题解 | #记负均正#__huawei-no.97-1

#include <iostream>
#include<numeric>
using namespace std;

int main() {
    int a ;
    cin >> a ;
    int sum = 0;
    int count_0 = 0;
    int count_1 = 0;
    int b = 0;
    for (int i = 0 ; i < a ; i++) {

        while (cin >> b) {
            if (b > 0) {
                sum = sum + b;
                count_1++;
            }
            if (b < 0) {
                count_0++;
            }
        }

    }
    printf("%d %.1f\n", count_0, count_1 == 0 ? 0.0 : static_cast<float>(float(sum) / float(count_1)));
    return 0;

}
// 64 位输出请用 printf("%lld") 这个题最难的是后面的输出