[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

这道题思路很简单，但是需要注意一处细节。

只需要算出给出的时间有多少天，然后转换成结果要的形式就行了。

比如设给出sumday天，sumday=年*365+月份*20+日+1，结果的年份是sumday%260，一处致命的细节是当给出的天数恰好是260时，输出结果应该是第一年的最后一天，也就是13 ahau 0。如果是260的倍数X*260，则应该是13 ahau X-1。

下面是AC代码

#include<iostream>using namespace std;int main(){    const string HAABMONTH[19] = {"pop","no","zip","zotz","tzec","xul","yoxkin","mol","chen","yax","zac","ceh","mac","kankin","muan","pax","koyab","cumhu","uayet"};    const string TZOLKINDAY[20] = {"imix", "ik","akbal","kan","chicchan","cimi","manik","lamat","muluk","ok","chuen","eb","ben","ix","mem","cib","caban","eznab","canac","ahau"};    int n;    cin>>n;    cout<<n<<endl;    int day,year;    string month;    char c;    while(n>0)    {        --n;        cin>>day>>c>>month>>year;        int sumday=year*365+day+1;        for(int i=0;i<19;++i)        {            if(month==HAABMONTH[i])            {                sumday+=i*20;                break;            }        }        year=sumday/260;        sumday%=260;        day=0;        int nameindex=-1;//相当于初始化为新一年的第0天，这天是无意义的，如果此时sumday==0，需要改写为前一年最后一天        while(sumday>0)//sumday模260的余数大于0时，需要算出在新一年第sumday个日子        {            --sumday;            if(++day==14) day=1;            if(++nameindex==20) nameindex=0;        }        if(nameindex==-1)//说明sumday恰好被260整除，需要设置为前一年最后一天而不是后一年第0天        {            day=13;            nameindex=19;            --year;        }        cout<<day<<' '<<TZOLKINDAY[nameindex]<<' '<<year<<endl;    }    return 0;}