B1052 卖个萌 (20 分| 字符串，附详细注释，逻辑分析)

写在前面

• 实现思路

• vector容器封装表情结果，​​vector<vector<string>> face;​​
• 循环字符读入
• 循环字符输出（非法示例输出）

• 问题点

• “Are you kidding me? @/@”的’\’是转义字符，想要输出’\’就要用’\’表示～

• 题目简单，15分钟a题

测试用例

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

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

ac代码

#include <iostream>
#include <vector>
using namespace std;

int main()
{
    vector<vector<string>> face;
    for(int i=0; i<3; i++)
    {
        vector<string> strs;
        string units;
        char ch;
        while(1)
        {
            scanf("%c", &ch);
            if(ch == '\n') break;
            if((ch==' ') || (ch=='['))
                continue;
            if(ch==']')
            {
                strs.push_back(units);
                units = "";
            }
            else
                units += ch;
        }
        face.push_back(strs);
    }

    int n;
    scanf("%d", &n);
    for(int i=0; i<n; i++)
    {
        int a, b, c, d, e;

        cin >> a >> b >> c >> d >> e;
        if(a > face[0].size() || b > face[1].size() || c > face[2].size() || d > face[1].size() || e > face[0].size() || a < 1 || b < 1 || c < 1 || d < 1 || e < 1)
        {
            printf("Are you kidding me? @\\/@\n");
            continue;
        }
        cout << face[0][a-1] << "(" << face[1][b-1] << face[2][c-1] << face[1][d-1] << ")" << face[0][e-1] << endl;
    }

    return 0;
}

​​学习代码​​

#include <iostream>
#include <vector>
using namespace std;
int main() {
    vector<vector<string> > v;
    for(int i = 0; i < 3; i++) {
        string s;
        getline(cin, s);
        vector<string> row;
        int j = 0, k = 0;
        while(j < s.length()) {
            if(s[j] == '[') {
                while(k++ < s.length()) {
                    if(s[k] == ']') {
                        row.push_back(s.substr(j+1, k-j-1));
                        break;
                    }
                }
            }
            j++;
        }
        v.push_back(row);
    }
    int n;
    cin >> n;
    for(int i = 0; i < n; i++) {
        int a, b, c, d, e;
        cin >> a >> b >> c >> d >> e;
        if(a > v[0].size() || b > v[1].size() || c > v[2].size() || d > v[1].size() || e > v[0].size() || a < 1 || b < 1 || c < 1 || d < 1 || e < 1) {
            cout << "Are you kidding me? @\\/@" << endl;
            continue;
        }
        cout << v[0][a-1] << "(" << v[1][b-1] << v[2][c-1] << v[1][d-1] << ")" << v[0][e-1] << endl;
    }
    return 0;
}