[C++知识库]hnu 暑期实训之Maya历法

题目

思路

思路就是模拟跟着走就可以了。注意在取模运算的时候，如果a mod b当a==b的时候得到的结果是0，但是这题显然我们并不想得到0，所以要判断一下。既然是两种值的判断，那就不如玩点好玩的：三目运算符。
a=(条件判断) ? 值1:值2.
跟python中的if-else很像，但是C++的比较抽象，如果条件判断为真，那么就选择值1，否则就选择值2.

AC代码

#include <bits/stdc++.h>
using namespace std;

//先要建立一个关于冬青年的map
map<int, string> Tzolkin = {
	{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"},
	{14, "ix"},
	{15, "mem"},
	{16, "cib"},
	{17, "caban"},
	{18, "eznab"},
	{19, "canac"},
	{20, "ahau"},
};

map<string, int> Haab = {
	{"pop", 1},
	{"no", 2},
	{"zip", 3},
	{"zotz", 4},
	{"tzec", 5},
	{"xul", 6},
	{"yoxkin", 7},
	{"mol", 8},
	{"chen", 9},
	{"yax", 10},
	{"zac", 11},
	{"ceh", 12},
	{"mac", 13},
	{"kankin", 14},
	{"muan", 15},
	{"pax", 16},
	{"koyab", 17},
	{"cumhu", 18},
};

int main() {
	int n;
	char month[10];
	int day = 0, year = 0, month_day_sum = 0;
	cin >> n;
	for (int i = 1; i <= n; ++i) {
		scanf("%d.%s %d", &day, month, &year);
		string month_str;
		for (int j = 0; month[j]; ++j) {
			month_str += month[j];
		}
		//cout << month_str << endl;
		if (month_str == "uayet") {
			month_day_sum = 18 * 20;
		} else {
			month_day_sum = (Haab[month_str] - 1) * 20;
		}
		int day_sum = year * 365 + month_day_sum + day + 1;
		//cout << day_sum << endl;
		int year_new = day_sum / 260;
		int a = (day_sum - year_new * 260) % 20;
		//cout << a << endl;
		a = (a == 0) ? 20 : a;
		//cout << a << endl;
		string day_new = Tzolkin[a];
		int num_new = (day_sum - year_new * 260) % 13;
		num_new = (num_new == 0) ? 13 : num_new;
		cout << num_new << " " << day_new << " " << year_new << endl;
	}
}

收获

在调试代码的时候，发现当我们用char p[]，定义一个数组的时候，cout<<p<<endl。是可以直接把p输出的；
另外就是注意取模运算的结果的0啦！