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/* 1.输入年月日，编写程序计算所输日期是当年的第几天 */
/* 2.已知列车隔日发车，且1/1/2006不发车(无ticket),如果所输入数据在此日期之后，则输出有没有车票，否则仅输出上一步结果。*/
/* month/date/year is which day of the year */
/* is there a ticket today(2006.01.01--yeath!)?? */
#include <stdio.h>
int year,month,date;
void input()
/* read input number and judge is it legal*/
{
puts("please input year,month,date:");
scanf("%d%d%d",&year,&month,&date);
while(date>31||date<1||month>12||month<1)
{
puts("month OR date Wrong!!!");
puts("please input year,month,date:");
scanf("%d%d%d",&year,&month,&date);
}
}
int isleapyear(int year)
/* is the year a leapyear */
{if(year%4==0&&year%100!=0||year%400==0) return(1);
else return(0);
}
int days(int year,int month,int date)
/* Calculate the days since Jan,1,year.(you input the "year") */
{int day;
switch(month)
{
case 1:day=date;break;
case 2:day=date+31;break;
case 3:day=date+59;break;
case 4:day=date+90;break;
case 5:day=date+120;break;
case 6:day=date+151;break;
case 7:day=date+181;break;
case 8:day=date+212;break;
case 9:day=date+243;break;
case 10:day=date+273;break;
case 11:day=date+304;break;
case 12:day=date+334;break;
default:printf("Data error!!!");
}
if(isleapyear(year)&&month>2) day+=1;
return(day);
}
int main(void)
{
int i,m=0,n;
/* "i" is a counter;"m" is the days since Jan,1,2006 ;"n"the days since Jan,1,year.*/
input();
n=days(year,month,date);
if(year>2005) /*is there a m ticket today which is later than 2006.01.20 */
{
for (i=2006;i<year;i++)
/* add the days from 2006 to year-1.(if it is a leapyear add 366,or add 365)*/
if(isleapyear(i)) m+=366;
else m+=365;
m=m+n-days(2006,1,1);
if(m%2==1)puts("Ticket:Yeath!!/n");
else puts("Ticket:Sorry!!!/n");
}
printf("%d/%d/%d day:%d/n",month,date,year,n);
puts("Press anykey to quit!");
getch();
return 0;
}
PS:1.本人初学，不求其他只愿代码能够尽量简洁高效，欢迎各位多多批评、多多挑毛病^_^。加我QQ 505011298。
2.文章全部为原创，过程中可能有些文章参考了某些文章的思路，但本人宗旨和目标只是：a.不如我之人可以向我学习b.希望高手可以对我指点一二，以求做到最好；c.资源共享；别无其他。
3.TC2.0下通过。当然，解法很多，如果哪位找到更好方法的话，如若可以分享我将很荣幸。
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