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c++ - 我不知道如何处理我的日历代码

转载 作者：行者123 更新时间：2023-11-28 04:55:57 26 4
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<分区>

我目前正在制作一个日历算法，并且有一个关于它的问题。想请教一下日历，关于在日历上更改月份的第一天。

这是我的日历代码:

#include "stdafx.h"
#include <stdio.h>
#include <conio.h>
int days(int year, int month)
{
    int days, i;
    for (i = 1; i<month; i++)
    {
        if (i == 1 || i == 3 || i == 5 || i == 7 || i == 8 || i == 10 || i == 12)
        {
            days = days + 31;
        }
        else if (i == 4 || i == 6 || i == 9 || i == 11) 
        {
            days = days + 30;
        }
        else if (i == 2) 
        {
            days = days + 28;
        }
    }
    days = days + 365 * (year - 1980);
   return days;
}

int leapyear(int year, int month)
{
    int i;
    i = (year - 1977) / 4;
    if (year % 4 == 0 && month>2)
    {
        i++;
    }
    return i;
}

int weekday(int days)
{
    int i;
    i = days % 7;
    return i;
}

void calendar(int year, int month, int week)
{
    int i, j, k, d, cal[42] = { 0 };
    char m[12][10] = { "January", "February", "March", "April", "May", "June", "July", "Agust", "September", "Octorber", "November", "December" };

    switch (month)
    {
       case 1: d = 31;
               break;
       case 2:
           {
               if ((year % 4) == 0) d = 29;
               else d = 28;
           }
           break;
       case 3: d = 31;
           break;
       case 4: d = 30;
           break;
       case 5: d = 31;
           break;
       case 6: d = 30;
           break;
       case 7: d = 31;
           break;
       case 8: d = 31;
           break;
       case 9: d = 30;
           break;
       case 10: d = 31;
           break;
       case 11: d = 30;
           break;
       case 12: d = 31;
           break;
    }

    printf("     < %d %s >\n", year, m[month - 1]);
    printf("===========================\n");
    printf("Sun Mon Tue Wed Thu Fri Sat");
    printf("\n---------------------------\n");
    j = 1;
    for (i = (week + 2) % 7; i<(d + (week + 2) % 7); i++)
    {
        cal[i] = j;
        j++;
    }
    k = 0;
    for (i = 0; i<6; i++)
    {
        for (j = 0; j<7; j++)
        {
            if (cal[k] == 0)
            {
                printf("    ");
            }
            else
            {
                printf("%3d ", cal[k]);
            }
            k++;
        }

        if (i<5)
        {
            printf("\n---------------------------\n");
        }
        else
        {
            printf("\n===========================\n");
        }
    }
}

void main()
{
    int year, month, total_days, week;
    char YN = 'Y';
    do
    {
        printf("entered year and month with 1980 ~ 2030.(EX) :2003 5)\n");
        scanf_s("%d %d", &year, &month);
        if (year >= 1980 && year <= 2030 && month >= 1 && month <= 12)
        {
            printf("Your entered %dYear %dMonth.\n", year, month);
            total_days = days(year, month) + leapyear(year, month);
            week = weekday(total_days);
            calendar(year, month, week);
        }
        printf("Retry It?(Y/N):");
        YN = _getche();
        printf("\n");
    } while (YN != 'N');
}

我希望这些结果出来:

 < 2003 May >

Sun Mon Tue Wed Thu Fri Sat    
                  1   2   3    
  4   5   6   7   8   9  10    
 11  12  13  14  15  16  17    
 18  19  20  21  22  23  24    
 25  26  27  28  29  30  31


 Retry It?(Y/N):

26 4 0
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