POJ 1008 Maya Calendar (模拟)

Maya Calendar

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 78576 Accepted: 24165

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

题意：玛雅人有两种历法，一种叫Haab，另一种叫Tzolkin。

Haab历法：一年365天 = 18个月*30天 + 1个月*5天

这19个月的名字分别是：pop, no, zip, zotz, tzec, xul, yoxkin, mol, chen, yax, zac, ceh, mac, kankin, muan, pax, koyab, cumhu, uayet.

天则直接用数字表示。

Tzolkin历法：一年260天 = 20*13，使用20个单词和13个数字来记录日期，

单词分别是imix, ik, akbal, kan, chicchan, cimi, manik, lamat, muluk, ok, chuen, eb, ben, ix, mem, cib, caban, eznab, canac, ahau

日期是这样表示的：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, （单词进入下一个循环） 8 imix, 9 ik, 10 akbal . . .

可以看到这种历法是用单词和数字各自循环然后组合，产生260种组合，也就对应了260天。

现在我们要做的就是把输入的Haab历法转换为Tzolkin历法输出

解题思路：用map<string,int>把所有Haab历法的月份转换为数字，然后计算 总天数=365*年份+30*月份+天数

然后用 总天数/260 就可以得到Tzolkin历法的年份，总天数%260就是这年的第多少天。为了方便起见，我们先把这260天所有的表示法都用map<int,string>计算出来。

注意：这里要稍微注意下的就是起始日期。

Haab: 0. pop 0

Tzolkin: 1 imix 0 

代码：

#include <iostream>#include <cmath>#include <cstdio>#include <map>#include <sstream>using namespace std;char m1[][10]={"pop","no","zip","zotz","tzec","xul","yoxkin","mol","chen","yax","zac","ceh","mac","kankin","muan","pax","koyab","cumhu","uayet"};char m2[][10]={"imix","ik","akbal","kan","chicchan","cimi","manik","lamat","muluk","ok","chuen","eb","ben","ix","mem","cib","caban","eznab","canac","ahau"};map<string,int> haab;map<int,string> tzo;string num2str(int i)//int转string{    stringstream ss;    ss<<i;    return ss.str();}int main(){    //把Haab历法的月份对应成0~19    for(int i=0;i<19;i++)    {        haab[m1[i]]=i;    }    //把Tzolkin历法260天每天对应的表示法计算出来    int a=0,b=1;    for(int i=0;i<260;i++,a++,b++)    {        if(a>=20)            a=0;        if(b>13)            b=1;        string data=num2str(b)+" "+(string)m2[a];        tzo[i]=data;    }    int n;    scanf("%d",&n);    printf("%d\n",n);    while(n--)    {        int d,y;        char m[10];        scanf("%d.%s%d",&d,m,&y);        int days=y*365+haab[m]*20+d;        cout<<tzo[days%260]<<' '<<days/260<<endl;    }    return 0;}