomorrow.year = today.year + 1;
    		}
    		else {
    			tomorrow.day = 1;
    			tomorrow.month = today.month + 1;
    			tomorrow.year = today.year;
    		}
    		n--;
    		today.day = tomorrow.day;
    		today.month = tomorrow.month;
    		today.year = tomorrow.year;
    	}
    	printf("%d %d %d\n", tomorrow.year, tomorrow.month, tomorrow.day);
    	return 0;
    }

对于第三个问题，就是加一个判断，判断是否第二天是后面一天的日期，如果不是，就让count++：

int main()
    {
    	struct date today, tomorrow, hou;
    	int count = 0;
    	scanf("%d %d %d", &today.year, &today.month, &today.day);
    	scanf("%d %d %d", &hou.year, &hou.month, &hou.day);
    	while (hou.year!=today.year || hou.month!=today.month || hou.day!=today.day) 
    	{
    		if (today.day != lastdays(today))
    		{
    			tomorrow.day = today.day + 1;
    			tomorrow.month = today.month;
    			tomorrow.year = today.year;
    		}
    		else if (today.month == 12) {
    			tomorrow.day = 1;
    			tomorrow.month = 1;
    			tomorrow.year = today.year + 1;
    		}
    		else {
    			tomorrow.day = 1;
    			tomorrow.month = today.month + 1;
    			tomorrow.year = today.year;
    		}
    		today.day = tomorrow.day;
    		today.month = tomorrow.month;
    		today.year = tomorrow.year;
    		count++;
    	}
    	printf("%d\n", count);
    	return 0;
    }

这便是一个简单的日期计算器。

[20180601224054376]: /images/20220523/e0702a2507ed42b392cbb3b4b03e7fc2.png