日历项目几乎用 C 语言完成答案

【问题标题】：Calendar Project Almost done in C日历项目几乎用 C 语言完成
【发布时间】：2016-03-15 17:59:29
【问题描述】：

我正在处理一个大约 4 天前到期的日历项目，但我一直在处理它，即使分数会大幅降低，因为我想尝试理解它。到目前为止，我已经达到了在调试中打印出日历以及日历中正确天数的程度。我唯一要解决的问题是，我想让每个月的第一天从正确的位置开始，所以可能是某种循环+“”，这样它就可以将 4 个空格循环到该月的第一天或其他东西像那样。我也想要这样当我输入月份、年份和 num 月份时，月份和年份将显示一个月的日历，而 nummonths 将显示接下来的月份。这对我来说是一个很难理解的概念。帮助很大！赞赏！我一直在研究这个 WAYYY 很长时间了。

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>

int getdaycode(int month, int year);
void printheader(int month, int year);
int getndim(int month, int year);

int main(void) {
    int  day, month, year, nummonths;
    printf("Enter Month, Year, and number of Months: ");
    if (scanf("%d %d %d", &month, &year, &nummonths) != 3
    ||  year < 0 || month < 1 || month > 12) {
        printf("invalid input\n");
        return 1;
    }

    for (int i = 0; i < nummonths; i++) {
        printheader(month, year);
        int numdays = getndim(month, year);
        int daycode = getdaycode(month, year);

        printf("%*s", daycode * 4, "");   /* print 4 spaces for each skipped day */
        for (day = 1; day <= numdays; day++) {
            printf("%3d", day);
            daycode = (daycode + 1) % 7;
            if (daycode != 0)
                printf(" ");
            else
                printf("\n");
        }
        if (daycode != 0)
            printf("\n");
        printf("\n");

        month = month + 1;
        if (month > 12) {
            month -= 12;
            year += 1;
        }
    }
    return 0;
}

int getdaycode(int month, int year)
{
	int numdays;
	{
		numdays = ((year - 1) * 365 + ((year - 1) / 4) - ((year - 1) / 100) + ((year - 1) / 400)); // how many days including exceptions

		if (((year % 4 == 0) && (year % 100 != 0)) || (year % 400 == 0))		//check if leapyear
		{
			if (month == 1)							// January 
				numdays = numdays;
			if (month == 2)							// February 
				numdays = numdays + 31;
			if (month == 3)							// March 
				numdays = numdays + 28 + 31 + 1;
			if (month == 4)							// April 
				numdays = numdays + 31 + 28 + 31 + 1;
			if (month == 5)							// May 
				numdays = numdays + 30 + 31 + 28 + 31 + 1;
			if (month == 6)							// June 
				numdays = numdays + 31 + 30 + 31 + 28 + 31 + 1;
			if (month == 7)							// July 
				numdays = numdays + 30 + 31 + 30 + 31 + 28 + 31 + 1;
			if (month == 8)							// August 
				numdays = numdays + 31 + 30 + 31 + 30 + 31 + 28 + 31 + 1;
			if (month == 9)							// September 
				numdays = numdays + 31 + 31 + 30 + 31 + 30 + 31 + 28 + 31 + 1;
			if (month == 10)							// October						
				numdays = numdays + 30 + 31 + 31 + 30 + 31 + 30 + 31 + 28 + 31 + 1;
			if (month == 11)							// November
				numdays = numdays + 31 + 30 + 31 + 31 + 30 + 31 + 30 + 31 + 28 + 31 + 1;
			if (month == 12)							// December
				numdays = numdays + 30 + 31 + 30 + 31 + 31 + 30 + 31 + 30 + 31 + 28 + 31 + 1;
		}
		else
		{
			if (month == 1)							// January 
				numdays = numdays;
			if (month == 2)							// February 
				numdays = numdays + 31;
			if (month == 3)							// March 
				numdays = numdays + 28 + 31;
			if (month == 4)							// April 
				numdays = numdays + 31 + 28 + 31;
			if (month == 5)							// May 
				numdays = numdays + 30 + 31 + 28 + 31;
			if (month == 6)							// June 
				numdays = numdays + 31 + 30 + 31 + 28 + 31;
			if (month == 7)							// July 
				numdays = numdays + 30 + 31 + 30 + 31 + 28 + 31;
			if (month == 8)							// August 
				numdays = numdays + 31 + 30 + 31 + 30 + 31 + 28 + 31;
			if (month == 9)							// September 
				numdays = numdays + 31 + 31 + 30 + 31 + 30 + 31 + 28 + 31;
			if (month == 10)							// October						
				numdays = numdays + 30 + 31 + 31 + 30 + 31 + 30 + 31 + 28 + 31;
			if (month == 11)							// November
				numdays = numdays + 31 + 30 + 31 + 31 + 30 + 31 + 30 + 31 + 28 + 31;
			if (month == 12)							// December
				numdays = numdays + 30 + 31 + 30 + 31 + 31 + 30 + 31 + 30 + 31 + 28 + 31;
		}

	}
	int daycode = (numdays + 1) % 7;
	return daycode;
}

void printheader(int month, int year)
	{
			printf("%14d %1d\n", month, year);
			printf("Sun ");
			printf("Mon ");
			printf("Tue ");
			printf("Wed ");
			printf("Thu ");
			printf("Fri ");
			printf("Sat\n");
		}

int getndim(int month, int year)
{
	int numdays;
	if (((year % 4 == 0) && (year % 100 != 0)) || (year % 400 == 0))		//check if leapyear
	{
		if (month == 1)							// January 
			numdays = 31;
		if (month == 2)							// February 
			numdays = 29;
		if (month == 3)							// March 
			numdays = 31;
		if (month == 4)							// April 
			numdays = 30;
		if (month == 5)							// May 
			numdays = 31;
		if (month == 6)							// June 
			numdays = 30;
		if (month == 7)							// July 
			numdays = 31;
		if (month == 8)							// August 
			numdays = 31;
		if (month == 9)							// September 
			numdays = 30;
		if (month == 10)							// October						
			numdays = 31;
		if (month == 11)							// November
			numdays = 30;
		if (month == 12)							// December
			numdays = 31;
	}
	else
	{
		if (month == 1)							// January 
			numdays = 31;
		if (month == 2)							// February 
			numdays = 28;
		if (month == 3)							// March 
			numdays = 31;
		if (month == 4)							// April 
			numdays = 30;
		if (month == 5)							// May 
			numdays = 31;
		if (month == 6)							// June 
			numdays = 30;
		if (month == 7)							// July 
			numdays = 31;
		if (month == 8)							// August 
			numdays = 31;
		if (month == 9)							// September 
			numdays = 30;
		if (month == 10)							// October						
			numdays = 31;
		if (month == 11)							// November
			numdays = 30;
		if (month == 12)							// December
			numdays = 31;
	}
	return numdays;
}

【问题讨论】：

嗯...我猜你做得很好...但是：谁说二月是请求年的 28 天？ ...您不允许使用mktime 或strptime 吗？他们提供struct tm，其中有一个字段tm_wday，您可以使用它来计算月初的空格...
if (((year % 4 == 0) && (year % 100 != 0)) || (year % 400 == 0))
表示如果是闰年 2 月是 29 天，如果不是则 28 天
我只需要一种方法将我的日期代码乘以“”四个空格
而且我从未听说过这些命令，所以我假设如果我在代码中不使用任何命令会很奇怪

标签： c printf scanf

【解决方案1】：

您的代码有几个问题：

您对 daycode 的计算结果差了 1。
您没有在main() 中使用daycode，而不是nummonths，导致所有月份似乎都是从星期日开始的。

这是一个建议：

int main(void) {
    int  day, month, year, nummonths;
    printf("Enter Month, Year, and number of Months: ");
    if (scanf("%d %d %d", &month, &year, &nummonths) != 3
    ||  year <= 0 || month < 1 || month > 12) {
        printf("invalid input\n");
        return 1;
    }

    for (int i = 0; i < nummonths; i++) {
        printheader(month, year);
        int numdays = getndim(month, year);
        int daycode = getdaycode(month, year);

        printf("%*s", daycode * 4, "");   /* print 4 spaces for each skipped day */
        for (day = 1; day <= numdays; day++) {
            printf("%3d", day);
            daycode = (daycode + 1) % 7;
            if (daycode != 0)
                printf(" ");
            else
                printf("\n");
        }
        if (daycode != 0)
            printf("\n");
        printf("\n");

        month = month + 1;
        if (month > 12) {
            month -= 12;
            year += 1;
        }
    }
    return 0;
}

您还需要修复getdaycode，方法是更改公式以从函数末尾的numdays 计算它：

            ...
        }
        int daycode = (numdays + 1) % 7;
        return daycode;
    }
}

如果你已经学过数组，你的代码可以大大简化：

static int const days[12] = { 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };
static int const leapdays[12] = { 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };

int getdaycode(int month, int year) {
    int numdays = 0;
    if (year > 0) {
        // how many days in previous years including exceptions
        numdays = (year - 1) * 365 + (year - 1) / 4 - (year - 1) / 100 + (year - 1) / 400;
    }
    if (month >= 1 && month <= 12) {
        if (month > 2 && (((year % 4 == 0) && (year % 100 != 0)) || (year % 400 == 0)))
            numdays += 1;  // count february 29
        for (int i = 0; i < month - 1; i++)
            numdays += days[i];
    }
    // need to offset by one so 1/1/1 falls on a Monday.
    // dates before 1753 are computed according to the
    // proleptic Gregorian calendar.
    return (numdays + 1) % 7;
}

int getndim(int month, int year) {
    if (month < 1 || month > 12)
        return 0;
    else
    if ((year % 4 == 0 && year % 100 != 0) || (year % 400 == 0))
        return leapdays[month - 1];
    else
        return days[month - 1];
}

【讨论】：

我不知道这是否有帮助，但是.. numdays = ((year - 1) * 365 + ((year - 1) / 4) - ((year - 1) / 100) + ((year - 1) / 400)) 应该是 numdays = ((year - 1) * 365 + ((year - 1) / 4) - ((year - 1) / 100) + ((year - 1 ) / 400)+天)
也许这弄乱了我的 daycode，因为我从 getdaycode 中删除了 int day
哇，它的工作非常接近我的目标
我将发布上面的代码，并附上输出的图片以及它应该是什么。
@tyooo：我更新了答案，以通过更改 daycode 的公式来向您展示我的意思。

【解决方案2】：

我最近做了类似的事情，因为我想知道 3 月、4 月和 5 月的星期二数量。所以我写了这段代码

#include <stdio.h>

int main( void )
{
    char *monthNames[] = { "March", "April", "May" };

    printf( "Enter starting day (0-6): " );
    int day;
    if ( scanf( "%d", &day ) != 1 )
        return 1;

    day = -(day % 7);
    for ( int m = 0; m < 3; m++ ) {
        printf( "\n%s\n", monthNames[m] );
        int maxday = (m == 1) ? 30 : 31;

        while ( day < maxday ) {
            for ( int col = 0; col < 7; col++ ) {
                if ( day < 0 )
                    printf( "   " );
                else if ( day < maxday )
                    printf( " %2d", day+1 );
                day++;
            }
            printf( "\n" );
        }

        if ( day > maxday )
            day = day - maxday - 7;
        else
            day = 0;
    }
}

一般的想法是变量day 以-6 和0 之间的数字开头。只要day 为负数，代码就会打印三个空格。当day 介于 0 和该月的天数之间时，代码会打印一个数字。在月底，day 的起始值将根据上个月结束的位置为下个月计算。

【讨论】：