Maya动画制作之 POJ-1008:Maya Calendar-职坐标

Maya动画制作之 POJ-1008:Maya Calendar

凌雪 2018-08-17 来源：网络阅读 815 评论 0

摘要：本文将带你了解Maya动画制作之 POJ-1008:Maya Calendar，希望本文对大家学Maya有所帮助

本文将带你了解Maya动画制作之 POJ-1008:Maya Calendar，希望本文对大家学Maya有所帮助

1008:Maya Calendar

时间限制:
1000ms
内存限制:
65536kB
描述
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.
输入
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.
输出
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.
样例输入
3
10. zac 0
0. pop 0
10. zac 1995

样例输出
3
3 chuen 0
1 imix 0
9 cimi 2801

问题的实质是一个查表问题，中间混合了简单的数学求余计算：
#include <string.h>#include <iostream>using namespace std; int main(){ char Haab_month_code[19][7] = {"pop", "no" , "zip", "zotz" , "tzec", "xul", "yoxkin", "mol", "chen" , "yax" , "zac", "ceh" , "mac", "kankin", "muan", "pax", "koyab" , "cumhu", "uayet" }; char Tzolkin_day_code[20][9] = {"imix" ,"ik" ,"akbal","kan" ,"chicchan", "cimi" ,"manik","lamat","muluk","ok" , "chuen","eb" ,"ben" ,"ix" ,"mem" , "cib" ,"caban","eznab","canac","ahau" }; int n; cin>>n; int *day = new int[n]; char **day_name = new char*[n]; int *year = new int[n]; for(int i=0; i<n; ++i) { int H_day; char dot; char H_month[10]; int H_year; cin>>H_day; cin>>dot; cin>>H_month; cin>>H_year; int month_int =0; for(int j=0; j<19; ++j) { if(!strcmp(H_month,Haab_month_code[j])) month_int = j; } int num = 0; if(month_int != 18) { num = H_year*365 + month_int*20 + H_day; } else { num = H_year*365 + 360 +H_day;} day_name[i] = new char[10]; day[i] = ((num%260)%13+1); day_name[i] = Tzolkin_day_code[(num%260)%20]; year[i] =( num/260); } cout<<n<<endl; for(int i=0; i<n; ++i) { cout<<day[i]<<" "; cout<<day_name[i]<<" "; cout<<year[i]<<endl; } return 0;}