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网上有关“怎么用C语言编写万年历程序”话题很是火热，小编也是针对怎么用C语言编写万年历程序寻找了一些与之相关的一些信息进行分析，如果能碰巧解决你现在面临的问题，希望能够帮助到您。

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万年历

#include "stdio.h" /* Required for MS-DOS use */

#define ENTER 0x1C0D /* Enter key */

int year, month, day;

static char *days[8] = {" ","Sunday ","Monday ","Tuesday ",

"Wednesday","Thursday ","Friday ","Saturday "};

struct TIMEDATE {

int year; /* year 1980..2099 */

int month; /* month 1=Jan 2=Feb, etc. */

int day; /* day of month 0..31 */

int hours; /* hour 0..23 */

int minutes; /* minute 0..59 */

int seconds; /* second 0..59 */

int hsecs; /* 1/100ths of second 0..99 */

char dateline[47]; /* date & time together */

};

static struct TIMEDATE today;

main()

{

char cmonth[3];

char cday[3];

char cyear[5];

double getdays();

double daynumb, numbnow;

int weekday, retcode, dayer, i;

dayer = datetime(&today);

clrscn();

for (i=0;i<3;++i)cmonth[i]='\0';

for (i=0;i<3;++i)cday[i]='\0';

for (i=0;i<5;++i)cyear[i]='\0';

putstr(5,8,14,"Enter date in MM DD YYYY format:");

while (retcode != ENTER)

{

retcode = bufinp(5,41,13,2,cmonth);

if (retcode != ENTER) retcode = bufinp(5,44,13,2,cday);

if (retcode != ENTER) retcode = bufinp(5,47,13,4,cyear);

}

year = atoi(&cyear);

month = atoi(&cmonth);

day = atoi(&cday);

daynumb = getdays(year, month, day);

numbnow = getdays(today.year, today.month, today.day);

weekday = weekdays(daynumb);

if (numbnow - daynumb == 0)

printf("\n\n%02d-%02d-%d is",month, day, year);

if (numbnow - daynumb > 0)

printf("\n\n%02d-%02d-%d was",month, day, year);

if (numbnow - daynumb < 0)

printf("\n\n%02d-%02d-%d will be",month, day, year);

printf(" a %s\n",days[weekday]);

} /* end MAIN */

/************************************************************

* GETDAYS - From integer values of year (YYYY), month *

* (MM) and day (DD) this subroutine returns a *

* double float number which represents the *

* number of days since Jan 1, 1980 (day 1). *

* This routine is the opposite of GETDATE. *

************************************************************/

double getdays(year, month, day)

int year, month, day;

{

int y,m;

double a,b,d, daynumb;

double floor(),intg();

/**********************************

** make correction for no year 0 **

**********************************/

if (year < 0) y = year + 1;

else y = year;

/*********************************************************

** Jan and Feb are months 13 and 14 in this calculation **

*********************************************************/

m = month;

if (month < 3)

{

m = m + 12;

y = y - 1;

}

/**************************

** calculate Julian days **

**************************/

d = floor(365.25 * y) + intg(30.6001 * (m + 1)) + day - 723244.0;

/**********************************************

** use Julian calendar if before Oct 5, 1582 **

**********************************************/

if (d < -145068.0) daynumb = d;

/*************************************

** otherwise use Gregorian calendar **

*************************************/

else

{

a = floor(y / 100.0);

b = 2 - a + floor(a / 4.0);

daynumb = d + b;

}

return(daynumb);

} /* end GETDAYS */

/********************************************************

* GETDATE - This routine takes a double float number *

* representing the number of days since Jan 1,*

* 1980 (day 1) and returns the year month and *

* day as pointer integers *

* This routine is the opposite of GETDAYS *

********************************************************/

getdate(numb)

double numb;

{

double a,aa,b,c,d,e,z;

double date;

date = numb;

z = intg(date + 2444239.0);

if (date < -145078.0) a = z;

else

{

aa = floor((z - 1867216.25) / 36524.25);

a = z + 1 + aa - floor(aa/4.0);

}

b = a + 1524.0;

c = intg((b - 122.1) / 365.25);

d = intg(365.25 * c);

e = intg((b - d) / 30.6001);

day = b - d - intg(30.6001 * e);

if (e > 13.5) month = e - 13.0;

else month = e - 1.0;

if (month > 2) year = c - 4716.0;

else year = c - 4715.0;

if (year < 1) --year;

return;

} /* end GETDATE */

/********************************************************

* WEEKDAYS - This routine takes a double float number *

* representing the number of days since Jan 1,*

* 1980 (day 1) and returns the day of the week*

* where 1 = Sunday, 2 = Tuesday, etc. *

********************************************************/

int weekdays(numb)

double numb;

{

double dd;

int day;

dd = numb;

while (dd > 28000.0) dd = dd - 28000.0;

while (dd < 0) dd = dd + 28000.0;

day = dd;

day = ((day + 1) % 7) + 1;

return(day);

}

/********************************************************

* FRACT - This routine takes a double float number *

* and returns the fractional part as a double *

* float number *

********************************************************/

double fract(numb)

double numb;

{

int inumb;

double fnumb;

while (numb < -32767) numb += 32767;

while (numb > 32767) numb -= 32767;

inumb = numb;

fnumb = inumb;

return(numb-fnumb);

} /* end FRACT */

/********************************************************

* FLOOR - This routine takes a double float number *

* and returns the next smallest integer *

********************************************************/

double floor(numb)

double numb;

{

double fract(), intg();

double out;

out = intg(numb);

if (numb < 0 && fract(numb) != 0) out -= 1.0;

return(out);

} /* end FLOOR */

/********************************************************

* INTG - This routine takes a double float number *

* and returns the integer part as a double *

* float number *

********************************************************/

double intg(numb)

double numb;

{

double fract();

return(numb - fract(numb));

} /* end INTG */

C语言程序设计万年历怎么编写

注意：下面的程序需要以命令行的格式运行，例如，把编绎好的可执行文件的名字为 filename，那要查看某2002/01/01就输入 filename 2002/02/01，要查看2002年1月份就输入filename 2002/01或filename2002/01/0。

/*** 万年历 ***/

/*** 命令行:

若查看某一天,例如 calen 2001/12/28

若查看某个月,例如 calen 2001/12/0 或 calen 2001/12

***/

/**************************************************

1. 以2000/01/01日为星期六为基准。

2. dy为该年 1月1号 到2000年1月1号的 " 星期差 "

3. dm为该月以前的月所引起的 " 星期差 "

4. m2为二月所引起的 " 星期差 "

**************************************************/

main(int x,char **date)

{

int year=0,month=0,day=0,week;

int d,i,dm,dy,m2;

char WEEK[9];

if (x==1)

{

printf ("\n ERROR! you forgot to enter the date you want to view\n");

exit (0);

}

i=0; d=-1;

while (date[1][i])

{

if ((date[1][i]=='/'||date[1][i]=='.')&&d==-1) { d=0; i++; continue; }

if ((date[1][i]=='/'||date[1][i]=='.')&&d==0) { d=1; i++; continue; }

if (d==-1) year=year*10+(date[1][i]-'0');

if (d==0) month=month*10+(date[1][i]-'0');

if (d==1) day=day*10+(date[1][i]-'0');

i++;

}

if (month<1||month>12)

{ printf ("\n ERROR! the entered MONTH is invalid\n"); exit (0); }

if (year==2000) { dy=0; m2=1; goto la_100; }

if (year>2000) d=(year-1-2000)/4-(year-1-2000)/100+(year-1-2000)/400+1;

else d=(year-2000)/4-(year-2000)/100+(year-2000)/400;

dy=(year-2000)+d; /*** 该年 1月1号 到2000年1月1号的 " 星期差 " ***/

if((year%4==0&&year%100!=0)||(year%100==0&&year%400==0))

m2=1; else m2=0; /*** 该年是否润 ***/

la_100: /**** la_100 ****/

/*** 该月以前的月所引起的 " 星期差 " ***/

switch (month)

{

case 1: dm=0; month=31; break; /*** month 在此存放该月天数 ***/

case 2: dm=3; month=d==1? 29:28; break;

case 3: dm=3+m2; month=31; break;

case 4: dm=6+m2; month=30; break;

case 5: dm=1+m2; month=31; break;

case 6: dm=4+m2; month=30; break;

case 7: dm=6+m2; month=31; break;

case 8: dm=2+m2; month=31; break;

case 9: dm=5+m2; month=30; break;

case 10: dm=m2; month=31; break;

case 11: dm=3+m2; month=30; break;

case 12: dm=5+m2; month=31; break;

}

if (day<0||day>month)

{ printf ("\n ERROR! the entered DAY is invalid\n"); exit (0); }

week=(dy+dm+day-1+6)%7; if(week<0) week+=7;

if (day>0) /*** 判断查看类型 ***/

{

switch (week)

{

case 0: strcpy (WEEK,"SUNDAY"); break;

case 1: strcpy (WEEK,"MONDAY"); break;

case 2: strcpy (WEEK,"TUESDAY"); break;

case 3: strcpy (WEEK,"WEDNESDAY"); break;

case 4: strcpy (WEEK,"THURSDAY"); break;

case 5: strcpy (WEEK,"FRIDAY"); break;

case 6: strcpy (WEEK,"SATURDAY"); break;

}

printf ("\n this day is %s \( %d \)\n\n OK!\n",WEEK,week);

}

else

{

week=++week%7;

printf ("\n the calender of this month as following\n");

printf ("\n *********************************\n");

printf (" SUN MON TUE WEN THU FRI STA\n");

for (i=0;i<week;i++) printf (" ");

for (i=1;i<=month;i++)

{ printf (" %2d ",i); week++; if (week%7==0&&i!=month) printf ("\n"); }

printf ("\n *********************************\n");

printf ("\n OK!\n");

}

}

#include <stdio.h>

int IsLeapYear(int);

main()

{

int i;

int day;

int year;

int temp;

int temp_i;

long int Year_days = 0;

int Year_Start = 1;

int Per_Year_Days;

int month_day[]={31,28,31,30,31,30,31,31,30,31,30,31,29};

printf("Please enter the year: ");

scanf("%d",&year);

while(Year_Start < year)

{

if( IsLeapYear( Year_Start ) )

Per_Year_Days = 366;

else

Per_Year_Days = 365;

Year_days = Year_days + Per_Year_Days;

Year_Start++;

}

for( temp = 1; temp <=12; temp++ )

{

switch( temp )

{

case 1:

printf(" January(%d)\n",year);

break;

case 2:

printf(" February(%d)\n",year);

break;

case 3:

printf(" March(%d)\n",year);

break;

case 4:

printf(" April(%d)\n",year);

break;

case 5:

printf(" May(%d)\n",year);

break;

case 6:

printf(" June(%d)\n",year);

break;

case 7:

printf(" July(%d)\n",year);

break;

case 8:

printf(" August(%d)\n",year);

break;

case 9:

printf(" September(%d)\n",year);

break;

case 10:

printf(" October(%d)\n",year);

break;

case 11:

printf(" November(%d)\n",year);

break;

case 12:

printf(" December(%d)\n",year);

break;

}

i = Year_days % 7;

printf("Mon Tue Wed Thu Fri Sat Sun\n");

if( i != 0 )

for( temp_i = 0; temp_i < i; temp_i++)

printf(" ");

day = 1;

if( IsLeapYear(year) && temp == 2)

while( day <= month_day[12] )

{

if( day >1 )

if( Year_days % 7 == 0 )

printf("\n");

if( day >= 10 )

printf("%d ",day);

else

printf("%d ",day);

Year_days++;

day++;

}

else

while (day <= month_day[temp-1])

{

if( day > 1 )

if( Year_days % 7 == 0 )

printf("\n");

if( day >=10 )

printf("%d ",day);

else

printf("%d ",day);

Year_days++;

day++;

}

printf("\n");

if( getch() == 'q' )

exit(0);

}

getch();

}

int IsLeapYear( int year )

{

if ((year %4 == 0) && (year % 100 != 0) ||

(year % 400 == 0) )

return 1;

else

return 0;

}

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