POJ 2965 玛雅历 解题报告
POJ 2965 玛雅历 解题报告
编号：2965
考查点：日期和时间处理
思路：在进行进制转换问题，包括日期和时间转换、数制转换的时候，一般需要先找到两种转换之间的公共点，这题里面就是逝去的天数，然后把天数按目的日历显示即可..
提交情况: 经过昨天POJ
2964的失败教训，我这次先整理出思路，然后写出框架函数，思路相当清晰，自己debug时因为天数变量未初始化出了点问题，不过只提交一次就直接AC了..享受..
Source Code：
//POJ Grids 2965
#include <string.h>
#include <iostream>
using namespace std;
char Haab[19][10] = {"pop","no", "zip", "zotz", "tzec", "xul", "yoxkin", "mol", "chen", "yax", "zac", "ceh","mac", "kankin", "muan", "pax", "koyab", "cumhu","uayet"};
char Tzolkin[20][10] = {"imix", "ik", "akbal", "kan", "chicchan", "cimi", "manik", "lamat", "muluk", "ok", "chuen", "eb", "ben", "ix", "mem", "cib", "caban", "eznab", "canac", "ahau"};
int type[13] = {1,2,3,4,5,6,7,8,9,10,11,12,13};
int days = 0;
void calday(char* day);
void calmonth(char* month);
void calyear(int year);
int main()
{
int n;
cin>>n;
cout<<n<<endl;
while (n--)
{
days = 0;
char day[10],month[10];
int year;
memset(day,0,10);
memset(month,0,10);
cin>>day>>month>>year;
calday(day);
calmonth(month);
calyear(year);
year = days/260;
days %= 260;
cout<<type[days%13];
cout<<" ";
cout<<Tzolkin[days%20];
cout<<" ";
cout<<year<<endl;
}
return 0;
}
void calday(char* day)
{
int temp = *day-'0';
while (*(++day)!='.')
{
temp = temp*10+*day-'0';
}
days += temp;
}
void calmonth(char* month)
{
for (int i=0;i<19;i++)
{
if (strcmp(month,Haab[i])==0)
{
days += i*20;
return;
}
}
}
void calyear(int year)
{
days += year*365;
}
总结：循环里用于每次统计的变量，尤其是全局变量，一定要记得每次循环先初始化，然后就是解决日期、时间转换问题要找出两种进制之间的公共点.
By Ns517
Time
