日历项目几乎在C中完成
我正在制作一个大约4天前发布的日历项目，但我一直在研究它，即使分数会大幅减少，因为我想尝试理解它。到目前为止，我已经将日历中的日历与日历中正确的天数一起打印出来。我想要解决的唯一问题是我想让每个月的第一天从正确的位置开始，所以也许某种循环+“”以便它可以循环4个空格到月的第一天或者其他什么东西像那样。我也想要它，这样当我输入月份，年份和月份时，月份和年份将显示一个月的日历，而nummonths将显示以下月份。这对我来说是一个非常难以理解的概念。帮助很大！不胜感激！我一直在为WAYYY做长期工作。
#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;
}
2 个答案:
答案 0 :(得分：3)
您的代码有几个问题：
- 您对
daycode的计算结果不合时宜。 - 您不会在
daycode中使用nummonths，而不是main()，导致所有月份都出现在星期日。
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;
}
您还需要修改getdaycode，方法是更改公式，以便在函数末尾从numdays计算它：
...
}
int daycode = (numdays + 1) % 7;
return daycode;
}
}
如果您已经学过数组，那么您的代码可以简化很多：
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];
}
答案 1 :(得分：1)
我最近做了类似的事情，因为我想知道三月，四月和五月的周二数。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和月份中的天数之间时，代码会打印一个数字。在月末，day的起始值将根据前一个月的结束时间计算下个月。
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