poj2080

Problem D

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 60000/30000K (Java/Other)
Total Submission(s) : 14   Accepted Submission(s) : 5
Problem 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
题目考查的是日期的调换问题,瑞年一年366天,2月29天,平年365天,2月28天,考虑好瑞年的计算方法,就可以实现了,只是在实现的时候需要注意细节的调试
代码如下
#include <cstdio> #include <cstring> using namespace std; int year[2]={365,366};    char week[7][10]={"Saturday","Sunday","Monday","Tuesday","Wednesday","Thursday","Friday"};    int month[2][12]={31,28,31,30,31,30,31,31,30,31,30,31,31,29,31,30,31,30,31,31,30,31,30,31}; int nian(int n) {     if((n%4==0&&n%100!=0)||n%400==0)     return 1;//瑞年         else     return 0;//平年 } int main() {    int day,i,j;    int dayofweek;     while(scanf("%d",&day)!=EOF&&day!=-1)     {         dayofweek=day%7;         for( i=2000;day>=year[nian(i)];i++)         day-=year[nian(i)];         for(j=0;day>=month[nian(i)][j];j++)             day-=month[nian(i)][j];         printf("%d-%02d-%02d %s\n",i,j+1,day+1,week[dayofweek]);     }     return 0; }
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