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java - 用户输入仅发生一次或在循环期间永不停止

转载 作者:行者123 更新时间:2023-12-02 07:17:19 24 4
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我正在尝试一个程序,它读取未指定数量的整数,查找总和、正数、负数和平均值。我的问题是,要么它只运行并允许输入一个整数,然后什么都不做,要么使用下面的代码,它永远不会停止让你输入数字,因此我无法通过。我的 number = 0 输出正确。

public class Compute {

// Count positive and negative numbers and compute the average of numbers
public static void main(String[] args) {
Scanner input = new Scanner(System.in);

int sum = 0;
positive = 0;
negative = 0;
total = 0;

System.out.println("Enter an integer, the input ends if it is 0: ");
int numbers = input.nextInt();

do {
if (numbers > 0) {
positive++;//add 1 to positive count
} else if (numbers < 0) {
negative++;//add 1 to negative count
}//end else if

sum += numbers; //add integer input to sum

numbers = input.nextInt();
total++;
} while (numbers != 0);

if (numbers == 0) {
System.out.println("No numbers are entered except " + numbers);
}//end if
}
}

最佳答案

尝试下面的代码。终止循环并在执行的任何时候将输出类型 0 视为输入。

import java.util.Scanner;

public class Compute {

// Count positive and negative numbers and compute the average of numbers
public static void main(String[] args) {
Scanner input = new Scanner(System.in);

int sum = 0;
int positive = 0;
int negative = 0;
int total = 0;

System.out.println("Enter an integer, the input ends if it is 0: ");
int numbers = input.nextInt();

do {

if (numbers > 0) {
positive++;// add 1 to positive count
sum += numbers; // add integer input to sum
}

else if (numbers < 0) {
negative++;// add 1 to negative count
sum -= numbers; // add integer input to sum
}

numbers = input.nextInt();
total++;

} while (numbers != 0);

System.out.println("The number of positives is \t " + positive);
System.out.println("The number of negatives is \t " + negative);
System.out.println("The total count of number is \t " + total);
System.out.println("The sum of all number is \t" + sum);
System.out.println("The average is \t"
+ ((double) sum / (positive + negative)));

}// end main
}// end Compute

关于java - 用户输入仅发生一次或在循环期间永不停止,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/14765363/

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