poj1008

玛雅历法

水题，简单的转换2种历法就可以了，要注意2个历法的最小公倍数

 1 #include<iostream>
 2 #include<string>
 3 using namespace std;
 4 
 5 string m[19]={"pop", "no", "zip", "zotz", "tzec", "xul", "yoxkin", "mol", "chen", "yax", "zac", "ceh", "mac", "kankin", "muan", "pax", "koyab", "cumhu","uayet"};
 6 string d[20]={"imix", "ik", "akbal", "kan", "chicchan", "cimi", "manik", "lamat", "muluk", "ok", "chuen", "eb", "ben", "ix", "mem", "cib", "caban", "eznab", "canac", "ahau"};
 7 
 8 int summons(string hm)
 9 {
10     for(int i=0;i<19;i++)
11     {
12         if(hm==m[i])
13         {
14             return i;
15         }
16     }
17     return -1;
18 }
19 
20 
21 
22 int main()
23 {
24     long n,i,j,k;
25     cin>>n;
26     cout<<n<<endl;
27     for(i=0;i<n;i++)
28     {
29         long hd;
30         string hm;
31         long hy;
32         cin>>hd>>hm>>hm>>hy;
33         long hm1=summons(hm);
34         long hdays=0;
35         hdays=hy*365+20*hm1+hd;
36         
37 
38 
39         
40         long y=hdays/260;
41         long  days=hdays%260;
42         
43         cout<<days%13+1<<" "<<d[days%20]<<" "<<y<<endl;
44     }
45     return 0;
46 }
posted @ 2012-10-26 23:53  zerojetlag  阅读(1077)  评论(0)  编辑  收藏  举报