本题的基本要求非常简单:给定 N 个实数,计算它们的平均值。但复杂的是有些输入数据可能是非法的。一个“合法”的输入是 [−1000,1000] 区间内的实数,并且最多精确到小数点后 2 位。当你计算平均值的时候,不能把那些非法的数据算在内。
输入格式:
输入第一行给出正整数 N(≤100)。随后一行给出 N 个实数,数字间以一个空格分隔。
输出格式:
对每个非法输入,在一行中输出 ERROR: X is not a legal number
,其中 X
是输入。最后在一行中输出结果:The average of K numbers is Y
,其中 K
是合法输入的个数,Y
是它们的平均值,精确到小数点后 2 位。如果平均值无法计算,则用 Undefined
替换 Y
。如果 K
为 1,则输出 The average of 1 number is Y
。
输入样例 1:
7
5 -3.2 aaa 9999 2.3.4 7.123 2.35
输出样例 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
输入样例 2:
2
aaa -9999
输出样例 2:
ERROR: aaa is not a legal number
ERROR: -9999 is not a legal number
The average of 0 num
代码实现:
import javax.xml.transform.Source;
import java.io.*;
/**
* @author yx
* @date 2022-07-22 9:18
*/
public class Main {
static PrintWriter out=new PrintWriter(System.out);
static BufferedReader ins=new BufferedReader(new InputStreamReader(System.in));
static StreamTokenizer in=new StreamTokenizer(ins);
public static void main(String[] args) throws IOException {
in.nextToken();
int n=(int) in.nval;
double sum=0;
int sum_i=0;
String[] split=ins.readLine().split(" ");
for (int i = 0; i < n; i++) {
if((split[i].charAt(0)>='0'&&split[i].charAt(0)<='9')){
int flag=0;int dd=0;
for (int j = 0; j < split[i].length(); j++) {
if(split[i].charAt(j)=='.') {
flag++;
dd=j+1;
}
}
if (flag==0){
double m=Double.parseDouble(split[i]);
if(m<=1000&&m>=-1000){
sum+=m;
sum_i++;
}else {
System.out.println("ERROR: "+split[i]+" is not a legal number");
}
}else if (flag==1){
if(split[i].length()-dd>2){
System.out.println("ERROR: "+split[i]+" is not a legal number");
}else {
double m=Double.parseDouble(split[i]);
if(m<=1000&&m>=-1000){
sum+=m;
sum_i++;
}else {
System.out.println("ERROR: "+split[i]+" is not a legal number");
}
}
}
else{
System.out.println("ERROR: "+split[i]+" is not a legal number");
}
} else if(split[i].charAt(0)=='-'&&(split[i].charAt(1)<='9'&&split[i].charAt(1)>='0')){
int flag=0;int dd=0;
for (int j = 0; j < split[i].length(); j++) {
if(split[i].charAt(j)=='.') {
flag++;
dd=j+1;
}
}
if (flag==0){
double m=Double.parseDouble(split[i]);
if(m<=1000&&m>=-1000){
sum+=m;
sum_i++;
}else {
System.out.println("ERROR: "+split[i]+" is not a legal number");
}
}else if (flag==1){
if(split[i].length()-dd>2){
System.out.println("ERROR: "+split[i]+" is not a legal number");
}else {
double m=Double.parseDouble(split[i]);
if(m<=1000&&m>=-1000){
sum+=m;
sum_i++;
}else {
System.out.println("ERROR: "+split[i]+" is not a legal number");
}
}
}
else{
System.out.println("ERROR: "+split[i]+" is not a legal number");
}
} else{
System.out.println("ERROR: "+split[i]+" is not a legal number");
}
}
if(sum_i==0){
System.out.println("The average of 0 numbers is Undefined");
}else if(sum_i==1) {
System.out.printf("The average of "+sum_i+" number is %.2f",sum);
}else {
System.out.printf("The average of "+sum_i+" numbers is %.2f",sum/sum_i);
}
}
}