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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#573040#9320. Find the Easiest ProblemfeifangeRE 0ms0kbC++142.0kb2024-09-18 17:09:122024-09-18 17:09:13

Judging History

你现在查看的是最新测评结果

  • [2024-09-18 17:09:13]
  • 评测
  • 测评结果:RE
  • 用时:0ms
  • 内存:0kb
  • [2024-09-18 17:09:12]
  • 提交

answer

#include <iostream>  
#include <cstring>  
  
using namespace std;  
  
int main() {  
    int t, n;  
    cin >> t;  // 读取测试用例数量  
    while (t--) {  
        cin >> n;  // 读取每个测试用例的提交数量  
          
        // 初始化问题统计数组,假设最多26个问题(A-Z)  
        int solvedCount[26] = {0};  
        char teamNames[100001][11];  // 存储所有团队名称,预留足够空间  
        bool teamSeen[100001] = {false};  // 标记团队是否已解决某问题  
          
        // 读取提交记录  
        for (int i = 0; i < n; i++) {  
            char team[11], problem;  
            string result;  
            cin >> team >> problem >> result;  
              
            // 将团队名称转换为整数索引(简单哈希,基于题目保证团队名不同)  
            int teamIndex = 0;  
            for (int j = 0; team[j] != '\0'; j++) {  
                teamIndex = teamIndex * 26 + (team[j] - 'A');  
            }  
              
            // 如果这是该团队第一次解决此问题,增加计数器  
            if (!teamSeen[teamIndex] && result == "accepted") {  
                teamSeen[teamIndex] = true;  
                solvedCount[problem - 'A']++;  
            }  
              
            // 存储团队名称,仅为了演示,实际代码中未使用  
            strcpy(teamNames[i], team);  
        }  
          
        // 寻找最易问题  
        char easiestProblem = 'Z' + 1;  // 初始化为大于'Z'的字符  
        int maxTeams = 0;  
        for (int i = 0; i < 26; i++) {  
            if (solvedCount[i] > maxTeams || (solvedCount[i] == maxTeams && easiestProblem > 'A' + i)) {  
                maxTeams = solvedCount[i];  
                easiestProblem = 'A' + i;  
            }  
        }  
          
        // 输出结果  
        cout << easiestProblem << endl;  
    }  
      
    return 0;  
}

详细

Test #1:

score: 0
Runtime Error

input:

2
5
teamA A accepted
teamB B rejected
teamC A accepted
teamB B accepted
teamD C accepted
4
teamA A rejected
teamB A accepted
teamC B accepted
teamC B accepted

output:


result: