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1054 求平均值

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:

tex
7
5 -3.2 aaa 9999 2.3.4 7.123 2.35
7
5 -3.2 aaa 9999 2.3.4 7.123 2.35

输出样例 1:

tex
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
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:

tex
2
aaa -9999
2
aaa -9999

输出样例 2:

tex
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
The average of 0 numbers is Undefined

Solution:

java
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {
	public static void main(String[] args) throws IOException {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(in.readLine());
		String[] a = in.readLine().split(" ");
		double sum = 0;
		double X = 0;
		int K = 0;
		for (String s : a) {
			try {
				X = Double.parseDouble(s);
				String str = String.valueOf(X);
				if (X >= -1000 && X <= 1000 && str.substring(str.indexOf(".")).length() <= 3) {
					sum += X * 100;
					K++;
				} else
					System.out.println("ERROR: " + s + " is not a legal number");
			} catch (NumberFormatException e) {
				System.out.println("ERROR: " + s + " is not a legal number");
			}
		}
		if (K >= 1)
			System.out.println("The average of " + K + (K == 1 ? " number is " : " numbers is ")
					+ String.format("%.2f", sum / K / 100));
		else
			System.out.println("The average of " + K + " numbers is Undefined");
	}
}
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {
	public static void main(String[] args) throws IOException {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(in.readLine());
		String[] a = in.readLine().split(" ");
		double sum = 0;
		double X = 0;
		int K = 0;
		for (String s : a) {
			try {
				X = Double.parseDouble(s);
				String str = String.valueOf(X);
				if (X >= -1000 && X <= 1000 && str.substring(str.indexOf(".")).length() <= 3) {
					sum += X * 100;
					K++;
				} else
					System.out.println("ERROR: " + s + " is not a legal number");
			} catch (NumberFormatException e) {
				System.out.println("ERROR: " + s + " is not a legal number");
			}
		}
		if (K >= 1)
			System.out.println("The average of " + K + (K == 1 ? " number is " : " numbers is ")
					+ String.format("%.2f", sum / K / 100));
		else
			System.out.println("The average of " + K + " numbers is Undefined");
	}
}

Released under the MIT License.