pat 甲级 1108. Finding Average

这个必须写， 想了我好多天，哼
1108. Finding Average (20)
时间限制
400 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
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 [-1000, 1000] 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 (<=100). 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
1108. Finding Average (20)
时间限制
400 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
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 [-1000, 1000] 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 (<=100). 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
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
感悟
1.这个题是我觉得pat甲级20分里面最简单的，但是也是想了好多天才能够得满分的题目。哼，好气！
2.题目不难懂，就是将符合标准的数字进行求平均值，不符合标准的直接输出错误。
3.看到题目，第一反应，正则表达式嘛！当然，正则表达式是可以做出来的，但是必须必须一定一定要注意一类数，像（3.）这样的数字也是double型的！！！！。我就是没有注意只一点才导致我，只能能17分。气， 但是归根到底是自己java基本功不够扎实，double类型中，明确说了3.这样的也是对的，还是要加强基本功的练习啊！正则表达式如下：“-{0,1}[0-9]+.[0-9]{1,2}” || “-{0,1}[0-9]+” ||s.matches”-{0,1}[0-9]+.”
4.除了正则表达式，还有另外一种简单的，不需要考虑像3.这样的情况，那就是try-catch。具体就是将不符合题目要求的输出。不符合的有三种情况。1：是double，但是不是3位。2：是double，是2位，但是不再【-1000,1000】 区间。3：不是double。请注意：怎么判断是不是几位呢？很简单那，只需要判断小数点的位置嘛。（1）没有小数点，用indexof结果就是-1.（2）有小数点，用字符串长度-小数点的下标>3.这个判断超级机智呢！
5.鉴于3和4，我建议用4。
6.正则表达式代码如下

import java.math.BigDecimal;import java.util.Scanner;public class Main {    public static void main(String[] args) {        Scanner sc = new Scanner(System.in);        int n;        n = sc.nextInt();        double sum = 0;        int k = 0;        String s = "";        for (int i = 0; i < n; i++) {            s = sc.next();            if(s.matches("-{0,1}[0-9]+.[0-9]{1,2}") || s.matches("-{0,1}[0-9]+") ||s.matches("-{0,1}[0-9]+.")){                double d = Double.valueOf(s);                //System.out.println(d);                if(d>=-1000 && d<=1000){                sum += Double.valueOf(s);                k++;                }                else{                    System.out.println("ERROR: "+s+" is not a legal number");                }            }            else{                System.out.println("ERROR: "+s+" is not a legal number");            }        }        if(k==0){            System.out.println("The average of 0 numbers is Undefined");        }        else if(k>1){            BigDecimal b = new BigDecimal(sum/(k*1.0));            b = b.setScale(2, BigDecimal.ROUND_HALF_UP);                System.out.println("The average of "+k+" numbers is "+b);        }        else{            BigDecimal b = new BigDecimal(sum/(k*1.0));            b = b.setScale(2, BigDecimal.ROUND_HALF_UP);            System.out.println("The average of 1 number is "+b);        }        sc.close();    }}

7.try-catch代码如下：

import java.math.BigDecimal;import java.util.Scanner;public class Main {    public static void main(String[] args) {        Scanner sc = new Scanner(System.in);        int n = sc.nextInt();        double sum = 0;        int k = 0;        String s = "";        for (int i = 0; i < n; i++) {            try {                s = sc.next();                double num = Double.parseDouble(s);                int pos = s.indexOf(".");                if ((num >= -1000 && num <= 1000) && pos == -1) {                    sum += num;                    k++;                } else if (s.length() - pos > 3 || num < -1000 || num > 1000) {                    System.out.println("ERROR: " + s + " is not a legal number");                } else {                    sum += num;                    k++;                }            } catch (Exception e) {                System.out.println("ERROR: " + s + " is not a legal number");            }        }        if(k==0){            System.out.println("The average of 0 numbers is Undefined");        }        else if(k==1){            BigDecimal b = new BigDecimal(sum/(k*1.0));            b = b.setScale(2, BigDecimal.ROUND_HALF_UP);            System.out.println("The average of 1 number is "+b);            }        else {            BigDecimal b = new BigDecimal(sum/(k*1.0));            b = b.setScale(2, BigDecimal.ROUND_HALF_UP);            System.out.println("The average of "+k+" numbers is "+b);        }        sc.close();    }}