求一个日历系统的C语言课程设计 C语言课程设计-万年历
#include <windows.h>
int week(int y,int m,int d);
void main()
{
int monthday[12]={31,28,31,30,31,30,31,31,30,31,30,31};
int y,w,i,m=1,d=1;
printf("请输入一个年份yyyy：\n");
scanf("%d",&y);
if (y%4==0&&y%100!=0) monthday[1]=29;
for(m=1;m<=12;m++)
{
printf("\n %d年,%d月\n",y,m);
printf("S M T W T F S \n");
for (d=1;d<=monthday[m-1];d++)
{
w=week(y,m,d);
if(d==1)
{
for (i=0;i<w;i++) printf(" ");
}
if(d<10) printf("%d ",d);
else printf("%d ",d);
if(w==6) printf(" \n");
}
}
scanf("%d",&y);
}
int week (int y,int m,int d)
{
int w;
if((m==1)||(m==2))
{
y--;
m+=12;
}
w=(d+2*m+3*(m+1)/5+y+y/4-y/100+y/400+1)%7;
return (w);
}
没有阴历
因为阴历没用什么固定规则
呵呵
急求： C语言课程设计 日历查询系统~
#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("
%02d-%02d-%d is",month, day, year);
if (numbnow - daynumb > 0)
printf("
%02d-%02d-%d was",month, day, year);
if (numbnow - daynumb < 0)
printf("
%02d-%02d-%d will be",month, day, year);
printf(" a %s
",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 */
1.系统封面设计
内容：题目名称（中英文）、进入(中英文)、作者：***、时间：****-**-**
2.输入界面
内容：（1）密码口令输入及容错（3次）
（2）年份的输入及容错（3次）
3.日历计算设计
（1）求某月某日实行奇迹的函数（邱每月一号的星期数）。算法：（y-1）+(y-1)/4-（y-1）/100+(y-1)/400+c 得出S (y为年份，c为某月某日是这一年的第几天，S为总的天数)
（2）邱某月某日至这一年第几天的函数（求c） 闰年的算法: y/400==0 ?或y/400==0&&y/100!=0
根据（1）（2）求出每个月的日期
4、日历格式显示设计
要求：每屏显示4个月。
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