topic:
Moemoe emoticons are usually composed of three main parts: "hand", "eye" and "mouth". For simplicity, we assume that an emoji is output in the following format:
[左手]([左眼][口][右眼])[右手]
Now give a set of optional symbols, please output emoticons according to the user's requirements.
Input format:
The input first gives the optional symbol sets of hands, eyes, and mouths in the first three lines. Each symbol is enclosed in a pair of square brackets []. The title guarantees that each set has at least one symbol and no more than 10 symbols; each symbol contains 1 to 4 non-null characters.
The next line gives a positive integer K, which is the number requested by the user. Followed by K lines, each line gives a user's symbol selection, the order is left hand, left eye, mouth, right eye, right hand - only the serial number of the symbol in the corresponding set (starting from 1) is given here, and the numbers are separated by spaces separated.
Output format:
For each user request, output the generated emoji on one line. If the serial number selected by the user does not exist, it is output Are you kidding me? @\/@.
Input sample:
[╮][╭][o][~\][/~] [<][>]
[╯][╰][^][-][=][>][<][@][⊙]
[Д][▽][_][ε][^] ...
4
1 1 2 2 2
6 8 1 5 5
3 3 4 3 3
2 10 3 9 3
Sample output:
╮(╯▽╰)╭
<(@Д=)/~
o(^ε^)o
Are you kidding me? @\/@
Code length limit 16 KB
time limit 400 ms
memory limit 64 MB
problem solving ideas
- Given the three optional character sets of "hand", "eye", and "mouth", we need to extract their valid characters in order and put them into a vector container (the The element type is string type.
- Then the serial number -1 of the input "hand", "eye" and "mouth" corresponds to the subscript in the vector, and it can be output; but if the serial number selected by the user does not exist, then output, because it is
Are you kidding me? @\/@a\transfer Escape characters cannot be output directly, so you need to\escape with another escape character to\\output backslashes.
AC code
#include <bits/stdc++.h>
using namespace std;
int main()
{
vector<string> v[3];
for (int i = 0; i < 3; i++)
{
string tmp;
//由于一行中可能出现空格，所以不能用cin进行输入，需要用getline进行输入一行
getline(cin, tmp);
int find1 = 0;
int find2 = 0;
while (1)
{
find1 = tmp.find('[', find1);
find2 = tmp.find(']', find2);
if (find1 == -1 && find2 == -1)
{
break;
}
v[i].push_back(tmp.substr(find1+1, find2 - find1 -1));
find1++;
find2++;
}
}
int n = 0;
cin >> n;
while (n--)
{
int a1, a2, a3, a4, a5 = 0;
cin >> a1 >> a2 >> a3 >> a4 >> a5;
a1 -= 1; a2 -= 1; a3 -= 1; a4 -= 1; a5 -= 1;
if (a1 >= v[0].size() || a2 >= v[1].size()
|| a3 >= v[2].size() || a4 >= v[1].size()
|| a5 >= v[0].size())
{
cout << "Are you kidding me? @\\/@" << endl;
}
else
{
cout << v[0][a1] << "(" << v[1][a2] << v[2][a3]
<< v[1][a4] << ")" << v[0][a5] << endl;
}
}
return 0;
}