/* 模擬題，細(xì)心點(diǎn)就好了，沒難度。 Habb歷一年365天 Tzolkin歷一年260天 先計(jì)算Habb歷從第0天到輸入日期的總天數(shù)sumday Sumday/day就是Tzolkin歷的年份 Tzolkin歷的天數(shù)Name每20一循環(huán)，先建立Tzolkin歷天數(shù)Name與1~20的映射， 因此Sumday %20+1就是Tzolkin歷的天數(shù)Name Tzolkin歷的天數(shù)ID每13一循環(huán)，且從1開始，則Sumday %13+1就是Tzolkin歷的天數(shù)ID */ #include<iostream> #include<stdio.h> #include<string.h>using namespace std; int main() { char H[19][10]={"pop","no", "zip", "zotz", "tzec", "xul", "yoxkin", "mol", "chen", "yax", "zac", "ceh", "mac", "kankin", "muan", "pax", "koyab", "cumhu","uayet"}; char T[20][10]={"imix", "ik", "akbal", "kan", "chicchan", "cimi", "manik", "lamat", "muluk", "ok", "chuen", "eb", "ben", "ix", "mem", "cib", "caban", "eznab", "canac", "ahau"}; int day,year; char month[10]; int n,i,j,days; cin>>n; cout<<n<<endl; for(i=0;i<n;i++) { scanf("%d. %s %d",&day,month,&year); for(j=0;j<19;j++) { if(!strcmp(month,H[j])) break; } days=year*365+j*20+day; cout<<1+days%13<<" "<<T[days%20]<<" "<<days/260<<endl;}return 0; }

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