1 条题解

  • 0
    @ 2025-3-22 16:54:02

    #include #include using namespace std; int main() { struct Student { string name; int chinese, m, e, sum; }student[1000]; int N; cin>>N; for (int i = 0; i < N; i++) { cin>>student[i].name>>student[i].chinese>>student[i].m>>student[i].e; student[i].sum = student[i].chinese + student[i].m + student[i].e; } for (int i = 0; i < N - 1; i++) { for (int j = i + 1; j < N; j++) { if (abs(student[i].chinese - student[j].chinese) <= 5 && abs(student[i].m - student[j].m) <= 5 && abs(student[i].e - student[j].e) <= 5 && abs(student[i].sum - student[j].sum) <= 10) { if (student[i].name < student[j].name) cout<<student[i].name << " " << student[j].name << "\n"; else cout<<student[j].name << " " << student[i].name << "\n"; } } } return 0; }

    【深基7.例10】旗鼓相当的对手 - 加强版

    信息

    ID
    938
    时间
    1000ms
    内存
    256MiB
    难度
    8
    标签
    (无)
    递交数
    15
    已通过
    5
    上传者