POJ calendar

Calendar http://poj.org/problem?id=2080

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 9858		Accepted: 3696

Description

A calendar is a system for measuring time, from hours and minutes, to months and days, and finally to years and centuries. The terms of hour, day, month, year and century are all units of time measurements of a calender system.
According to the Gregorian calendar, which is the civil calendar in use today, years evenly divisible by 4 are leap years, with the exception of centurial years that are not evenly divisible by 400. Therefore, the years 1700, 1800, 1900 and 2100 are not leap years, but 1600, 2000, and 2400 are leap years.
Given the number of days that have elapsed since January 1, 2000 A.D, your mission is to find the date and the day of the week.

Input

The input consists of lines each containing a positive integer, which is the number of days that have elapsed since January 1, 2000 A.D. The last line contains an integer −1, which should not be processed.
You may assume that the resulting date won’t be after the year 9999.

Output

For each test case, output one line containing the date and the day of the week in the format of "YYYY-MM-DD DayOfWeek", where "DayOfWeek" must be one of "Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday" and "Saturday".

Sample Input

1730
1740
1750
1751
-1

Sample Output

2004-09-26 Sunday
2004-10-06 Wednesday
2004-10-16 Saturday
2004-10-17 Sunday

个人代码：

#include<stdio.h>
int main()
{
 int m;
 while(scanf("%ld",&m)&&m!=-1)
 {
  int n,year=2000,month=1,date=1;
  n=m%7;
  while(m>=365)
  { 
   if(leap(year)==1&&m>365)
   {
    m-=366;
    year++;
   }
   else if(leap(year)==0)
   {
    m-=365;
    year++;
   }
   else
    break;
  }
  if(leap(year)==1)
  {
   if(month==1&&m>30)
   {
    m-=31;
    month++;
   }
   if(month==2&&m>28)
   {
    m-=29;
    month++;
   }
   if(month==3&&m>30)
   {
    m-=31;
    month++;
   }
   if(month==4&&m>29)
   {
    m-=30;
    month++;
   }
   if(month==5&&m>30)
   {
    m-=31;
    month++;
   }
   if(month==6&&m>29)
   {
    m-=30;
    month++;
   }
   if(month==7&&m>30)
   {
    m-=31;
    month++;
   }
   if(month==8&&m>30)
   {
    m-=31;
    month++;
   }
   if(month==9&&m>29)
   {
    m-=30;
    month++;
   }
   if(month==10&&m>30)
   {
    m-=31;
    month++;
   }
   if(month==11&&m>29)
   {
    m-=30;
    month++;
   }
   date=m+1;
  }
  if(leap(year)==0)
  {
   if(month==1&&m>30)
   {
    m-=31;
    month++;
   }
   if(month==2&&m>27)
   {
    m-=28;
    month++;
   }
   if(month==3&&m>30)
   {
    m-=31;
    month++;
   }
   if(month==4&&m>29)
   {
    m-=30;
    month++;
   }
   if(month==5&&m>30)
   {
    m-=31;
    month++;
   }
   if(month==6&&m>29)
   {
    m-=30;
    month++;
   }
   if(month==7&&m>30)
   {
    m-=31;
    month++;
   }
   if(month==8&&m>30)
   {
    m-=31;
    month++;
   }
   if(month==9&&m>29)
   {
    m-=30;
    month++;
   }
   if(month==10&&m>30)
   {
    m-=31;
    month++;
   }
   if(month==11&&m>29)
   {
    m-=30;
    month++;
   }
   date=m+1;
  }
  if(month<10&&date<10)
   printf("%d-0%d-0%d ",year,month,date);
  else if(month<10)
   printf("%d-0%d-%d ",year,month,date);
  else if(date<10)
   printf("%d-%d-0%d ",year,month,date);
  else
   printf("%d-%d-%d ",year,month,date);
  switch(n)
  {
   case 1:printf("Sunday\n");break;
   case 2:printf("Monday\n");break;
   case 3:printf("Tuesday\n");break;
   case 4:printf("Wednesday\n");break;
   case 5:printf("Thursday\n");break;
   case 6:printf("Friday\n");break;
   default:printf("Saturday\n");
  }
 }
 return 0;
}
int leap(int y)
{
 if(y%400==0||y%100!=0&&y%4==0)
  return 1;
 else
  return 0;
}

有点小麻烦，我先判断年份，在作判断年份时，我以前忘记编写break语句，致使结果一直不对。后面对小于365或小于366的数据进行处理。我是逐月减去天数，最后获得一个m值而后判断日期。我同窗的代码以下：

#include<string.h>
#include<stdio.h>
int main()
{
 int y,m,d,n,a[13],year[10000],i,c,w;
 a[1]=31;a[3]=31;a[4]=30;a[5]=31;a[6]=30;a[7]=31;a[8]=31;a[9]=30;a[10]=31;a[11]=30;a[12]=31;
  for(i=0;i<10000;i++){
   if(((i+2000)%4==0&&(i+2000)%100!=0)||(i+2000)%400==0)
       year[i]=366;
   else
    year[i]=365;}
 while(scanf("%d",&n)&&n!=-1)
 {
  n+=1;
  for(y=2000,i=0;n>year[i];i++)
  {n-=year[i];y+=1;}
  if((y%4==0&&y%100!=0)||y%400==0)
   a[2]=29;
  else
   a[2]=28;
  for(i=1,m=1;n>a[i];i++)
  {
   n-=a[i];
   m+=1;
  }
  printf("%d-",y);
  if(m>=10)
   printf("%d-",m);
  if(m<10)
   printf("0%d-",m);
  if(n>=10)
   printf("%d ",n);
  if(n<10)
   printf("0%d ",n);
  if(m<3)
  {m+=12;y-=1;}
  d=n;
  c=y/100;
  y=y%100;
  w=((c/4)-2*c+y+(y/4)+(13*(m+1)/5)+d-1)%7;
  while(w<0)
   w+=7;
  switch(w)
  {
  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;}
  case 6:{printf("Saturday\n");break;}
  }
 }
 return 0;
}

比个人简单多了。

另外一个网上的：

#include<stdio.h>  
int main(){  
    char w[7][10]={"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday" ,"Saturday"};  
    int m[2][13]={0,31,28,31,30,31,30,31,31,30,31,30,31,  
                  0,31,29,31,30,31,30,31,31,30,31,30,31};  
    int yd[2]={365,366};  
    long day;  
    int year,month,week;  
    int i,j,flag;  
    while(scanf("%ld",&day)&&-1!=day){  
        week=(day+6)%7;//获得星期几   
        year=2000;  
        flag=(0==year%4&&year%100!=0)||0==year%400;//flag=1为闰年   
        ++day;//题目说通过多少天，因此在这里先加1   
        for(;day>yd[flag];){//获得年份、剩余天数   
            day-=yd[flag];   
            year++;  
            flag=(0==year%4&&year%100!=0)||0==year%400;  
        }  
        for(month=1;day>m[flag][month];++month){//获得月份和对应天数   
            day-=m[flag][month];  
        }  
        printf("%d-%02d-%02d %sn",year,month,day,w[week]);//%02d很方便          
    }  
}