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  • POJ 1008 Maya Calendar

    链接:http://poj.org/problem?id=1008
    Maya Calendar
    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 71745   Accepted: 22083

    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

    题意:有两种日历: 第一个为 Haab   一年365天,共19个月,前18个月的名字pop, no, zip, zotz, tzec, xul... 每个月20天 ,0 --- 19,最后的名字是uayet, 这个月只有5天, 0 --- 5。
                                第二个为Tzolkin 一年260天,他们用20个英文和13个数字组合来表示每一天。20个英文依次是:imix, ik, akbal,……。年初几天的描述如下:1 imix, 2 ik, 3 akbal……。
     思路:可先将两个日历的名字存起来,然后求第一个日历的天数,转换为第二个即可;
    #include<iostream>
    #include<algorithm>
    #include<stdio.h>
    #include<stdlib.h>
    #include<string.h>
    #include<math.h>
    using namespace std;
    char s1[20][10]={"pop","no","zip","zotz","tzec","xul","yoxkin","mol","chen","yax","zac","ceh","mac","kankin","muan","pax","koyab","cumhu","uayet"};
    char s2[20][10]={"imix","ik","akbal","kan","chicchan","cimi","manik","lamat","muluk","ok","chuen","eb","ben","ix","mem","cib","caban","eznab","canac","ahau"};
    int main()
    {
        int t,day,year,ans,sum,num;
        char mon[10];
        scanf("%d",&t);
        printf("%d
    ",t);
        for(int xx=0;xx<t;xx++)
        {
            ans=sum=0;
            scanf("%d. %s %d",&day,mon,&year);
            for(int i=0;i<20;i++)
            {
                if(strcmp(s1[i],mon)==0)
                {
                    ans=year*365+day+i*20;
                     break;
                }
    
            }
            year=ans/260;
            num=ans%260;
            day=num%13;
            day+=1;
            printf("%d %s %d
    ",day,s2[num%20],year);
        }
        return 0;
    }

     

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  • 原文地址:https://www.cnblogs.com/cn-blog-cn/p/4936587.html
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