poj 1008:Maya Calendar(模拟题,玛雅日历转换)
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 64795 | Accepted: 19978 |
Description
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
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
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
Habb历一年365天
Tzolkin历一年260天
先计算Habb历从第0天到输入日期的总天数sumday
Sumday/day就是Tzolkin历的年份
Tzolkin历的天数Name每20一循环,先建立Tzolkin历天数Name与1~20的映射,
因此Sumday %20+1就是Tzolkin历的天数Name
Tzolkin历的天数ID每13一循环,且从1开始,则Sumday %13+1就是Tzolkin历的天数ID
year = sumday/;
dayName = sumday%; //这只是下标,最后输出对应的天数名字
dayId = sumday%+;
最后输出:
cout<<dayId<<' '<<Tzolkin[dayName]<<' '<<year<<endl;
代码:
#include <iostream>
#include <string.h>
using namespace std;
char Habb[][] =
{"pop", "no", "zip", "zotz", "tzec",
"xul", "yoxkin", "mol", "chen", "yax",
"zac", "ceh", "mac", "kankin", "muan",
"pax", "koyab", "cumhu","uayet"};
char Tzolkin[][] =
{"imix", "ik", "akbal", "kan", "chicchan",
"cimi", "manik", "lamat", "muluk", "ok",
"chuen", "eb", "ben", "ix", "mem",
"cib", "caban", "eznab", "canac", "ahau"
};
int Name2month(char name[]) //返回该名字对应的月份
{
for(int i=;i<;i++){ //依次比较
if(strcmp(Habb[i],name)==)
return i+;
}
return -;
}
int GetHabbSumday(int day,int month,int year) //获得总天数
{
int sumday = ;
sumday += year*;
sumday += *(month-);
sumday += day;
return sumday;
}
int main()
{
int n,Number,Year;
char c,Name[];
cin>>n;
cout<<n<<endl;
while(n--){
cin>>Number>>c>>Name>>Year; //输入
int sumday = GetHabbSumday(Number,Name2month(Name),Year); //获得总天数
//计算结果
int year = sumday/;
int dayName = sumday%;
int dayId = sumday%+;
cout<<dayId<<' '<<Tzolkin[dayName]<<' '<<year<<endl;
}
return ;
}
Freecode : www.cnblogs.com/yym2013
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