标签：des c style class blog code
Maya Calendar
| Time Limit: 1000MS |
|
Memory Limit: 10000K |
| Total Submissions: 64795 |
|
Accepted: 19978 |
Description
During his last sabbatical, professor M. A. Ya
made a surprising discovery about the old Maya calendar. From an old knotted
message, professor discovered that the Maya civilization used a 365 day long
year, called Haab, which had 19 months. Each of the first 18 months was 20 days
long, and the names of the months were pop, no, zip, zotz, tzec, xul, yoxkin,
mol, chen, yax, zac, ceh, mac, kankin, muan, pax, koyab, cumhu. Instead of
having names, the days of the months were denoted by numbers starting from 0 to
19. The last month of Haab was called uayet and had 5 days denoted by numbers 0,
1, 2, 3, 4. The Maya believed that this month was unlucky, the court of justice
was not in session, the trade stopped, people did not even sweep the
floor.
For religious purposes, the Maya used another calendar in
which the year was called Tzolkin (holly year). The year was divided into
thirteen periods, each 20 days long. Each day was denoted by a pair consisting
of a number and the name of the day. They used 20 names: imix, ik, akbal, kan,
chicchan, cimi, manik, lamat, muluk, ok, chuen, eb, ben, ix, mem, cib, caban,
eznab, canac, ahau and 13 numbers; both in cycles.
Notice that each
day has an unambiguous description. For example, at the beginning of the year
the days were described as follows:
1 imix, 2 ik, 3 akbal, 4 kan, 5
chicchan, 6 cimi, 7 manik, 8 lamat, 9 muluk, 10 ok, 11 chuen, 12 eb, 13 ben, 1
ix, 2 mem, 3 cib, 4 caban, 5 eznab, 6 canac, 7 ahau, and again in the next
period 8 imix, 9 ik, 10 akbal . . .
Years (both Haab and Tzolkin)
were denoted by numbers 0, 1, : : : , where the number 0 was the beginning of
the world. Thus, the first day was:
Haab: 0. pop
0
Tzolkin: 1 imix 0
Help professor M. A. Ya and write a
program for him to convert the dates from the Haab calendar to the Tzolkin
calendar.
Input
The date in Haab is given in the following
format:
NumberOfTheDay. Month Year
The first line of the
input file contains the number of the input dates in the file. The next n lines
contain n dates in the Haab calendar format, each in separate line. The year is
smaller then 5000.
Output
The date in Tzolkin should be in the following
format:
Number NameOfTheDay Year
The first line of the
output file contains the number of the output dates. In the next n lines, there
are dates in the Tzolkin calendar format, in the order corresponding to the
input dates.
Sample Input
3
10. zac 0
0. pop 0
10. zac 1995
Sample Output
3
3 chuen 0
1 imix 0
9 cimi 2801
Source
水题，模拟题。
题意真让人无语，把我这个英语渣渣虐得死去活来……好吧，鄙视我吧
题意：
Habb历一年365天
Tzolkin历一年260天
先计算Habb历从第0天到输入日期的总天数sumday
Sumday/day就是Tzolkin历的年份
Tzolkin历的天数Name每20一循环，先建立Tzolkin历天数Name与1~20的映射，
因此Sumday %20+1就是Tzolkin历的天数Name
Tzolkin历的天数ID每13一循环，且从1开始，则Sumday %13+1就是Tzolkin历的天数ID
另外Habb历是一共有19个月，前18个月每月20天，最后一月只有5天，加起来一年一共365天；
Tzolkin历一共有13个月，每月20天，一年一共260天。尽管有20个名字，正好对应每月20天，但是Tzolkin历的天数编号有些坑。例如第一个月20天不是从1到20编号，而是从1到13再从1到7，即第一个月编号是这样的：1、2、3、4、5、6、7、8、9、10、11、12、13、1、2、3、4、5、6、7。下一个月再从8开始。
输入时Habb历的天数，月份名字，年份。
最后让你输出Tzolkin历的天数编号，对应的天数名字，年份。（不要求输出月份）
思路：
先根据输入的Habb历的天数，月份（需要将名字转换成数字），年份，求出从0开始度过的总天数sumday。
然后转换成Tzolkin历的天数编号dayId，天数名字dayName，年份year。
求法分别为：
year = sumday/260;
dayName = sumday%20; //这只是下标，最后输出对应的天数名字
dayId = sumday%13+1;
最后输出：
cout<<dayId<<‘ ‘<<Tzolkin[dayName]<<‘ ‘<<year<<endl;
代码：
1 #include <iostream>
2 #include <string.h>
3 using namespace std;
4 char Habb[19][20] =
5 {"pop", "no", "zip", "zotz", "tzec",
6 "xul", "yoxkin", "mol", "chen", "yax",
7 "zac", "ceh", "mac", "kankin", "muan",
8 "pax", "koyab", "cumhu","uayet"};
9 char Tzolkin[20][20] =
10 {"imix", "ik", "akbal", "kan", "chicchan",
11 "cimi", "manik", "lamat", "muluk", "ok",
12 "chuen", "eb", "ben", "ix", "mem",
13 "cib", "caban", "eznab", "canac", "ahau"
14 };
15 int Name2month(char name[]) //返回该名字对应的月份
16 {
17 for(int i=0;i<19;i++){ //依次比较
18 if(strcmp(Habb[i],name)==0)
19 return i+1;
20 }
21 return -1;
22 }
23 int GetHabbSumday(int day,int month,int year) //获得总天数
24 {
25 int sumday = 0;
26 sumday += year*365;
27 sumday += 20*(month-1);
28 sumday += day;
29 return sumday;
30 }
31 int main()
32 {
33 int n,Number,Year;
34 char c,Name[20];
35 cin>>n;
36 cout<<n<<endl;
37 while(n--){
38 cin>>Number>>c>>Name>>Year; //输入
39 int sumday = GetHabbSumday(Number,Name2month(Name),Year); //获得总天数
40 //计算结果
41 int year = sumday/260;
42 int dayName = sumday%20;
43 int dayId = sumday%13+1;
44 cout<<dayId<<‘ ‘<<Tzolkin[dayName]<<‘ ‘<<year<<endl;
45 }
46 return 0;
47 }
Freecode : www.cnblogs.com/yym2013
poj 1008:Maya Calendar（模拟题，玛雅日历转换）,布布扣,bubuko.com
poj 1008:Maya Calendar（模拟题，玛雅日历转换）
标签：des c style class blog code
原文地址：http://www.cnblogs.com/yym2013/p/3774566.html
