摘要:本文将带你了解Maya动画制作之 acm.pku 1008 玛雅历,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 acm.pku 1008 玛雅历,希望本文对大家学Maya有所帮助
题目:Maya Calendar
Time Limit:
1000MS
Memory Limit:
10000K
Total Submissions:
33760
Accepted:
10300
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
3
10. zac 0
0. pop 0
10. zac 1995
Sample Output
3
3 chuen 0
1 imix 0
9 cimi 2801
Source
Central Europe 1995
编程环境: GCC
解法:[cpp] view plain copy print?#include <stdio.h> #include <conio.h> #include <stdlib.h> #include <string.h> int main(){ char Haab[19][7]={"pop", "no", "zip", "zotz", "tzec", "xul", "yoxkin","mol","chen","yax","zac","ceh","mac","kankin","muan","pax","koyab","cumhu","uayet"}; char Tzolkin[20][10]={"imix","ik","akbal","kan","chicchan","cimi","manik","lamat","muluk","ok","chuen","eb","ben","ix","mem","cib","caban","eznab","canac","ahau"}; char temp[7]; int m, *days, d, y, day; int total,times; int t,i; scanf("%d",&total); days= calloc(total,sizeof(int)); for(times=0;times<total;times++){ day=0; scanf("%d.%s%d",&d,temp,&y); for(m=0;strcmp(Haab[m],temp) && m<19 ;m++); day+=1+d+m*20+y*365; days[times]=day; } printf("%d/n",total); for(times=0;times<total;times++){ day=days[times]; y=(day-1)/260; day-=y*260; d=day%13; if(d==0) d=13; m=day%20; if(m==0) m=20; printf("%d %s %d/n",d,Tzolkin[m-1],y); } getch(); return 0; } #include <stdio.h>
#include <conio.h>
#include <stdlib.h>
#include <string.h>
int main(){
char Haab[19][7]={"pop", "no", "zip", "zotz", "tzec", "xul", "yoxkin","mol","chen","yax","zac","ceh","mac","kankin","muan","pax","koyab","cumhu","uayet"};
char Tzolkin[20][10]={"imix","ik","akbal","kan","chicchan","cimi","manik","lamat","muluk","ok","chuen","eb","ben","ix","mem","cib","caban","eznab","canac","ahau"};
char temp[7];
int m, *days, d, y, day;
int total,times;
int t,i;
scanf("%d",&total);
days= calloc(total,sizeof(int));
for(times=0;times<total;times++){
day=0;
scanf("%d.%s%d",&d,temp,&y);
for(m=0;strcmp(Haab[m],temp) && m<19 ;m++);
day+=1+d+m*20+y*365;
days[times]=day;
}
printf("%d/n",total);
for(times=0;times<total;times++){
day=days[times];
y=(day-1)/260;
day-=y*260;
d=day%13; if(d==0) d=13;
m=day%20; if(m==0) m=20;
printf("%d %s %d/n",d,Tzolkin[m-1],y);
}
getch();
return 0;
}
这是一道比较平常的日期转化题目,经查证,那两个历法:Haab 和Tzolkin 都是实际存在的,这也让我更多了解了点历史,很不错的一道题目。
开始想过用 enum 做,但个人对于enum 掌握的并不是很好,试验了下原来它仅仅是一个 (个人理解) 让程序员更方便记住数字的代表,并不能提供 数字 到 文字 的转化。 如果需要转化,上一些论坛搜索,人们提供的是这个解答方案:加上一段比如 switch 代码。
其中有一处很容易出现思维漏洞,就是最后那一处,要考虑Tzolkin最后一天的问题。感谢pku 里面discussion 某牛人提供了测试数据,让我及时发现了问题 。(测试数据:4. uayet 259 ,应该输出 年份最后一位是4, 不是5.)
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