php做动漫网站,网络上如何推广网站,关于酒店网站规划建设方案书,个人主页模板设计/* 1.输入年月日#xff0c;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#xff0c;且1/1/2006不发车(无ticket),如果所输入数据在此日期之后#xff0c;则输出有没有车票#xff0c;否则仅输出上一步结果。*/ /* month/date/year is which day of the ye… /* 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(date31||date1||month12||month1) { 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%40year%100!0||year%4000) 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:daydate;break; case 2:daydate31;break; case 3:daydate59;break; case 4:daydate90;break; case 5:daydate120;break; case 6:daydate151;break; case 7:daydate181;break; case 8:daydate212;break; case 9:daydate243;break; case 10:daydate273;break; case 11:daydate304;break; case 12:daydate334;break; default:printf(Data error!!!); }if(isleapyear(year)month2) day1;return(day);} int main(void){ int i,m0,n; /* i is a counter;m is the days since Jan,1,2006 ;nthe days since Jan,1,year.*/ input(); ndays(year,month,date); if(year2005) /*is there a m ticket today which is later than 2006.01.20 */ { for (i2006;iyear;i) /* add the days from 2006 to year-1.(if it is a leapyear add 366,or add 365)*/ if(isleapyear(i)) m366; else m365; mmn-days(2006,1,1); if(m%21)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。 本文来自CSDN博客转载请标明出处http://blog.csdn.net/perforce/archive/2007/04/26/1585191.aspx 转载于:https://www.cnblogs.com/kevinzhwl/archive/2010/01/11/3878968.html
