接受一系列的数字？ for循环

Question

我是编程新手，所以我不确定如何编写一个程序来接受一系列数字，并且只有在用户输入 0 时才会退出，然后显示在用户之前输入了多少数字输入 0。

我的代码：

import java.util.Scanner;
class Sample {
    public static void main(String[]args) {
        Scanner x = new Scanner(System.in);
        int y;
        int count = 0;

        System.out.print("Enter any number to continue. Enter 0 to stop.");
        num = x.nextInt();
        for(int n = 0; n<=num; n++) { //is this for loop right?
            y = x.nextInt();
            count += y;
        }
        if(num == 0){
            System.out.printf("You entered %d numbers,count");
        }
    }
} // i know my code is missing alot.

输出应该是这样的。

Enter any number to continue. Enter 0 to stop 
12
11
5
6
1
0
You entered 6 numbers.
2 even
3 odd // i know my code is far from what i want my output to be.
EDIT :
if the user enters 2 then 0 the output should be
You Entered an even number // same with odd. how?

Answer 1

一些线索：

你的线路

num = x.nextInt();

可能会导致编译错误，因为您还没有声明变量num。

此外，您的循环应该继续 while 数字大于 0。为此，请使用 while 循环：

int num = 1; //some non-zero starting number
while (num != 0) {
    num = x.nextInt();
    count++;
}

Answer 2

您需要重写您的逻辑，如两种方式处理插入的数字或在插入 0 后处理所有下面是插入过程的示例虽然（真） { 读行检查是否为零
休息 ; 别的检查是否奇数增加奇数别的增加 evens 计数 } 显示你想要的

第二种情况 --

虽然（真） { 读行检查是否为零
休息 ; 别的添加否到列表 }

对于列表中的每个条目 { 检查是否奇数增加奇数别的增加 evens 计数 } 显示你想要的

Answer 3

System.out.println("Enter any number to continue. Enter 0 to stop.");
while (true) {
    int num = x.nextInt() ;

    if (num == 0) {
       break;
    }

    if (num % 2 == 0) {
       even++;
    } else {
       odd++;
    }

    count++;
}

Answer 4

代码是自描述的

public class Sample {
    public static void main(String[] args) {
        Scanner x = new Scanner(System.in);
        int odd = 0, even = 0, num = 0;
        System.out.println("Enter any number to continue. Enter 0 to stop.");
        while ((num = x.nextInt()) != 0) {

            if (num % 2 == 0) {
                even++;
            } else {
                odd++;
            }

        }
        System.out.printf("\nYou entered %d numbers\n", even + odd);
        System.out.printf("even %d\n", even);
        System.out.printf("odd %d\n", odd);

    }
}

Answer 5

Working demo here

你很接近，只需获取每个计数的变量，并在 while 循环中检查它：

public static void main(String[] args) {
    Scanner x = new Scanner(System.in);
    int num = 0; // to store user's enter
    int odd = 0, count = 0, even = 0; // to store all counts

    System.out.print("Enter any number to continue. Enter 0 to stop.");
    // ask for first number
    num = x.nextInt();
    
    // loop until 0 is entered
    while (num != 0) {
        // each time sum 1 because new number entered
        count ++;
        // if num / 2 modulus is 0 then even number 
        if (num % 2 == 0) {
            even ++;
        // if num / 2 modulus is not 0 then odd number 
        } else {
            odd ++;
        }

        // ask for a new number
        num = x.nextInt();
    }

    // print the results
    System.out.printf("You entered %d numbers\n",count);
    System.out.printf("You entered %d even numbers\n",even);
    System.out.printf("You entered %d odd numbers\n",odd);
}

输出

Enter any number to continue. Enter 0 to stop.
12
11
5
6
1
0

You entered 6 numbers
You entered 2 even numbers
You entered 3 odd numbers

接受一系列的数字？ for循环

accept a series of number? for loop

java

loops

for-loop

numbers

Working demo here