题解 | #记负均正#

#include <iomanip>
#include <ios>
#include <iostream>
using namespace std;

int main() {
    int n;
    cin >> n;

    double sum = 0;
    int fnum = 0;
    int num0;

    for (int i = 0; i < n; i++) {
        int m;
        cin >> m;
        if (m < 0)
            fnum++;
        else if (m == 0)
            num0++;
        else
            sum += m;
    }

    if ( n - num0 - fnum == 0) {
        cout << 0 << ' ' << "0.0";
        return 0;
    }
    cout << fnum << ' ' << fixed << setprecision(1) << sum / (double)(
             n - num0 - fnum);
}
// 64 位输出请用 printf("%lld")