学步园

//一道很不错的题目。忽略细节WA了3次<br /> //要注意特殊情况<br /> //如果是一年的最后一天，注意年份必须减1<br /> //总结：在做关于取模运算时，注意考虑当结果为0时是否是你要的答案<br /> #include<iostream><br /> #include<map><br /> #include<string><br /> using namespace std;<br /> map<string,int> stoi;<br /> map<int,string> itos;//利用map关联容器实现月份与数字的对应以及数字与月份的对应<br /> int n,dd,mm,yy;<br /> int DD,MM,YY,sumDay;<br /> char d[10];<br /> string m;<br /> int main()<br /> {<br /> stoi["pop"] = 1;stoi["no"] = 2;stoi["zip"] = 3;stoi["zotz"] = 4;stoi["tzec"] = 5;stoi["xul"] = 6;<br /> stoi["yoxkin"] = 7;stoi["mol"] = 8;stoi["chen"] = 9;stoi["yax"] = 10;stoi["zac"] = 11;stoi["ceh"] = 12;<br /> stoi["mac"] = 13;stoi["kankin"] = 14;stoi["muan"] = 15;stoi["pax"] = 16;stoi["koyab"] = 17;stoi["cumhu"] = 18;stoi["uayet"] = 19;<br /> itos[1] = "imix";itos[2] = "ik";itos[3] = "akbal";itos[4] = "kan";itos[5] = "chicchan";<br /> itos[6] = "cimi";itos[7] = "manik";itos[8] = "lamat";itos[9] = "muluk";itos[10] = "ok";<br /> itos[11] = "chuen";itos[12] = "eb";itos[13] = "ben";itos[14] = "ix";itos[15] = "mem";<br /> itos[16] = "cib";itos[17] = "caban";itos[18] = "eznab";itos[19] = "canac";itos[20] = "ahau";<br /> cin >> n;<br /> cout << n << endl;<br /> while(n--)<br /> {<br /> cin >> d >> m >> yy;<br /> dd = atoi(d) + 1;//注意加1,用atoi即可实现简单的转化<br /> mm = stoi[m];<br /> sumDay = yy * 365 + (mm != 19 ? (mm-1)*20 + dd : 18*20 + dd);<br /> YY = sumDay / 260;<br /> sumDay %= 260;<br /> if(sumDay == 0)<br /> {<br /> sumDay = 260;<br /> YY--;<br /> }<br /> DD = sumDay % 13;<br /> DD = (DD == 0 ? 13 : DD);//对取模运算的细节考虑<br /> MM = sumDay % 20;<br /> MM = (MM == 0 ? 20 : MM);//对取模运算的细节考虑<br /> cout << DD << " " << itos[MM] << " " << YY << endl;<br /> }<br /> return 0;<br /> }