我以为N比原题多开了一位是要卡我cin、cout，结果被单个负号卡了，原题判定没给这个测试点，佩服。

思路：

分别判断带字母的、小数点超过一个的、小数点一个但是后面超过两位的等非法字符串，合法字符串统计个数，并加到总和里。

代码：

#include <bits/stdc++.h>
using namespace std;

int N, sz = 0;
string s;
double sum = 0;

int main()
{
    cin >> N;
    for(int i = 0; i < N; i++)
    {
        cin >> s;
        int len = s.length(), cnt = 0, count = 0, flag = 0;
        for(int j = 0; j < len; j++)
        {
            if(s[j] == '-')
            {
                if(len == 1)
                    flag = 1;
                continue;
            }   
            if(s[j] == '.')
                cnt++;
            else if(!isdigit(s[j]))
                flag = 1;
            if(cnt == 1)
                count++;
        }
        if(flag || cnt >= 2 || count > 3)
            cout << "ERROR: " << s << " is not a legal number" << endl;
        else if(stod(s) < -1000 || stod(s) > 1000)
            cout << "ERROR: " << s << " is not a legal number" << endl;
        else
        {
            sz++;
            sum += stod(s);
        }   
    }
    if(sz > 1)
        cout << fixed << setprecision(2) << "The average of " << sz << " numbers is " << sum / sz;
    else if(sz == 1)
        cout << fixed << setprecision(2) << "The average of " << 1 << " number is " << sum / sz;
    else
        cout << "The average of 0 numbers is Undefined";

    return 0;
}