用·c语言求输入未来某天的日期，输出距今天还有多少天？是星期几？

#define_CRT_SECURE_NO_WARNINGS
#include
#include
usingnamespacestd;
//判断星期几
intgetWeekDay(constintyear,constintyday)
{
	intn=year-1+(int)((year-1)/4)-(int)((year-1)/100)+(int)((year-1)/400)+yday;
	returnn%7;
}
//是否闰年
boolisLeapYear(constintyear)
{
	return(year%4==0&&year%100!=0)||year%400==0;
}
//获取当天是一年中的第几天
intgetYearDay(constintyear,constintmonth,constintmday)
{
	intyday=0;
	for(inti=1;i	{
		switch(i)
		{
		case1:
		case3:
		case5:
		case7:
		case8:
		case10:
		case12:
			yday+=31;
			break;
		case4:
		case6:
		case9:
		case11:
			yday+=30;
			break;
		case2:
			yday+=(isLeapYear(year)?29:28);
			break;
		}
	}
	yday+=mday;
	returnyday;
}
//计算两个日期的天数差（假设date1小于date2）
intgetDayCount(conststructtm&date1,conststructtm&date2)
{
	intnday=0;
	intyday1=getYearDay(date1.tm_year+1900,date1.tm_mon+1,date1.tm_mday);
	intyday2=getYearDay(date2.tm_year+1900,date2.tm_mon+1,date2.tm_mday);
	for(inti=date1.tm_year;i	{
		if(isLeapYear(i+1900))
			nday+=366;
		elsenday+=365;
	}
	nday+=(yday2-yday1);
	returnnday;
}
//公历节日，随便弄15个。
constchar*festivals[]={
	"1,1","元旦",
	"2,14","情人节",
	"3,8","妇女节",
	"3,12","植树节",
	"3,15","消费者权益日",
	"4,1","愚人节",
	"4,5","清明节",
	"5,1","劳动节",
	"5,4","青年节",
	"6,1","儿童节",
	"7,1","建党节",
	"8,1","建军节",
	"9,10","教师节",
	"10,1","国庆节",
	"12,25","圣诞节"
};
char*weekdays[]={"日","一","二","三","四","五","六"};
voidsetDate(structtm&date,constintyear,constintmonth,constintday)
{
	date.tm_year=year;
	date.tm_mon=month;
	date.tm_mday=day;
}
voidtestDate1()
{
	intmonth,year,day;
	cout<<"请输入你想知道的年月日（用逗号隔开）：";
	scanf("%d,%d,%d",&year,&month,&day);
	//cin>>year>>month>>day;
	//距离元旦的天数
	intyday=getYearDay(year,month,day);
	//将日期赋值到tm结构体中，方便函数调用
	structtmdate;
	setDate(date,year-1900,month-1,day);
	//取得今天的时间，并转换为tm结构体
	time_tnow=time((time_t*)NULL);
	structtm*today=localtime(&now);
	intnday;
	//将年月日换算成如20160412的整数然后进行比较，简单直观
	if(year*10000+(month-1)*100+day>=
		(today->tm_year+1900)*1000+today->tm_mon*100+today->tm_mday)
		nday=getDayCount(*today,date);
	elsenday=getDayCount(date,*today);
	printf("%d,%d,%d离今天有%d天！\n",year,month,day,nday);
	intwday=getWeekDay(year,yday);
	cout<<"是星期"<	//看看今天是什么节日
	charszDate[10]={'\0'};
	sprintf(szDate,"%d,%d",month,day);
	for(inti=0;i<30;i+=2)
	{
		if(strcmp(szDate,festivals[i])==0)
		{
			cout<<"是"<			break;
		}
	}
}
voidtestDate2()
{
	structtmdate1900;
	setDate(date1900,0,0,1);
	intyear,month,day;
	cout<<"请输入你想知道的年月日（用逗号隔开）：";
	scanf("%d,%d,%d",&year,&month,&day);
	structtmdate;
	setDate(date,year-1900,month-1,day);
	//求出所输入日期和1900年1月1日的日期差就能判断
	intndays=getDayCount(date1900,date);
	//三天打渔两天晒网，即每5天一个周期。
	intcycle=ndays%5;
	//对5求模，则结果是0-3的话就是打渔，否则就是晒网
	if(cycle<3)
		cout<<"这天是在打渔！"<	else
		cout<<"这天是在晒网！"<}
intmain()
{
	cout<<"1.离今天还有多少天！\n"<<
		"2.是打渔还是晒网！\n"<<
		"3.退出！"<	intop;
	do
	{
		cout<<"请选择1，2，3！\n";
		cin>>op;
		if(op<1||op>3)
			continue;
		switch(op)
		{
		case1:
			testDate1();
			break;
		case2:
			testDate2();
			break;
		default:
			break;
		}
	}while(op!=3);
	system("pause");
	return0;
}

用c语言设定一个纪念日时间，输入一个日期，显示当天距离纪念日的天数。。大佬们求帮忙，小弟明天有急用

#include
#include
#defineYear2018
#defineMonth1
#defineDay8//纪念日时期，可自行更改
intyear_alldays(intyear);
intyear_sumday(intyear,intmonth);
intget_alldays(intyear,intmonth,intday);
intmain()
{
intyear,month,day;
printf("纪念日日期为：%d-%02d-%02d\n",Year,Month,Day);

printf("输入一个日期(如2017-03-21)：");
scanf("%d-%d-%d",&year,&month,&day);

intday1=get_alldays(year,month,day);
intday2=get_alldays(Year,Month,Day);

printf("距离纪念日天数为：%d\n",abs(day1-day2));
return0;
}
//获取公历年的天数
intyear_alldays(intyear)
{
if((year%4==0&&year%100!=0)||year%400==0)return366;elsereturn365;
}
//获取公历年初至某整月的天数
intyear_sumday(intyear,intmonth)
{
intsum=0;
intrui[12]={31,29,31,30,31,30,31,31,30,31,30,31};
intping[12]={31,28,31,30,31,30,31,31,30,31,30,31};
intruiflag=0;
if((year%4==0&&year%100!=0)||year%400==0)ruiflag=1;
for(intindex=0;index{
if(ruiflag==1)sum+=rui[index];elsesum+=ping[index];
}
returnsum;
}
//获取从公历1800年1月25日至当前日期的总天数
intget_alldays(intyear,intmonth,intday)
{
inti=1800,days=-24;
while(i{
days+=year_alldays(i);
i++;
}
intdays2=year_sumday(year,month);
returndays+days2+day;
}

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