# 题目

[左手]([左眼][口][右眼])[右手]

[╮][╭][o][~\][/~] [<][>]
[╯][╰][ ^][-][=][>][<][@][⊙]
[Д][▽][_][ε][^] …
4
1 1 2 2 2
6 8 1 5 5
3 3 4 3 3
2 10 3 9 3

╮(╯▽╰)╭
<(@Д=)/~
o(^ε^)o
Are you kidding me? @\/@

# 分析

1.getline(cin,string)之后需要清空cin，方法如下

cin.sync();
cin.clear();


2.题目说“这里只给出符号在相应集合中的序号（从1开始）”
3.别忘记给眼睛的左右两边加英文括号

# 源代码

//C/C++实现
#include <iostream>
#include <string>
#include <vector>

using namespace std;

int main(){
string hands, eyes, mouths;
vector<string> hands_v(1), eyes_v(1), mouths_v(1); //先放一个占位置
//input hands
int i = 0;
getline(cin, hands);
int end = hands.find(']', i);
while(end != -1){
int start = hands.find('[', i);
hands_v.push_back(hands.substr(start + 1, end - start - 1));
i = end + 1;
end = hands.find(']', i);
}
cin.sync();
cin.clear();
//input eyes
i = 0;
getline(cin, eyes);
end = eyes.find(']', i);
while(end != -1){
int start = eyes.find('[', i);
eyes_v.push_back(eyes.substr(start + 1, end - start - 1));
i = end + 1;
end = eyes.find(']', i);
}
cin.sync();
cin.clear();
//input mouths
i = 0;
getline(cin, mouths);
end = mouths.find(']', i);
while(end != -1){
int start = mouths.find('[', i);
mouths_v.push_back(mouths.substr(start + 1, end - start - 1));
i = end + 1;
end = mouths.find(']', i);
}
cin.sync();
cin.clear();
int n;
scanf("%d", &n);
int tmp;
for(int j = 0; j < n; ++j){
string result = "";
//左手
scanf("%d", &tmp);
if(tmp > 0 && tmp < hands_v.size()){
result += hands_v[tmp];
}
else{
printf("Are you kidding me? @\\/@\n");
continue;
}
//左眼
scanf("%d", &tmp);
if(tmp > 0 && tmp < eyes_v.size()){
result += ("(" + eyes_v[tmp]);
}
else{
printf("Are you kidding me? @\\/@\n");
continue;
}
//口
scanf("%d", &tmp);
if(tmp > 0 && tmp < mouths_v.size()){
result += mouths_v[tmp];
}
else{
printf("Are you kidding me? @\\/@\n");
continue;
}
//右眼
scanf("%d", &tmp);
if(tmp > 0 && tmp < eyes_v.size()){
result += (eyes_v[tmp] + ")");
}
else{
printf("Are you kidding me? @\\/@\n");
continue;
}
//右手
scanf("%d", &tmp);
if(tmp > 0 && tmp < hands_v.size()){
result += hands_v[tmp];
}
else{
printf("Are you kidding me? @\\/@\n");
continue;
}
cout << result << endl;
}
return 0;
}