1108 Finding Average (20分)

The basic task is simple: given N real numbers, you are supposed to calculate their average. But what makes it complicated is that some of the input numbers might not be legal. A legal input is a real number in [−] and is accurate up to no more than 2 decimal places. When you calculate the average, those illegal numbers must not be counted in.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (≤). Then N numbers are given in the next line, separated by one space.

Output Specification:

For each illegal input number, print in a line ERROR: X is not a legal number where X is the input. Then finally print in a line the result: The average of K numbers is Y where K is the number of legal inputs and Y is their average, accurate to 2 decimal places. In case the average cannot be calculated, output Undefined instead of Y. In case K is only 1, output The average of 1 number is Y instead.

Sample Input 1:

7
5 -3.2 aaa 9999 2.3.4 7.123 2.35

 

Sample Output 1:

ERROR: aaa is not a legal number
ERROR: 9999 is not a legal number
ERROR: 2.3.4 is not a legal number
ERROR: 7.123 is not a legal number
The average of 3 numbers is 1.38

 

Sample Input 2:

2
aaa -9999

 

Sample Output 2:

ERROR: aaa is not a legal number
ERROR: -9999 is not a legal number
The average of 0 numbers is Undefined

这题考察如下的自动机，就按照如下的自动机执行即可判断，考察编译原理中识别数字。

 

 

 

#include <iostream>
using namespace std;
int N, cnt = 0; string tmp;
double sum = 0.0;
// 自动机解决数字判断
bool isLegal(string& s) {
    int i = 0;
    if(s[i] == '-') {
        i++; // 第一位是-
        if(i >= s.length() || !isdigit(s[i])) return false; // 第二位不是数字
    }
    while(i < s.length() && isdigit(s[i])) i++; // 过滤掉小数点前数字
    if(i == s.length()) return true; // 结束1
    if(s[i] == '.') i++; // 过滤. 
    int cnt = 2;
    while(cnt--) if(i < s.length() && isdigit(s[i])) i++; // 过滤掉小数点后2位数字
    return i == s.length(); // 如果到了最后，则返回true，否则返回false
}
int main() {
    cin >> N;
    while(N--) {
        cin >> tmp;
        if(isLegal(tmp) && stod(tmp) >= -1000 && stod(tmp) <= 1000) {
            cnt++;
            sum += stod(tmp);
        } else printf("ERROR: %s is not a legal number\n", tmp.c_str());
    }
    if(cnt == 0) printf("The average of 0 numbers is Undefined\n");
    else if(cnt == 1) printf("The average of 1 number is %.2lf\n", sum);
    else printf("The average of %d numbers is %.2lf\n", cnt, sum / (double)cnt);
    return 0;
}