c语言简单万年历程序设计

#include <stdio.h>#include <stdlib.h>#include <ctype.h>int leap (int year) ;//判断是否为闰年int days_month (int month,int year) ;//判断这个月的天数int menu_select();/*void f(){int day,month,year,sum,leap,S;printf("\n请输入年月日\n");scanf("%d%d%d",&year,&month,&day);switch(month){case 1:sum=0;break;case 2:sum=31;break;case 3:sum=59;break;case 4:sum=90;break;case 5:sum=120;break;case 6:sum=151;break;case 7:sum=181;break;case 8:sum=212;break;case 9:sum=243;break;case 10:sum=273;break;case 11:sum=304;break;case 12:sum=334;break;default:printf("data error");break;}sum=sum+day;if(year%400==0||(year%4==0&&year%100!=0))leap=1;else leap=0;if(leap==1&&month>2)sum++;S=(year-1+(year-1)/4-(year-1)/100+(year-1)/400+sum)%7; //X表示年份,C是该年份元旦开始到这一日的天数//S/7的系数就是星期数//01printf("%d",S);switch(S){case 1:printf("星期一\n");break;case 2:printf("星期二\n");break;case 3:printf("星期三\n");break;case 4:printf("星期四\n");break;case 5:printf("星期五\n");break;case 6:printf("星期六\n");break;case 0:printf("星期日\n");break;}}*/int Leap(int year){ int leap;if(year%400==0||(year%4==0&&year%100!=0))leap=1;else leap=0;return leap;}int SU(int sum,int year){int S;S=(year-1+(year-1)/4-(year-1)/100+(year-1)/400+sum)%7;return S;}int Sum(int year,int month,int day=1){int sum,leap;switch(month){case 1:sum=0;break;case 2:sum=31;break;case 3:sum=59;break;case 4:sum=90;break;case 5:sum=120;break;case 6:sum=151;break;case 7:sum=181;break;case 8:sum=212;break;case 9:sum=243;break;case 10:sum=273;break;case 11:sum=304;break;case 12:sum=334;break;default:printf("data error");break;}sum=sum+day;leap=Leap(year);if(leap==1&&month>2)sum++;return sum;}void Print(int S){switch(S){case 1:printf("星期一\n");break;case 2:printf("星期二\n");break;case 3:printf("星期三\n");break;case 4:printf("星期四\n");break;case 5:printf("星期五\n");break;case 6:printf("星期六\n");break;case 0:printf("星期日\n");break;}}void Fun1(){int day,month,year,sum,S;printf("\n请输入年月日\n");scanf("%d%d%d",&year,&month,&day);sum=Sum(year,month,day);S=SU(sum,year);Print(S);}void Fun2(){int i,j=1,k=1,a,b,month,year,sum;printf("\n 输入年月:\n");scanf("%d%d",&year,&month);b=days_month(month,year);sum=Sum(year,month);a=SU(sum,year);printf("*************************************\n");printf(" Sun Mon Tue Wed Thu Fri Sat \n");if(a==7){for(i=1;i<=b;i++){printf("%4d",i);if(i%7==0){printf("\n");}}}if(a!=7){while (j<=4*a){printf(" ");j++;}for(i=1;i<=b;i++){printf("%4d",i);if(i==7*k-a){printf("\n");k++;}}}printf("\n*************************************\n");printf("\n");}int days_month (int month,int year){if(month==1||month==3||month==5||month==7||month==8||month==10||month==12) return 31;if(month==4||month==6||month==9||month==11)return 30;if(month==2&&Leap(year)==1) return 29;else return 28;}void main(){for(;;){switch(menu_select()){case 1:printf("输入年月日计算该月第一天是星期几\n");Fun1();break;case 2:printf("打印任意日历\n");Fun2();break;case 3:printf("谢谢使用!\n");exit(0);}}}int menu_select(){ system("cls");char s;int cn;printf("1. 输入年月日计算该月第一天是星期几\n");printf("2. 打印任意日历\n");printf("3.谢谢使用! \n");printf("input 1-3:");do {s=getchar();cn=(int)s-48;}while(cn<0||cn>6);return cn;}。

#include"stdio.h"int mon_day[]={31,28,31,30,31,30,31,31,30,31,30,31};int judge(int year,int month){if(month==1||month==3||month==5||month==7||month==8||month==10||month ==12)return(1);else if(month==2){if(year%4!=0||year%100==0&&year%400!=0)return(2);else return(3);}else return(4);}void show2(){int year,i,j,a,n,m,k;char ** p;char * week[]={"Sun","Mon","Tue","Wed","Thu","Fri","Sat"};char*month[]={"January","February","March","April","May","June","July","Augu st","September","October","November","December"};printf("please inter the year:");scanf("%d",&year);printf("\n");printf("the calendar of the year%d.",year);printf("\n");a=(year+(year-1)/4-(year-1)/100+(year-1)/400)%7;for(i=0;i<12;i++){n=judge(year,i+1);p=month+i;printf("%s\n",*p);printf("\n");for(j=0;j<7;j++){p=week+j;printf("%6s",*p);}printf("\n");for(k=0;k<a;k++)printf(" ");for(m=1;m<32;m++){printf("%6d",m);if((a+m)%7==0)printf("\n");if(n==1&&m==31) break;else if(n==2&&m==28) break;else if(n==3&&m==29) break;else if(n==4&&m==30) break;}a=(a+m)%7;/*计算出下个月的第一天是星期几*/printf("\n");printf("**************************************************");printf("\n");}}void show(){printf("******************************\n");printf("1.某年某月某日是星期几\n2.是否闰年\n3.某年某月的最大天数\n4.打印该月日历\n5.退出\n");printf("******************************\n");}int show1(int y,int m,int d){int i,R,sum=0;for(i=0;i<m-1;i++)sum=sum+mon_day[i];sum=sum+d;if((y%4 == 0 &&y%100 != 0)||(y%400 == 0)&&m>2)sum=sum+1;R=(5+(y+1)+(y-1)/4-(y-1)/100+(y-1)/400+sum)%7;if(R==0)R=7;return R;}int leap(int y){if((y%4 == 0 &&y%100 != 0)||(y%400 == 0))return 1;elsereturn 0;}int max_day(int y,int m){int X;if(leap(y)==1&&m==2)X=29 ;//printf("该年该月最大天数:29");elseX=mon_day[m-1];//printf("%d",mon_day[m-1]);return X;}void main(){int year,month,day,i;F:show();printf("请选择：");scanf("%d",&i);switch(i){case 1:printf("请输入年、月，日：");scanf("%d%d%d",&year,&month,&day);printf("%d-%d-%d是星期%d\n",year,month,day,show1(year,month,day));break;case 2:printf("请输入年份：");scanf("%d",&year);if(leap(year)==1)printf("该年是闰年.\n");elseprintf("该年不是闰年\n");break;case 3:printf("请输入年、月：");scanf("%d%d",&year,&month);printf("该月最大天数：%d",max_day(year,month)); printf("\n");break;case 4:show2();break;case 5:return 0;break;}goto F;}。

这是当时我做的一个小小的课题，希望对你有所帮助#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 */。

万年历的C语言编程实现源程序：#include<stdio.h>#include<stdlib.h>int Isleapyear(int y)/*判断是否是闰年*/{if((y%4==0&&y%100!=0)||(y%400==0))return 1;elsereturn 0;}int Getdaysofmonth(int y,int m)/*确定某个月的天数*/{int months[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};if(Isleapyear(y))/*如果是闰年则2月份的天数改变为29*/ months[2]=29;return months[m];}int Gettotaldays(int y,int m)/*计算从1901年1月开始到给定年月的天数*/{intmonths[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};int days=0;int i,j;if((Isleapyear(y)))months[2]=29;for(i=1901;i<y;i++)/*从1901年1月开始计算，累加y年之前的天数*/{if(Isleapyear(i))days+=366;elsedays+=365;}for(j=1;j<m;j++)/*累加m月之后的天数*/days+=months[j];return days;}int Getfirstdayintable(int y,int m)/*计算给定年月的第一天在日历表中的位置*/{int d;d=Gettotaldays(y,m);d=d+3;/*1901年1月1日是星期二，在日历中星期二是第三个位置*/d=d%7;return d;}void Showdate(int y,int m)/*显示日历*/{int first;int d;int i,k;first=Getfirstdayintable(y,m);/*得到第一天在日历中的第一个位置*/d=Getdaysofmonth(y,m);/*得到这个月的天数*/k=0;printf("-------------------------------------\n");printf(" SUN MON TUE WEN THU FRI STA\n");for(i=1;i<first;i++)/*打印第一天之前的空格*/{printf(" ");k++;/*k用于确定日历中是否回车*/}for(i=1;i<=d;i++){printf("%5d",i);/*每个具体的日期占5个字符的宽度*/k++;if(k==7)/*每行打印7个日期数字后回车*/{printf("\n");k=0;}}printf("\n-------------------------------------\n"); }int main(){int y,m;printf("Input year:");scanf("%d",&y);printf("Input month:");scanf("%d",&m);Showdate(y,m);return 0;}欢迎您的下载，资料仅供参考！致力为企业和个人提供合同协议，策划案计划书，学习资料等等打造全网一站式需求。

C语言编写万年历【要求】：1．程序运行后，首先在屏幕上显示主菜单：2．查询某年某月某日是星期几3．查询某年是否是闰年4．打印某年的全年日历5．退出2．在主菜单中输入1后，显示：“请输入年月日（XXXX，XX，XX）”运行后输出：XXXX年XX月XX日是星期X，是否继续查询（Y/N）？如果输入Y，则重新显示“请输入年月日（XXXX，XX，XX）”，否则回到主菜单。

3．在主菜单中输入2后，显示：“请输入要查哪一年？（XXXX）”运行后输出：XXXX年是（否）是闰年，是否继续查询（Y/N）？如果输入Y，则重新显示，“请输入要查哪一年？（XXXX）”，否则回到主菜单。

4．在主菜单中输入3后，显示：“请输入要打印的年份（XXXX）”运行后输出XXXX年的日历，格式为：XXXXX（月数.奇） X（月数.）0 1 2 3 4 5 6 0 1 2 3 4 5 6 S M T W T F S S M T W T F S x x x x x x x x x x xx x x xx xx xx xx x x x xx xx xx xx xx xx xx xx xx xx xx x x x xx xx xx xx xx xx xx xx xx xx xx xx xx xx xx xx xx xx xx xx xx xx xx xx xx xx xx xx 运行完后显示：“是否继续打印（Y/N）？”如果输入Y，则重新显示，“请输入要打印的年份（XXXX）”，否则回到主菜单。

5．在主菜单中输入4后，显示：“是否要真的退出（Y/N）？”如果输入Y，结束程序运行，否则重新显示主菜单。

代码来自东篱下，Turbo C&C++3.0编译调试通过，已经验证输出结果正确：#include <stdio.h>#include <stdlib.h>#include <conio.h>#define X " Sun Mon Tue Wed Thu Fri Sat"void menu(){system("cls");printf("\nTHIS IS THE MENU OF THE PROGRAM !");printf("\nYOU CAN CHOOSE THE NUMBER FOR THE FUNCTIOM:"); printf("\n\n");printf("1 Find the day by year-month-date.\n");printf("2 Find the year you input if the leap year.\n");printf("3 Print the calendar of the year you input.\n");printf("4 Exit.\n\n");printf("Input your choice:");}/*函数定义，将在主函数中调用*/int getday(int year,int month,int date){int flag,s,i;int a[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};int cont=0;flag=isleap(year);if(flag==1)/*修改二月份天数*/a[2]++;for(i=1;i<month;i++){cont=cont+a[i];}cont=cont+date;s=year+1+(year-1)/4+(year-1)/100+(year-1)/400+cont; return s%7;/*利用公式求星期几*/}int isleap(int year){if(year%4==0&&year%100||year%400==0)return 1;elsereturn 0;}/*判定闰年*/void print(int n){int i;for(i=0;i<n;i++) printf(" ");}/*打印空格，排版使用*/int day(int year){int a,b;if(year<=2000){a=2000-year;b=6-(a+a/4-a/100+a/400)%7;return b;}else{a=year-2000;b=(a+1+(a-1)/4-(a-1)/100+(a-1)/400)%7+6;return b%7;}}/*打印日历函数*/void printcalendar(int year){int i,j,k,m,n,f1,f2,d;int a[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};printf("\nThe calendar of the year %d.\n\n",year);d=day(year);if(isleap(year)==1)a[2]++;for(i=1;i<=12;i+=2){m=0; n=0; f1=0; f2=0;switch(i){case 1:printf(" Januray 1 ");break;case 3:printf(" March 3 ");break;case 5:printf(" May 5 ");break;case 7:printf(" July 7 "); break;case 9:printf(" September 9 ");break;case 11:printf(" Nevember 11 ");break;}print(21); /*打印空格，隔开奇偶月份名*/switch(i+1){case 2:printf(" February 2 "); break;case 4:printf(" April 4 "); break;case 6:printf(" June 6 "); break;case 8:printf(" August 8 "); break;case 10:printf(" October 10 "); break;case 12:printf(" December 12"); break;}printf("\n");printf(X); /*宏调用，打印星期字符串*/print(6);/*打印空格，隔开奇偶月份数据*/printf(X);printf("\n");for(j=0;j<6;j++){if(j==0){print(d*4); /*每个日期数采用个空格的宽度，即4字节。

C语言课程设计计算机科学与技术专业****班学号：******姓名：******目录序言 (1)说明 (2)流程图 (3)源代码 (4)小结 (5)序言1. 经过上一个学期对《C程序设计》的学习，同学们都认为真正的学到了有用知识，可能有一些人学的不够理想，但无论如何，这些知识都为我们的下一步学习打下了坚实的基础。

做这么一个课程设计，一方面是为了检查我们一个学期来我们学习的成果，另一方面也是为了让我们进一步的掌握和运用它，同时也让我们认清自己的不足之处和薄弱环节，加以弥补和加强。

本程序是一个万年历程序，可以输出公元一年一月以后任意月份的天数及每一天是星期几，只要在运行程序时按照要求输入年份再输入和月份，程序就会把这一月份的日历显示在屏幕上，使人一目了然。

该程序能与目前天文学研究中所采用的各种日历程序所能计算的范围相一致。

2 . 课程设计题目：万年历要求：输入年份和月份，自动输出该月的日历，清楚的显示每一天是星期几。

指导老师：翟海霞说明一．算法分析说明：1. 总天数的算法：首先用if语句判断定义年到输入年之间每一年是否为闰年，是闰年，该年的总天数为366，否则，为355。

然后判断输入的年是否为定义年，若是，令总天数S=1，否则，用累加法计算出定义年到输入年之间的总天数，再把输入年的一月到要输出的月份之间的天数累加起来，若该月是闰年中的月份并且该月还大于二月，再使总天数加1，否则，不加，既算出从定义年一月一日到输出年的该月一日的总天数。

2. 输出月份第一天为星期几的算法：使总天数除以7取余加2得几既为星期几，若是7，则为星期日。

3. 算出输出月份第一天为星期几的算法：算出输出月份第一天为星期几后，把该日期以前的位置用空格补上，并总该日起一次输出天数直到月底，该月中的天数加上该月一日为星期几的数字再除以7得0换行，即可完整的输出该月的日历。

二．要用到的函数和语句1.<stdio.h>,<conio.h>,<math.h> /*头文件*/2.main() /*主函数*/3.printf(),4.scanf()5.textbackground(),textcolor() /*定义背景和字体颜色*/6.if 语句7.for 语句8.printstar() /*调用函数*/9.int day_year() /*定义函数*/10.goto /*循环语句*/流程图1.图一输入的年份判断从定义年到输入的年份中每个年份是否为闰年是否每年为366天每年为365天返回该年的天数2.图二输入的年月判断输入年是否为定义年是否总天数S=1 总天数S为定义年到输入年的总天数把S加上该年一月到输入月份的天数，再加1判断输入的月份是否为闰年中的月份是否该月是否大于二月是否把S加上该年一月到输入月份的天数把S再加上该把S加上该年年一月到输入一月到输入月的月份的天数份的天数再加1总天数为S判断S的最后一天为星期几从这一天起依次输出输入月份的天数直到该月底为止源代码#include<conio.h>#include<stdio.h>#include <math.h>main(){long s=1;int a[14]={0,0,31,28,31,30,31,30,31,31,30,31,30,31}; int y,m,p,i,j,k,n,leap;int day_year(int y);textbackground(RED);textcolor(YELLOW);clrscr();printf("please input the year:");scanf("%d",&y);printf("\n");printf("please input the month:");scanf("%d",&m);printf("\n");textbackground(3);clrscr();printf("\tYear:%d Month:%d\n",y,m);printf("\n");if(y==1)s=1;for(i=1;i<y;i++)s=s+day_year(y-1);if(y%4==0&&y%100!=0||y%400==0)leap=1;else leap=0;for(i=0;i<=m;i++)s=s+a[i];if(leap==1&&m>2)s=s+1;else s=s+0;p=s%7+3;printf(" Sun Mon Tue Wed Thu Fri Sat\n"); printstar();if(p<=7)n=4*(p-1);else n=4*(p-8);for(j=1;j<=n;j++)printf(" ");if(leap==1&&m==2)a[3]=a[3]+1;else a[3]=a[3]+0;for(j=1;j<=a[m+1];j++){printf("%4d",j);if((j+n/4)%7==0)printf("\n");}printf("\n");printstar();printstar();s=1;printf("do you want contintue(y/n)");scanf("%c",&q);for(;;){int k;k=bioskey(0);if(k==0x316e) break;if(k==0x1579) goto loop;}}printstar(){printf("******************************\n"); }int day_year(int y){if(y%4==0&&y%100!=0||y%400==0)return(366);else return(365);}小结经历了这次课程设计的设计和制作的整个过程，我才发现我知识的贫乏和知识面的狭窄，原本以为不就是一个小小的课程设计，小case，很容易就可以搞定，可是到了后来实际的操作的时候可是花费了我九牛二虎之力，不说其他的，只说时间上面，我在宿舍里面熬了三个通宵，更不用说白天的时间了。