Question

我需要编写一个接受以下条件的程序：

它必须是一个奇数，
有一个奇数位，
所有数字都必须是奇数和
数字必须介于101和1000001之间。

我目前正试图检查它是否有奇数位数。任何帮助，将不胜感激。更新：感谢所有人的帮助。这真的很有帮助！

pred_score

Answer 1

这个怎么样？

此伪代码不是解决方案或尝试解决方案。

//this should probably be outside and before your while loop.
int length = Integer.toString(userInput).length();
if (length % 2 == 0) {
  return false; //even number of digits
}

...

//this should be inside your while loop
while(userInput != 0) {
  remainder = userInput % 10;
  userInput = userInput / 10;
  if (remainder % 2 != 0) {
    return false; //One of the digit is even
  }
  //if every digit is odd, it has to be an odd number, so checking if origina userInput is an odd number is not necessary. 
}

以此伪代码为出发点。

Answer 2

尝试将userInput的值传递给字符串，并检查其长度是否为奇数。

String total = "";

do{
        System.out.println("please enter a positive whole number between"
                + "" + MIN + " and " + MAX + ". Enter " + EXIT + " "
                + "when you are done entering numbers.");
        userInput = stdin.nextInt();
        total += String.valueOf(userInput);
    }   
    while(userInput != EXIT && userInput >= MIN && userInput <= MAX);
    System.out.println(total);
    if(total.length() % 2 == 1){  
          System.out.println("This is an odd number");
        } 
        else{
            System.out.println("This is not an odd number");
        }

Answer 3

这段代码将检查每个数字是否为奇数，以及长度是否也为奇数。

    int test = 6789;
    int a = test;
    int length =0;
    boolean hasEachDigitOdd = true;
    while (a!= 0) {
        int remaider = a % 10;
        if (remaider % 2 == 0) {
            hasEachDigitOdd = false;
        }
        a = a / 10;
        length++;
    }
    System.out.println(length);
    System.out.println(hasEachDigitOdd);

Answer 4

if (num >= 101 && num <= 1000001 && num % 2 == 0)

％符号为您提供除法运算的剩余部分。

Answer 5

请在下面查看 - 满足您所有需求的解决方案。甚至你输入的方式都需要修复。

public static void main(String[] args) {
    Scanner stdin = new Scanner(System.in);
    int userInput;
    final int EXIT = -1;
    final int MIN = 101;
    final int MAX = 1000001;
    do {
        System.out.println("please enter a positive whole number between"
                + "" + MIN + " and " + MAX + ". Enter " + EXIT + " "
                + "when you are done entering numbers.");
        userInput = stdin.nextInt();

        if(userInput==EXIT){
            System.out.println("done");
            break;
        }

        if (userInput % 2 == 1) {
            System.out.println("This is an odd number");
        } else {
            System.out.println("This is not an odd number");
        }

        if(String.valueOf(userInput).length()%2==1){
            System.out.println("Has odd number of digits");
        } else {
            System.out.println("Has even number of digits");
        }
        boolean[] allOdds = {true};

        String.valueOf(userInput).chars().forEach(i -> {
            if(Integer.parseInt(String.valueOf((char)i))%2==0){
                allOdds[0] = false;
            }
        });
        if(allOdds[0]){
            System.out.println("all digits are odds");
        }else{
            System.out.println("all digits are not odds");
        }

    } while (userInput >= MIN && userInput <= MAX);
}

如何检查Java中的奇数位数？

5 个答案: