新生训练赛round2--G. Maya Calendar(模拟)

G. Maya Calendar

Time Limit: 1000ms

Memory Limit: 10000KB

64-bit integer IO format: %lld      Java class name: Main

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

---------------------------

赛后看的题目解得题。
首先看到这道题的题目就有点恐惧。。。那么多数据，代码肯定不短。
然后写了很多if，不停地复制粘贴过去，短的就两个两个的一起手打，然后导致后面WA一个致命错误——输错了。。。

还有一个WA是第一次WA时debug到的，
换算过去的周期13，我写的（x+1）%13会跳过13这个数字。正确应该是x%13后+1。

code：

#include
#include
using namespace std;



int Haab_month(char* s)
{
    if(!strcmp(s,"pop")) return 0;
    if(!strcmp(s,"no")) return 1;
    if(!strcmp(s,"zip")) return 2;
    if(!strcmp(s,"zotz")) return 3;
    if(!strcmp(s,"tzec")) return 4;
    if(!strcmp(s,"xul")) return 5;
    if(!strcmp(s,"yoxkin")) return 6;
    if(!strcmp(s,"mol")) return 7;
    if(!strcmp(s,"chen")) return 8;
    if(!strcmp(s,"yax")) return 9;
    if(!strcmp(s,"zac")) return 10;
    if(!strcmp(s,"ceh")) return 11;
    if(!strcmp(s,"mac")) return 12;
    if(!strcmp(s,"kankin")) return 13;
    if(!strcmp(s,"muan")) return 14;
    if(!strcmp(s,"pax")) return 15;
    if(!strcmp(s,"koyab")) return 16;
    if(!strcmp(s,"cumhu")) return 17;
    if(!strcmp(s,"uayet")) return 18;
}



int main()
{
    int sum;
    int n;
    scanf("%d",&n);
    printf("%d\n",n);
    while( n-- )
    {
        sum = 0;
        int m_d,m_y;
        char m_m[10];
        scanf("%d",&m_d);
        sum += m_d;
        getchar();
        getchar();
        scanf("%s",m_m);
        sum += Haab_month(m_m) * 20;
        scanf("%d",&m_y);
        sum += m_y * 365;


        int t_y = sum / 260;
        int t_d = sum % 260;
        int t_mm = t_d % 20;
        char t_m[10];
        if(t_mm == 0) strcpy(t_m,"imix");
        if(t_mm == 1) strcpy(t_m,"ik");
        if(t_mm == 2) strcpy(t_m,"akbal");
        if(t_mm == 3) strcpy(t_m,"kan");
        if(t_mm == 4) strcpy(t_m,"chicchan");
        if(t_mm == 5) strcpy(t_m,"cimi");
        if(t_mm == 6) strcpy(t_m,"manik");
        if(t_mm == 7) strcpy(t_m,"lamat");
        if(t_mm == 8) strcpy(t_m,"muluk");
        if(t_mm == 9) strcpy(t_m,"ok");
        if(t_mm == 10) strcpy(t_m,"chuen");
        if(t_mm == 11) strcpy(t_m,"eb");
        if(t_mm == 12) strcpy(t_m,"ben");
        if(t_mm == 13) strcpy(t_m,"ix");
        if(t_mm == 14) strcpy(t_m,"mem");
        if(t_mm == 15) strcpy(t_m,"cib");
        if(t_mm == 16) strcpy(t_m,"caban");
        if(t_mm == 17) strcpy(t_m,"eznab");
        if(t_mm == 18) strcpy(t_m,"canac");
        if(t_mm == 19) strcpy(t_m,"ahau");
        t_d = (t_d % 13) + 1;   //  之前错误的表示(t_d + 1)% 13 这种错误的表示不会输出"13"这个数字
        printf("%d %s %d\n",t_d,t_m,t_y);
    }
    return 0;
}