本文目录一览：
能不能帮我用c语言写下准确公历的计算程序
if(month==4||6||9||11)
if(month==1||3||5||7||8||10||12)
这种写法是错误的。
应该这样写
if(month==4||month==6||month==9||month==11)
if(month==1||month==3||month==5||month==7||month==8||month==10||month==12)
另外，其实你可以不这么写，太多了，可以用
switch..case
switch(month)
case 1:
case 3:
case 5:
case 7:
case 8:
case 10:
case 12:
day=31;break;
case 4:
case 6:
case 9:
case 11:
day=30;break;
case 2:
if()...
（c语言实现）打印指定年份的公历表和农历表
#include stdlib.h
#include stdio.h
#include conio.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("输入年份: "); /*从键盘输入年份*/
scanf("%d",year);
while(Year_Start year)
{
if( IsLeapYear( Year_Start ) ) /*判断输入的年份是否为闰年*/
Per_Year_Days = 366;
else
Per_Year_Days = 365;
Year_days = Year_days + Per_Year_Days;
Year_Start++;
} /*这个while循环计算从1到输入的年份共有多少天*/
for( temp = 1; temp =12; temp++ )
{ /*这个for循环打印出该年每个月的日历*/
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) /*如果是闰年的2月*/
while( day = month_day[12] )
{
if( day 1 )
if( Year_days % 7 == 0 )
printf("\n");
if( day = 10 )
printf("%-4d",day);
else
printf("%-4d",day);
Year_days++;
day++;
}
else /*不是闰年*/
while (day = month_day[temp-1])
{
if( day 1 )
if( Year_days % 7 == 0 )
printf("\n");
if( day =10 )
printf("%-4d",day);
else
printf("%-4d",day);
Year_days++;
day++;
}
printf("\n");
if( getch() == 'q' ) /*输入Q结束*/
exit(0);
}
getch();
}
int IsLeapYear( int year ) /*判断是否为闰年*/
{
if ((year %4 == 0) (year % 100 != 0) ||
(year % 400 == 0) )
return 1;
else
return 0;
}
用C语言怎么将公历日期转化为农历日期?
其实很简单，你的程序中定义一些数组，保存一些重要参数即可，下面是JAVASCRIPT的，看懂了你就可以用C来写：
var lunarInfo=new Array(
0x04bd8,0x04ae0,0x0a570,0x054d5,0x0d260,0x0d950,0x16554,0x056a0,0x09ad0,0x055d2,
0x04ae0,0x0a5b6,0x0a4d0,0x0d250,0x1d255,0x0b540,0x0d6a0,0x0ada2,0x095b0,0x14977,
0x04970,0x0a4b0,0x0b4b5,0x06a50,0x06d40,0x1ab54,0x02b60,0x09570,0x052f2,0x04970,
0x06566,0x0d4a0,0x0ea50,0x06e95,0x05ad0,0x02b60,0x186e3,0x092e0,0x1c8d7,0x0c950,
0x0d4a0,0x1d8a6,0x0b550,0x056a0,0x1a5b4,0x025d0,0x092d0,0x0d2b2,0x0a950,0x0b557,
0x06ca0,0x0b550,0x15355,0x04da0,0x0a5d0,0x14573,0x052d0,0x0a9a8,0x0e950,0x06aa0,
0x0aea6,0x0ab50,0x04b60,0x0aae4,0x0a570,0x05260,0x0f263,0x0d950,0x05b57,0x056a0,
0x096d0,0x04dd5,0x04ad0,0x0a4d0,0x0d4d4,0x0d250,0x0d558,0x0b540,0x0b5a0,0x195a6,
0x095b0,0x049b0,0x0a974,0x0a4b0,0x0b27a,0x06a50,0x06d40,0x0af46,0x0ab60,0x09570,
0x04af5,0x04970,0x064b0,0x074a3,0x0ea50,0x06b58,0x055c0,0x0ab60,0x096d5,0x092e0,
0x0c960,0x0d954,0x0d4a0,0x0da50,0x07552,0x056a0,0x0abb7,0x025d0,0x092d0,0x0cab5,
0x0a950,0x0b4a0,0x0baa4,0x0ad50,0x055d9,0x04ba0,0x0a5b0,0x15176,0x052b0,0x0a930,
0x07954,0x06aa0,0x0ad50,0x05b52,0x04b60,0x0a6e6,0x0a4e0,0x0d260,0x0ea65,0x0d530,
0x05aa0,0x076a3,0x096d0,0x04bd7,0x04ad0,0x0a4d0,0x1d0b6,0x0d250,0x0d520,0x0dd45,
0x0b5a0,0x056d0,0x055b2,0x049b0,0x0a577,0x0a4b0,0x0aa50,0x1b255,0x06d20,0x0ada0)
//====================================== 传回农历 y年的总天数
function lYearDays(y) {
var i, sum = 348
for(i=0x8000; i0x8; i=1) sum += (lunarInfo[y-1900] i)? 1: 0
return(sum+leapDays(y))
}
//====================================== 传回农历 y年闰月的天数
function leapDays(y) {
if(leapMonth(y)) return((lunarInfo[y-1900] 0x10000)? 30: 29)
else return(0)
}
//====================================== 传回农历 y年闰哪个月 1-12 , 没闰传回 0
function leapMonth(y) {
return(lunarInfo[y-1900] 0xf)
}
//====================================== 传回农历 y年m月的总天数
function monthDays(y,m) {
return( (lunarInfo[y-1900] (0x10000m))? 30: 29 )
}
C语言编写 年历显示程序
此题不是很难，我会前三个
关键是算那一年的第一天是周几
给你个基本算法，一年是365天，闰年366天
364是7的倍数，这样一年多出一天，中间多少个闰年，在加上相应的天数即可
400年多出来的天数是400+97=497也是7的倍数，那么400年一轮回，这样就算400年以内的即可
一个子函数，返回这一年的第一天是周几，很好做
int ye(int year)
{
int y=year%400;
int day=y;
int i;
for(i=0;iy;i++)
if(闰年判断)
day++;
return day%7;
}
做一个月份天数的全局数组
int mo[12]={31,28,31,30,31,30,31,31,30,31,30,31};
需要时用一个if判断闰年改变m0[1]的值是28还是29
打印子函数，有一个小技巧，就是每个月前面的空缺部分，这个1月份是ye子函数的返回值，后面的你看和前面月份是不是重合啊，吧这个记录下来在下一月份用即可
1 2
3 4
这样月份排列更好打印一些，你的那个截图有点麻烦
int pr(int year)
{
int k=ye(year);
int i,j,;
if(闰年判断)
mo[1]=29;
else
mo[1]=28;
for(i=0;i12;i++)
{
printf("周几英文的打印");
for(j=0;jk;j++)
printf("一般是四个空格，就是月份前面空缺的打印");
for(j=1;j=mo[i];j++)
{
printf("%4d",j);
k++;
if((k+j)%7==0)
printf("\n");
}
k%=7;
if(k!=0)
printf("\n");
}
}
如果像你那样，感觉需要赋值一个大一点的数组，在吧数组打印出来更好，呵呵
c语言 万年历的程序设计
#include windows.h
#include winnt.h
#includeiostream
#includeiomanip
using namespace std;
int week(int,int,int); //根据年月日判断星期几
int leap_year(int); //判断闰年
void display_year(int ); //显示某年日历
void demand_day(int,int,int); //查询某天
int main()
{
int y,m,d,es=1;
while(es)
{
HANDLE consolehwnd;
consolehwnd = GetStdHandle(STD_OUTPUT_HANDLE);
SetConsoleTextAttribute(consolehwnd,12);
cout"请选择操作：\n1→显示某年日历\
\n2→查询某天\n0→退出"endl;
char tp[20];cintp;
if(tp[1]!='\0'||tp[0]'2'||tp[0]'0'){cout"输入有误"endl;continue;}
switch(tp[0]-48)
{
case 1:{cout"请输入年份：";ciny;system("cls");display_year(y);break;}
case 2:{cout"请输入年、月、日，以空格分开：";cinymd;system("cls");
demand_day(y,m,d);break;}
case 0:{es=0;break;}
}
}
return 0;
}
//-----根据年月日判断星期几-------------------------
int week(int y,int m, int d)
{
int week1,yy=y;
if(m==1) {m=13;yy--;}
if(m==2) {m=14;yy--;}
week1=(d+2*m+3*(m+1)/5+yy+yy/4-yy/100+yy/400)%7;
int s;
switch (week1)
{
case 0: s=1; break;
case 1: s=2; break;
case 2: s=3; break;
case 3: s=4; break;
case 4: s=5; break;
case 5: s=6; break;
case 6: s=0; break;
}
return s;
}
//----判断闰年-------------------------------------
int leap_year(int y)
{
int i;
if((y%4==0y%100!=0)||y%400==0)i=1;
else i=0;
return i;
}
//--------显示某年日历------------------------
void display_year(int y)
{
int n1,n2,i,j,a[13],c,d;
HANDLE consolehwnd;
consolehwnd = GetStdHandle(STD_OUTPUT_HANDLE);
SetConsoleTextAttribute(consolehwnd,5);
coutsetw(38)y"年"endl;
coutsetw(28)"*********";
for(i=1;i=27;i++)cout'*';
coutendl;
a[1]=a[3]=a[5]=a[7]=a[8]=a[10]=a[12]=31;//
a[4]=a[6]=a[9]=a[11]=30; //确定每月天数
if(leap_year(y))a[2]=29;
else a[2]=28; //
for(i=1;i=11;i+=2) //六次循环
{
SetConsoleTextAttribute(consolehwnd,1);
coutsetw(14)i"月"setw(42)i+1"月"endl;
SetConsoleTextAttribute(consolehwnd,2);
coutsetw(4)"日"setw(4)"一"setw(4)"二"setw(4)"三"setw(4)\
"四"setw(4)"五"setw(4)"六";
coutsetw(16)' ';
coutsetw(4)"日"setw(4)"一"setw(4)"二"setw(4)"三"setw(4)\
"四"setw(4)"五"setw(4)"六"endl;
SetConsoleTextAttribute(consolehwnd,7);
n1=week(y,i,1);n2=week(y,i+1,1);
if(n1) //-----------
{
for(j=1;j=n1;j++) //
coutsetw(4)' ';
}
for(j=1;j=7-n1;j++)
coutsetw(4)j;
coutsetw(16)' ';
if(n2)
{ //-----输出每次循环的第一行---
for(j=1;j=n2;j++)
coutsetw(4)' ';
}
for(j=1;j=7-n2;j++)
coutsetw(4)j;
coutendl; //--------------
c=8-n1;d=8-n2;
for(int m=1;m6;m++) //每月日历最多占六行
{
if(ca[i])coutsetw(4*7)' ';//若ca[i]，则该月的这一行全部输出空格
for(j=c;j=a[i];j++)
{
coutsetw(4)j;
if((j-c+1)%7==0){c=j+1;break;}
if(j==a[i]){coutsetw((6-week(y,i,a[i]))*4)' ';c=j+1;break;}
//如果j是该月最后一天，该行剩下的全部补空格
}
coutsetw(16)' ';
if(da[i+1])coutsetw(4*7)' ';
for(j=d;j=a[i+1];j++)
{ //
coutsetw(4)j;
if((j-d+1)%7==0){d=j+1;break;}
if(j==a[i+1]){coutsetw((6-week(y,i+6,a[i+1]))*4)' ';d=j+1;break;}
}
coutendl;
}
coutendl;
}
coutendl;
}
//--------查询某天------------
void demand_day(int y,int m,int d)
{
int n;
HANDLE consolehwnd;
consolehwnd = GetStdHandle(STD_OUTPUT_HANDLE);
SetConsoleTextAttribute(consolehwnd,5);
n=week(y,m,d);
switch(n)
{
case 1:couty"年"m"月"d"日"",""星期一"endl;break;
case 2:couty"年"m"月"d"日"",""星期二"endl;break;
case 3:couty"年"m"月"d"日"",""星期三"endl;break;
case 4:couty"年"m"月"d"日"",""星期四"endl;break;
case 5:couty"年"m"月"d"日"",""星期五"endl;break;
case 6:couty"年"m"月"d"日"",""星期六"endl;break;
case 0:couty"年"m"月"d"日"",""星期日"endl;break;
default:break;
}
coutendl;
}
