POJ1008-Maya Calendar

Maya Calendar

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 81342		Accepted: 25016

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. 

链接

一、题意

        第一种日历叫Haab，一年有365天，19个月，前18个月为每个月20天，最后一个月为5天。年份从0开始编号，月份有名字，天数从0开始编号。

        第二种日历叫做Tzolkin，一年有260天，每天由两个部分组成，第一部分为数字，从1到13，第二部分为名字，两个部分各自循环出现。

        现在要求从Haab转换到Tzolkin

二、思路

        先将总天数算出来，从0开始。记Haab的年月日分别为hy, hm, hd，则总天数total = 365 * hy + 20 * hm + hd

        在转化为Tzolkin的对应日期，ty = total / 260；tm = total % 13 + 1；td = total % 20

三、代码

#include <iostream>
#include <sstream>
#include <iomanip>
#include <string>
#include <numeric>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <utility>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>

using namespace std;
typedef long long ll;
const int MAXN = 100100;
const int MOD7 = 1000000007;
const int MOD9 = 1000000009;
const int INF = 2000000000;//0x7fffffff
const double EPS = 1e-9;
const double PI = 3.14159265358979;
const int dir_4r[] = { -1, 1, 0, 0 };
const int dir_4c[] = { 0, 0, -1, 1 };
const int dir_8r[] = { -1, -1, -1, 0, 0, 1, 1, 1 };
const int dir_8c[] = { -1, 0, 1, -1, 1, -1, 0, 1 };

string str1[] = { "pop", "no", "zip", "zotz", "tzec",
					"xul", "yoxkin", "mol", "chen", "yax", 
					"zac", "ceh", "mac", "kankin", "muan",
					"pax", "koyab", "cumhu", "uayet" };

string str2[] = { "imix", "ik", "akbal", "kan", "chicchan",
					"cimi", "manik", "lamat", "muluk", "ok",
					"chuen", "eb", "ben", "ix", "mem",
					"cib", "caban", "eznab", "canac", "ahau" };

map<string, int> habbM;

int main() {
	int n, hy, hm, hd, ty, tm, td;
	char str[10];

	for (int i = 0; i < 19; ++i)
		habbM[str1[i]] = i;

	scanf("%d\n", &n);
	printf("%d\n", n);
	for (int i = 0; i < n; ++i) {
		scanf("%d. %s %d", &hd, str, &hy);
		hm = habbM[str];
		int total = 365 * hy + 20 * hm + hd;

		ty = total / 260;
		tm = total % 13;
		td = total % 20;

		printf("%d %s %d\n", tm + 1, str2[td].c_str(), ty);
	}

	//system("pause");
	return 0;
}