模拟日历计算 poj1008
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 69932 | Accepted: 21524 |
Description
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
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 09 cimi 2801
题意：看了我近半小时Orz 英语太弱了
玛雅人有两种纪年：haab 和 holly
haab ： 365天
19个月 最后一个uayet月 仅仅有5天
holly
260天
13个月
每月都是20天
总的天数sum%13+1 为日期
sum%20+1 为月份
sum/260为年份
#include <iostream> #include <stdio.h> #include <string.h> #include <algorithm> using namespace std; int init_Haab(char *s) { if(strcmp(s,"pop")==0) return 1; else if(strcmp(s,"no")==0) return 2; else if(strcmp(s,"zip")==0) return 3; else if(strcmp(s,"zotz")==0) return 4; else if(strcmp(s,"tzec")==0) return 5; else if(strcmp(s,"xul")==0) return 6; else if(strcmp(s,"yoxkin")==0) return 7; else if(strcmp(s,"mol")==0) return 8; else if(strcmp(s,"chen")==0) return 9; else if(strcmp(s,"yax")==0) return 10; else if(strcmp(s,"zac")==0) return 11; else if(strcmp(s,"ceh")==0) return 12; else if(strcmp(s,"mac")==0) return 13; else if(strcmp(s,"kankin")==0) return 14; else if(strcmp(s,"muan")==0) return 15; else if(strcmp(s,"pax")==0) return 16; else if(strcmp(s,"koyab")==0) return 17; else if(strcmp(s,"cumhu")==0) return 18; else if(strcmp(s,"uayet")==0) return 19; } int sumday(int d,int m,int y) { int sum=0; for(int i=1;i<=y;i++) sum+=365; for(int i=1;i<m;i++)//千万注意 sum+=20; return sum+d; } // imix, ik, akbal, kan, chicchan, cimi, manik, lamat, muluk, ok, chuen, eb, ben, ix, mem, cib, caban, eznab, canac, ahau char str[25][10]={"0","imix","ik","akbal","kan","chicchan","cimi","manik","lamat","muluk","ok","chuen","eb","ben","ix","mem","cib","caban","eznab","canac","ahau"}; int main() { int t; scanf("%d",&t); printf("%d\n",t); while(t--) { int d,y; char s[10]; scanf("%d. %s %d",&d,s,&y); int sum = sumday(d,init_Haab(s),y); int hd=sum%13+1; int hm=sum%20+1; int hy=sum/260; printf("%d %s %d\n",hd,str[hm],hy); } }