题目链接:传送门
思路:数的大小小于100,所以直接判输输入的是字母还是数字分情况处理即可,代码又写丑了,对两种情况都采用了循环来处理(但其实最多两位数,直接判断就行了。。)
代码:
#include <bits/stdc++.h>
using namespace std;
string f[2][13] = {{"tret" , "jan", "feb", "mar" , "apr", "may" , "jun" , "jly" , "aug" , "sep" , "oct" , "nov" , "dec"} ,
{"tret" , "tam" , "hel" , "maa" , "huh" , "tou" , "kes" , "hei" , "elo", "syy" , "lok" , "mer", "jou"}};
int main() {
int n;
cin >> n;
cin.get();
for(int i = 0 ; i < n ; i++) {
string s;
getline(cin , s);
if(isdigit(s[0])) {
vector <string> v;
int t = 0;
for(int j = 0 ; j < s.length() ; j++) {
t = t * 10 + s[j] - '0';
}
int tag = 0;
if(t >= 13)tag = 1;
if(!t) {
cout << "tret\n";
continue;
}
while(t) {
int x = t % 13;
if(x == 0) {
t /= 13; continue;
}
int t1 = 0;
if(t < 13 && tag)t1 = 1;
v.push_back(f[t1][x]);
t /= 13;
}
for(int j = v.size() - 1 ; j >= 0 ; j--) {
if(j < v.size() - 1)cout << " ";
cout <<v[j];
}
cout << "\n";
}
else {
stringstream ss(s);
string q;
int tag = 0 , ans = 0;
while(ss >> q) {
tag = 0;
for(int k = 0 ; k < 2 && !tag; k++) {
for(int j = 0 ; j < 13 ; j++) {
if(f[k][j] == q) {
if(k == 1)ans += j * 13;
else ans += j;
tag = 1;
break;
}
}
}
}
cout << ans << "\n";
}
}
return 0;
}