hdu 2133 日期的格式转化

What day is it

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2338    Accepted Submission(s): 687

Problem Description

Today is Saturday, 17th Nov,2007. Now, if i tell you a date, can you tell me what day it is ?

 

Input

There are multiply cases.
One line is one case.
There are three integers, year(0<year<10000), month(0<=month<13), day(0<=day<32).

 

Output

Output one line.
if the date is illegal, you should output "illegal". Or, you should output what day it is.

 

Sample Input

   
   
   
   

    
    
    
    2007 11 17
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    Saturday
   
   
   
   

这里要考虑闰年的2月这种特殊的情况

题目中还有个bug 要注意的是 当月份或者日子 （month ==0|| day==0）  输出  illegal

#include<stdio.h>
int year[10010],month[15]={0,31,59,90,120,151,181,212,243,273,304,334,365};
int main()
{
   year[0]=0;
   int i;
   for( i=1; i<10000 ;i++)
   {
      if( (i%4==0&&i%100!=0) || (i%400==0) ) year[i]+=year[i-1]+366; //累积年份所包含的日子
      else year[i]+=year[i-1]+365;     
      
   }
   int ye,mo,day,sum,judge;
   while( scanf("%d %d %d",&ye,&mo,&day)!=EOF )
   {
          
          if( ye<=0 || ye>=10000 || mo<=0 || mo>12 || day<=0 || day>31)  //限制条件
          {
              printf("illegal\n");
              continue;
          }
          if(mo==4||mo==6||mo==9||mo==11){   //也月份为30天的限制条件
                              if(day>30)
                               {
                                 printf("illegal\n");
                                 continue;
                               }  
                             }  
          if( (ye%4==0&&ye%100!=0) || (ye%400==0) )    //判断是否为闰年2月
          {   
                  if(mo==2 && day>29) {printf("illegal\n");continue;}      //闰年2月的限制条件
          }
          else {
                 if(mo==2 && day>28) {                                  //平年2月的限制条件
                                         printf("illegal\n");
                                         continue;
                                      }
               }
       
          sum=0;
          sum=year[ye-1]+month[mo-1]+day;    //累积日子数来警醒判定  要注意的是 1year  1month 1day 为星期六 
          if( (ye%4==0&&ye%100!=0) || (ye%400==0)  ){ if(mo>=3)sum+=1;} 
          judge=sum%7;
          switch(judge)
            {
               case 6:printf("Saturday\n");break;
               case 0:printf("Sunday\n");break;
               case 1:printf("Monday\n");break;
               case 2:printf("Tuesday\n");break;
               case 3:printf("Wednesday\n");break;
               case 4:printf("Thursday\n");break;
               case 5:printf("Friday\n");break;
               
            }
          
          
   }
    
    
}