C程序找到给定日期的星期几

Question

有没有一种方法可以在一行C代码中找出给定日期的星期几？

例如

由于19-05-2011（DD-MM-YYYY）给我周四

Answer 1

一行代码的可能性不大，但strptime function可以用来解析您的日期格式和struct tm参数可以用于查询其tm_wday成员在系统上自动修改这些字段（例如一些glibc实现）。

int get_weekday(char * str) { 
    struct tm tm; 
    memset((void *) &tm, 0, sizeof(tm)); 
    if (strptime(str, "%d-%m-%Y", &tm) != NULL) { 
    time_t t = mktime(&tm); 
    if (t >= 0) { 
     return localtime(&t)->tm_wday; // Sunday=0, Monday=1, etc. 
    } 
    } 
    return -1; 
} 

或者你也可以编码这些规则做一些算术很长的一行：

• 1900年1月1日是个周一。
• 九月，四月，六月和十一月有三十天;其余的人都有三十一人，单是二月就能拯救，二十八岁，有雨，二十八岁，闰年二十九岁。发生
• 闰年上任何一年能被4整除，但不能在一个世纪，除非它是由400

编辑整除：请注意，此解决方案仅适用于UNIX新纪元后日期（ 1970-01-01T00：00：00Z）。

Answer 2

不在一行代码，没有什么关于处理C标准库的日期。尽管如此，编写一个基于Doomsday算法的函数或类似的函数会相当简单。

Answer 3

我认为你可以发现，在巧舌如簧： http://developer.gnome.org/glib/unstable/glib-Date-and-Time-Functions.html#g-date-get-day

问候

Answer 4

下面是基于wikipedia's article about Julian Day

#include <stdio.h> 

const char *wd(int year, int month, int day) { 
    /* using C99 compound literals in a single line: notice the splicing */ 
    return ((const char *[])           \ 
      {"Monday", "Tuesday", "Wednesday",      \ 
      "Thursday", "Friday", "Saturday", "Sunday"})[   \ 
     (               \ 
      day              \ 
     + ((153 * (month + 12 * ((14 - month)/12) - 3) + 2)/5) \ 
     + (365 * (year + 4800 - ((14 - month)/12)))    \ 
     + ((year + 4800 - ((14 - month)/12))/4)    \ 
     - ((year + 4800 - ((14 - month)/12))/100)    \ 
     + ((year + 4800 - ((14 - month)/12))/400)    \ 
     - 32045             \ 
    ) % 7]; 
} 

int main(void) { 
    printf("%d-%02d-%02d: %s\n", 2011, 5, 19, wd(2011, 5, 19)); 
    printf("%d-%02d-%02d: %s\n", 2038, 1, 19, wd(2038, 1, 19)); 
    return 0; 
} 

C99的版本，通过在WD（）函数，它可以压缩到286字符单行:)

去除的 return线的拼接和空间

---

编辑：版本2（3行）---避免可能出现的有关通过使自动数组元素返回静态的复杂性。

#include <stdio.h> 

const char *wd(int year, int month, int day) { 
    static const char *weekdayname[] = {"Monday", "Tuesday", 
     "Wednesday", "Thursday", "Friday", "Saturday", "Sunday"}; 
    size_t JND =              \ 
      day              \ 
     + ((153 * (month + 12 * ((14 - month)/12) - 3) + 2)/5) \ 
     + (365 * (year + 4800 - ((14 - month)/12)))    \ 
     + ((year + 4800 - ((14 - month)/12))/4)    \ 
     - ((year + 4800 - ((14 - month)/12))/100)    \ 
     + ((year + 4800 - ((14 - month)/12))/400)    \ 
     - 32045; 
    return weekdayname[JND % 7]; 
} 

int main(void) { 
    printf("%d-%02d-%02d: %s\n", 2011, 5, 19, wd(2011, 5, 19)); 
    printf("%d-%02d-%02d: %s\n", 2038, 1, 19, wd(2038, 1, 19)); 
    return 0; 
} 

Answer 5

这一个工程：我把2006年1月作为参考。（这是一个星期日）

enter code here 

INT isLeapYear（INT年）{

 if(((year%4==0)&&(year%100!=0))||((year%400==0))) 
     return 1; 
    else 
     return 0; 

} 

INT isDateValid（INT DD，INT毫米，INT YYYY）{

int isValid=-1; 
    if(mm<0||mm>12) { 
     isValid=-1; 
    } 
    else { 

    if((mm==1)||(mm==3)||(mm==5)||(mm==7)||(mm==8)||(mm==10)||(mm==12)) { 
      if((dd>0)&&(dd<=31)) 
      isValid=1; 
    } else if((mm==4)||(mm==6)||(mm==9)||(mm==11)) { 
     if((dd>0)&&(dd<=30)) 
      isValid=1; 
    } else { 
      if(isLeapYear(yyyy)){ 
       if((dd>0)&&dd<30) 
        isValid=1; 
      } else { 
        if((dd>0)&&dd<29) 
        isValid=1; 
      } 
     } 



    } 
       return isValid; 



} 

INT calculateDayOfWeek（ int dd，int mm，int yyyy）{

   if(isDateValid(dd,mm,yyyy)==-1) { 
        return -1; 
       } 

       int days=0; 
        int i; 

        for(i=yyyy-1;i>=2006;i--) { 
         days+=(365+isLeapYear(i)); 

        } 
        printf("days after years is %d\n",days); 

        for(i=mm-1;i>0;i--) { 

         if((i==1)||(i==3)||(i==5)||(i==7)||(i==8)||(i==10)) { 

          days+=31; 
         } 
         else if((i==4)||(i==6)||(i==9)||(i==11)) { 
          days+=30; 
         } else { 

          days+= (28+isLeapYear(i)); 

         } 


        } 
         printf("days after months is %d\n",days); 

        days+=dd; 

        printf("days after days is %d\n",days); 

        return ((days-1)%7);     


       }` 

Answer 6

T他是我的执行。它很短，包括错误检查。如果你想在01-01-1900之前的日期，你可以很容易地改变锚到公历日期的开始日期。

#include <stdio.h> 

int main(int argv, char** arv) { 
    int month[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; 
    char* day[] = {"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"}; 
    int d, m, y, i; 

    printf("Fill in a date after 01-01-1900 as dd-mm-yyyy: "); 
    scanf("%d-%d-%d", &d, &m, &y); 

    // correction for leap year 
    if (y % 4 == 0 && (y % 100 != 0 || y % 400 == 0)) 
     month[1] = 29; 

    if (y < 1900 || m < 1 || m > 12 || d < 1 || d > month[m - 1]) { 
     printf("This is an invalid date.\n"); 
     return 1; 
    } 

    for (i = 1900; i < y; i++) 
     if (i % 4 == 0 && (i % 100 != 0 || i % 400 == 0)) 
      d += 366; 
     else 
      d += 365; 

    for (i = 0; i < m - 1; i++) 
     d += month[i]; 

    printf("This is a %s.\n", day[d % 7]); 
    return 0; 
} 

Answer 7

#include<stdio.h> 
static char day_tab[2][13] = { 
     {0,31,28,31,30,31,30,31,31,30,31,30,31}, 
     {0,31,29,31,30,31,30,31,31,30,31,30,31} 
     }; 
int main() 
{ 
    int year,month; 
    scanf("%d%d%d",&year,&month,&day); 
    printf("%d\n",day_of_year(year,month,day)); 
    return 0; 
} 
int day_of_year(int year ,int month,int day) 
{ 
     int i,leap; 
     leap = year%4 == 0 && year%100 != 0 || year%400 == 0; 
     if(month < 1 || month >12) 
     return -1; 
     if (day <1 || day > day_tab[leap][month]) 
     return -1; 
     for(i= 1;i<month ; i++) 
     { 
       day += day_tab[leap][year]; 
     } 
     return day; 
} 

Answer 8

作为还报道通过Wikipedia，1990年迈克尔Keith和汤姆Craver发表的表达以最小化进入一个自包含的功能用于转换公历日期到的数值一天所需的击键次数星期。

表达确实保持既不y也不d，并返回一个从零开始的索引表示一天，星期日开始，即，如果日期是星期一表达式返回1。

其使用表达的代码示例如下：

int d = 15 ; //Day  1-31 
int m = 5 ; //Month 1-12` 
int y = 2013 ; //Year 2013` 

int weekday = (d += m < 3 ? y-- : y - 2, 23*m/9 + d + 4 + y/4- y/100 + y/400)%7; 

表达式使用逗号操作，如在this answer讨论。

享受！;-)

Answer 9

/* 
Program to calculate the day on a given date by User 
*/ 

#include<stdio.h> 
#include<conio.h> 
#include<process.h> 
void main() 
{ 
int dd=0,mm=0,i=0,yy=0,odd1=0,todd=0;//variable declaration for inputing the date 
int remyr=0,remyr1=0,lyrs=0,oyrs=0,cyr=0,upyr=0,leap=0;//variable declaration for calculation of odd days 
int montharr[12]={31,28,31,30,31,30,31,31,30,31,30,31};//array of month days 
clrscr(); 
printf("Enter the date as DD-MM-YY for which you want to know the day\t:"); 
scanf("%d%d%d",&dd,&mm,&yy); //input the date 
/* 
check out correct date or not? 
*/ 
if(yy%100==0) 
    { 
    if(yy%400==0) 
     { 
     //its the leap year 
     leap=1; 
     if(dd>29&&mm==2) 
      { 

      printf("You have entered wrong date"); 
      getch(); 
      exit(0); 
      } 
     } 
    else if(dd>28&&mm==2) 
     { 
     //not the leap year 
     printf("You have entered wrong date"); 
     getch(); 
     exit(0); 

     } 
    } 
else if(yy%4==0) 
    { 
    //again leap year 
    leap=1; 
    if(dd>29&mm==2) 
     { 
     printf("You have entered wrong date"); 
     getch(); 
     exit(0); 

     } 
    } 
else if(dd>28&&mm==2) 
    { 
    //not the leap year 
    printf("You have entered wrong date"); 
    getch(); 
    exit(0); 

    } 
//if the leap year feb month contains 29 days 
if(leap==1) 
{ 
montharr[1]=29; 

} 
//check date,month,year should not be beyond the limits 

if((mm>12)||(dd>31)|| (yy>5000)) 
    { 
    printf("Your date is wrong"); 
    getch(); 
    exit(0); 

    } 
//odd months should not contain more than 31 days 
if((dd>31 && (mm == 1||mm==3||mm==5||mm==7||mm==8||mm==10||mm==12))) 
    { 
    printf("Your date is wrong"); 
    getch(); 
    exit(0); 

    } 
//even months should not contains more than 30 days 

if((dd>30 && (mm == 4||mm==6||mm==9||mm==11))) 
    { 
    printf("Your date is wrong"); 
    getch(); 
    exit(0); 

    } 

//logic to calculate odd days..... 
printf("\nYou have entered date: %d-%d-%d ",dd,mm,yy); 
remyr1=yy-1; 
remyr=remyr1%400; 

cyr=remyr/100; 
if(remyr==0) 
    { 
    oyrs=0; 
    } 
else if(cyr==0 && remyr>0) 
    { 
    oyrs=0; 
    } 


else if(cyr==1) 
    { 
    oyrs=5; 
    } 
else if(cyr==2) 
    { 
    oyrs=3; 
    } 
else if(cyr==3) 
    { 
    oyrs=1; 
    } 

    upyr=remyr%100; 
    lyrs=upyr/4; 
    odd1=lyrs+upyr; 
    odd1=odd1%7; 
    odd1=odd1+oyrs; 
    for(i=0;i<mm-1;i++) 
     { 
     odd1=odd1+montharr[i]; 
     } 
    todd=odd1+dd; 
    if(todd>7) 
     todd=todd%7; //total odd days gives the re quired day.... 
    printf("\n\nThe day on %d-%d-%d :",dd,mm,yy); 

    if(todd==0) 
     printf("Sunday"); 
    if(todd==1) 
     printf("Monday"); 
    if(todd==2) 
     printf("Tuesday"); 
    if(todd==3) 
     printf("Wednesday"); 
    if(todd==4) 
     printf("Thrusday"); 
    if(todd==5) 
     printf("Friday"); 
    if(todd==6) 
     printf("Saturday"); 
getch(); 
} 

Answer 10

我想出了答案：

const int16_t TM_MON_DAYS_ACCU[12] = { 
    0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334 
}; 

int tm_is_leap_year(unsigned year) { 
    return ((year & 3) == 0) && ((year % 400 == 0) || (year % 100 != 0)); 
} 

// The "Doomsday" the the day of the week of March 0th, 
// i.e the last day of February. 
// In common years January 3rd has the same day of the week, 
// and on leap years it's January 4th. 
int tm_doomsday(int year) { 
    int result; 
    result = TM_WDAY_TUE; 
    result += year;  // I optimized the calculation a bit: 
    result += year >>= 2; // result += year/4 
    result -= year /= 25; // result += year/100 
    result += year >>= 2; // result += year/400 
    return result; 
} 

void tm_get_wyday(int year, int mon, int mday, int *wday, int *yday) { 
    int is_leap_year = tm_is_leap_year(year); 
    // How many days passed since Jan 1st? 
    *yday = TM_MON_DAYS_ACCU[mon] + mday + (mon <= TM_MON_FEB ? 0 : is_leap_year) - 1; 
    // Which day of the week was Jan 1st of the given year? 
    int jan1 = tm_doomsday(year) - 2 - is_leap_year; 
    // Now just add these two values. 
    *wday = (jan1 + *yday) % 7; 
} 

这些定义（匹配struct tm的time.h）：

#define TM_WDAY_SUN 0 
#define TM_WDAY_MON 1 
#define TM_WDAY_TUE 2 
#define TM_WDAY_WED 3 
#define TM_WDAY_THU 4 
#define TM_WDAY_FRI 5 
#define TM_WDAY_SAT 6 

#define TM_MON_JAN 0 
#define TM_MON_FEB 1 
#define TM_MON_MAR 2 
#define TM_MON_APR 3 
#define TM_MON_MAY 4 
#define TM_MON_JUN 5 
#define TM_MON_JUL 6 
#define TM_MON_AUG 7 
#define TM_MON_SEP 8 
#define TM_MON_OCT 9 
#define TM_MON_NOV 10 
#define TM_MON_DEC 11 

Answer 11

#include<stdio.h> 
#include<math.h> 
#include<conio.h> 
int fm(int date, int month, int year) { 
int fmonth, leap; 
if ((year % 100 == 0) && (year % 400 != 0)) 
leap = 0; 
    else if (year % 4 == 0) 
    leap = 1; 
else 
    leap = 0; 

    fmonth = 3 + (2 - leap) * ((month + 2)/(2 * month))+ (5 * month + month/9)/2; 

fmonth = fmonth % 7; 

    return fmonth; 
} 
int day_of_week(int date, int month, int year) { 

    int dayOfWeek; 
    int YY = year % 100; 
    int century = year/100; 

    printf("\nDate: %d/%d/%d \n", date, month, year); 

    dayOfWeek = 1.25 * YY + fm(date, month, year) + date - 2 * (century % 4); 

    //remainder on division by 7 
    dayOfWeek = dayOfWeek % 7; 

    switch (dayOfWeek) { 
     case 0: 
     printf("weekday = Saturday"); 
     break; 
     case 1: 
     printf("weekday = Sunday"); 
     break; 
     case 2: 
     printf("weekday = Monday"); 
     break; 
     case 3: 
     printf("weekday = Tuesday"); 
     break; 
     case 4: 
     printf("weekday = Wednesday"); 
     break; 
     case 5: 
     printf("weekday = Thursday"); 
     break; 
     case 6: 
     printf("weekday = Friday"); 
     break; 
     default: 
     printf("Incorrect data"); 
    } 
    return 0; 
} 
int main() { 
    int date, month, year; 

    printf("\nEnter the year "); 
    scanf("%d", &year); 

    printf("\nEnter the month "); 
    scanf("%d", &month); 

    printf("\nEnter the date "); 
    scanf("%d", &date); 

    day_of_week(date, month, year); 

    return 0; 
} 

OUTPUT： 输入年月日2012

输入月02

输入日期29

日期：29/2/2012

平日=周三

Answer 12

超高效的班轮C代码为周查找日

int dayOfWeek(int y, int m, int d) { 
    return ((y-=m<3)+y/4-y/100+y/400+"-bed=pen+mad."[m]+d)%7; 
} 

来源-here ↗

详细信息 - here ↗

Answer 13

#include<stdio.h> 
int main(void) { 
int n,y; 
int ly=0; 
int mon; 
printf("enter the date\n"); 
scanf("%d",&n); 
printf("enter the month in integer\n"); 
scanf("%d",&mon); 
mon=mon-1; 
printf("enter year\n"); 
scanf("%d",&y); 
int dayT; 
dayT=n%7; 
if((y%4==0&&y%100!=0)|(y%4==0&&y%100==0&&y%400==0)) 
{ 
    ly=y; 
    printf("the given year is a leap year\n"); 
} 
char a[12]={6,2,2,5,0,3,5,1,4,6,2,4}; 
if(ly!=0) 
{ 
    a[0]=5; 
    a[1]=1; 
} 
int m,p; 
m=a[mon]; 
int i,j=0,t=1; 
for(i=1600;i<=3000;i++) 
{ 
i=i+99; 
    if(i<y) 
    { 

     if(t==1) 
     { 
     p=5;t++; 
     } 
     else if(t==2) 
     { 
      p=3; 
      t++; 
     } 
     else if(t==3) 
     { 
      p=1; 
      t++; 
     } 
     else 
     { 
      p=0; 
      t=1; 
     } 

}} 
int q,r,s; 
q=y%100; 
r=q%7; 
s=q/4; 
int yTerm; 
yTerm=p+r+s; 
int w=dayT+m+yTerm; 
w=w%7; 
if(w==0) 
printf("SUNDAY"); 
else if(w==1) 
printf("MONDAY"); 
else if(w==2) 
printf("TUESDAY"); 
else if(w==3) 
printf("WEDNESDAY"); 
else if(w==4) 
printf("THURSDAY"); 
else if(w==5) 
printf("FRIDAY"); 
else 
printf("SATURDAY"); 
return 0; 

}