POJ2080简单模拟

Calendar Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 13574   Accepted: 4838 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 Source Shanghai 2004 Preliminary

    这个题目是个简单的模拟题，但是我WA了六次，六次，六次。。。开始忽略了2000-01-01这一天是星期几，后来又发现题目数据的范围很大……不过最后还是AC了，也没什么难的，一到水题而已，只是要注意很多细节。以下是我的AC代码

#include <stdio.h>

int main()
{
    //freopen("in.txt","r",stdin);
    int n,year,month,day,t;
    int p[12]= {31,28,31,30,31,30,31,31,30,31,30,31};
    int r[12]= {31,29,31,30,31,30,31,31,30,31,30,31};
    int y[10005];
    for(int i=0; i<10005; i++)
    {
        if((i%4==0&&i%100!=0)||i%400==0) y[i]=366;
        else y[i]=365;
    }
    while(~scanf("%d",&n) &&n!=-1)
    {
        n++;
        t=n%7;
        for(int i=0; i<10005; i++)     //判断年
        {
            if(n<=y[i])
            {
                year=i;
                break;
            }
            n-=y[i];
        }
        if(!((year%4==0&&year%100!=0)||year%400==0))
        {
            for(int i=0; i<12; i++)     //判断月
            {
                if(n<=p[i])
                {
                    month=i+1;
                    break;
                }
                n-=p[i];
            }
        }
        else
        {
            for(int i=0; i<12; i++)
            {
                if(n<=r[i])
                {
                    month=i+1;
                    break;
                }
                n-=r[i];
            }
        }
        day=n;
        if(t==3)
            printf("%d-%02d-%02d Monday\n",year+2000,month,day);
        else if(t==4)
            printf("%d-%02d-%02d Tuesday\n",year+2000,month,day);
        else if(t==5)
            printf("%d-%02d-%02d Wednesday\n",year+2000,month,day);
        else if(t==6)
            printf("%d-%02d-%02d Thursday\n",year+2000,month,day);
        else if(t==0)
            printf("%d-%02d-%02d Friday\n",year+2000,month,day);
        else if(t==1)
            printf("%d-%02d-%02d Saturday\n",year+2000,month,day);
        else if(t==2)
            printf("%d-%02d-%02d Sunday\n",year+2000,month,day);
    }
    return 0;
}

测试数据：

输入：

1730
1740
1750
1751
0
1
2
365
366
367

输出：

2004-09-26 Sunday
2004-10-06 Wednesday
2004-10-16 Saturday
2004-10-17 Sunday
2000-01-01 Saturday
2000-01-02 Sunday
2000-01-03 Monday
2000-12-31 Sunday
2001-01-01 Monday
2001-01-02 Tuesday