【POJ】1008 Maya Calendar

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

310. zac 00. pop 010. zac 1995

Sample Output

33 chuen 01 imix 09 cimi 2801

一个日期转换问题，并不难。

用一个长整形存储从第一天开始到输入的haab历是多少天，

再转换成hooly历。

要注意的是haab历中的表达是【日. 月 年】，和hooly历是不一样的，

holly历的表达没有月份这一说，第一个数字和字符串组合表示这一年中的第几天。

haab历一年365天，而holly历一年只有260天。

在holly历中，第一个数字【0-13】和字符串是独立循环的，也就是说，数字循环20次，字符串循环13次。

怒贴代码：

#include<stdio.h>#include<iostream>#include<algorithm>#include<string.h>using namespace std;char haab[19][10]={"pop","no","zip","zotz","tzec","xul","yoxkin","mol","chen","yax","zac","ceh","mac","kankin","muan","pax", "koyab","cumhu","uayet"};char holly[20][10]={"imix","ik","akbal","kan","chicchan","cimi","manik","lamat","muluk","ok","chuen","eb","ben","ix","mem","cib","caban","eznab","canac","ahau"};int main(){    int t;    while(scanf("%d",&t)!=EOF)    {        printf("%d\n",t);        while(t--)        {            long long int day=0;            int dath,month,year1,year2;            char str[10];            scanf("%d. %s %d",&dath,str,&year1);            day+=year1*365+dath;            for(int i=0;i<19;i++)            {                if(strcmp(str,haab[i])==0)                {                    day+=i*20;                    break;                }            }            year2=day/260;            day=day%260;            month=1+day%13;            day=day%20;            printf("%d %s %d\n",month,holly[day],year2);        }    }    return 0;}