解题思路

1.分别用数组模拟Haab和Tzolkin
2.把给的输入，转化成离世界开始有多少天
3.计算Tzolkin

AC代码

//计算输入的日子离世界开始的天数 days //年*365 + yue*20 + day	//计算年份 year  = days / 260    days -= year*260
// 计算月份  前面的数字num+1  num= days % 13      holly[days%20] 
#include<stdio.h>
#include<string.h>
int main(){char haab[19][10] = {"pop","no","zip","zotz","tzec","xul","yoxkin","mol","chen","yax","zac","ceh","mac","kankin","muan","pax","koyab","cumhu","uayet"};  //一年365 char holly[20][10]={"imix", "ik", "akbal", "kan","chicchan", "cimi", "manik", "lamat","muluk", "ok", "chuen", "eb", "ben", "ix","mem", "cib", "caban", "eznab", "canac", "ahau"};//一年260天int n;int date,year,days;char month[10];scanf("%d",&n);printf("%d\n",n);for(int i=0;i<n;i++){scanf("%d. %s %d",&date,month,&year); //学习了 int count;for(count=0;count<19;count++) //学习了 if(!strcmp(haab[count],month))break;//计算输入的日子离世界开始的天数 days days = year*365+count*20+date;//计算Tzolkinint tyear = days / 260;days -= tyear*260;int num = days %13;int mon = days %20;printf("%d %s %d\n",num+1,holly[mon],tyear);} return 0;				   
}

总结

1）用数组模拟
2）两个日历的转换，使用中间公共的日子，例如离开始日子多少天
3）格式化输入

scanf("%d. %s %d",&date,month,&year)

4）找字符串出现在第几个

for(count=0;count<19;count++) if(!strcmp(haab[count],month))break;