超时了！~~
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.
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.
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
month = 1; day = 1; yd = 29; md = 31; for(i = 1; i <= x; i++){ day++; if(day == md + 1) { month++; if( month == 1 || month == 3 || month == 5 || month == 7 || month == 8 || month == 10 || month == 12 ) { md = 31; day = 1; } else if( month == 4 || month == 6 || month == 9 || month == 11 ) { md = 30; day = 1; } else { md = yd; day = 1; } if(month == 13) { month = 1; md = 31; year++; if( judge(year) ) { yd = 29; } else { yd = 28; } } } //switch((x+1) % 7)#1 switch(x % 7) { case 6: strcpy(s, "Friday"); break; case 0: strcpy(s, "Saturday"); break; case 1: strcpy(s, "Sunday"); break; case 2: strcpy(s, "Monday"); break; case 3: strcpy(s, "Tuesday"); break; case 4: strcpy(s, "Wednesday"); break; case 5: strcpy(s, "Thursday"); } } printf("%d-", year); if(month <= 9) printf("0%d-", month); else printf("%d-", month); if(day <= 9) printf("0%d ", day); else printf("%d ", day); printf("%s\n", s); //printf("%d\n", i - 1); } return 0; }
note：
#1：x是多少天，所以在switch中设置星期几时应以多少天后为标准，即几天后是星期几。
感想：思路是一天天比较（从日到月再到年），超时的原因是天数多时我的方法就显得笨重了。