1 条题解

  • 0
    @ 2025-11-6 22:16:27

    按题意直接模拟即可。

    AC code:

    #include <bits/stdc++.h>
    using namespace std;
    
    int main() {
        int n, k, has_ex;
        cin >> n >> k >> has_ex;
        map<int, set<int>> problem_map;
        for (int i = 0; i < n; ++i) {
            int id, hard;
            cin >> id >> hard;
            problem_map[id].insert(hard);
        }
        for (int id = 1; id <= k; ++id) {
            char char_id = 'A' + id - 1;
            set<int>& hard_set = problem_map[id];
    
            if (hard_set.size() == 1) {
                cout << char_id << endl;
            } else {
                int idx = 1;
                for (int hard : hard_set) {
                    cout << char_id << idx << endl;
                    idx++;
                }
            }
        }
        if (has_ex == 1) {
            cout << "Ex" << endl;
        }
        return 0;
    }
    
    • 1

    信息

    ID
    141
    时间
    1000ms
    内存
    128MiB
    难度
    2
    标签
    递交数
    7
    已通过
    3
    上传者