7 条题解

  • 1
    @ 2026-3-29 13:11:48
    //使用方法二
    #include <iostream>
    #include <iomanip>
    using namespace std;
    double a[10001],num;
    int n,maxx = -1;
    int main()
    {
        cin >> n;
        for (int i = 1;i <= n;i++)
        {
            cin >> num;
            a[(short)(num * 100)]++;
            if (a[(short)(num * 100)] > maxx)
                maxx = a[(short)(num * 100)];
        }
        for (int i = 1;i < 10001;i++)
        {
            if (a[i] == maxx)
                cout << fixed << setprecision(2) << i / 1.0 / 100.0 << ' ' << setprecision(0) << a[i] << endl;
        }
        return 0;
    }
    

    信息

    ID
    999
    时间
    1000ms
    内存
    128MiB
    难度
    5
    标签
    递交数
    403
    已通过
    157
    上传者