1052 卖个萌

[左手]([左眼][口][右眼])[右手]

Are you kidding me? @\/@。

输入样例：

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

输出样例：

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

C/C++ (stl) (基础解法看下方)

#include
using namespace std;
int main()
{
    vectorchar>> flag[3],result;  // 0手1眼2嘴,答案输出
    vector<char> box;                     // 辅助多个字符存入
    int N=0,n;
    string s;
 
    for(int z=0;z<3;z++){
        getline(cin,s,'\n'); // 这里考虑了这样的套娃情况[[-]],不过貌似没有这个测试点
        for(char x:s){
            if(x==']')  N--;
            if(N>0) box.push_back(x);
            if(N==0 && !box.empty()){
                flag[z].push_back(box);
                box.clear();
            }
            if(x=='[') N++;
        }
    }
 
    cin >> N;
    while (N--){
        bool f = true;  // 判断编号是否合理
        for(int z=0;z<5;z++){
            cin >> n;
            if(z%4==0)
            {
                if(n<=flag[0].size() && n>0) result.push_back(flag[0][n-1]);
                else f = false;
            }
            else if(z==2)
            {
                if(n<=flag[2].size() && n>0) result.push_back(flag[2][n-1]);
                else f = false;
            }
            else
            {
                if(n<=flag[1].size() && n>0) result.push_back(flag[1][n-1]);
                else f = false;
            }
        }
 
        if(!f) cout << "Are you kidding me? @\\/@" << endl;
        else
        {
            for(int z=0;z<5;z++) {
                for(char x:result[z]) cout << x;
                if(z==0) cout << "(";
                if(z==3) cout << ")";
            }
            cout << endl;
        }
        result.clear();
    }
 
    return 0;
}

C (基础解法)

#include
#include
int main(void) {
    char f[3][12][6], input[1001];
    int fNum[3] = {1, 1, 1}, N, a, b, c, d, e;
 
    for (int z = 0; z < 3; z++) {
        gets(input);
        int get = 0, len = 0;
        for (int z1 = 0; z1 < strlen(input); z1++) {
            if (input[z1] == ']') {
                f[z][fNum[z]][len] = '\0';
                fNum[z]++;
                get = len = 0;
            }
            if (get) f[z][fNum[z]][len++] = input[z1];
            if (input[z1] == '[') get = 1;
        }
    }
 
    scanf("%d", &N);
    while (N--) {
        int ok = 1;
        scanf("%d %d %d %d %d", &a, &b, &c, &d, &e);
        if (a < 1 || b < 1 || c < 1 || d < 1 || e < 1) ok = 0;
        if (a >= fNum[0] || b >= fNum[1] || c >= fNum[2] || d >= fNum[1] || e >= fNum[0]) ok = 0;
        if (ok==1) printf("%s(%s%s%s)%s\n", f[0][a], f[1][b], f[2][c], f[1][d], f[0][e]);
        else printf("Are you kidding me? @\\/@\n");
    }
 
    return 0;
}