题解 | #记负均正II#

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

int main() {
    int num;
    int negCount = 0;    // 负数个数
    int posCount = 0;    // 非负数个数
    double posSum = 0;   // 非负数之和
    
    while (cin >> num) {
        if (num < 0) {
            negCount++;
        } else {
            posCount++;
            posSum += num;
        }
    }
    
    cout << negCount << endl;
    if (posCount == 0) {
        cout << "0.0" << endl;
    } else {
        cout << fixed << setprecision(1) << posSum / posCount << endl;
    }
    
    return 0;
}

import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int negCount = 0;    // 负数个数
        int posCount = 0;    // 非负数个数
        double posSum = 0;   // 非负数之和
        
        while (sc.hasNextInt()) {
            int num = sc.nextInt();
            if (num < 0) {
                negCount++;
            } else {
                posCount++;
                posSum += num;
            }
        }
        
        System.out.println(negCount);
        if (posCount == 0) {
            System.out.println("0.0");
        } else {
            System.out.printf("%.1f%n", posSum / posCount);
        }
    }
}

neg_count = 0    # 负数个数
pos_count = 0    # 非负数个数
pos_sum = 0      # 非负数之和

while True:
    try:
        num = int(input())
        if num < 0:
            neg_count += 1
        else:
            pos_count += 1
            pos_sum += num
    except:
        break

print(neg_count)
if pos_count == 0:
    print("0.0")
else:
    print(f"{pos_sum/pos_count:.1f}")