Java 后面判定是否为0结尾的两位数

发布网友发布时间：2022-05-05 15:59

共1个回答

热心网友时间：2022-06-27 17:45

你前面的代码逻辑没动（变量名sum改为了sum1）

在你的代码后来添加了你要的判断逻辑，最终代码如下

import java.util.Scanner;

public class A2 {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int num = scanner.nextInt();
        String[] digitals = String.valueOf(num).split("");
        int sum1 = 0;
        int sum2 = 0;
        for (int i = 1; i <= digitals.length; i++) {
            if (i % 2 != 0) {
                sum1 += Integer.parseInt(digitals[i - 1]);
            } else {
                int n = Integer.parseInt(digitals[i - 1]) * 2;
                sum2 += n / 10 + n % 10;
            }
        }
        System.out.println("sum of odd numbers " + sum1);
        System.out.println("sum of even numbers " + sum2);

        // 判断是否合理
        int sum = sum1 + sum2;
        if (sum < 10) {
            // 一位数，不合理
            System.out.println("sum1 + sum2 = " + sum + "，需要加上" + (10 - sum));
        } else if (sum > 99) {
            // 三位数，不合理
            System.out.println("sum1 + sum2 = " + sum + "，需要减去" + (sum - 90));
        } else if (sum % 10 != 0) {
            // 尾数不是0，不合理
            int remainder = sum % 10;
            if (sum > 90) {
                System.out.println("sum1 + sum2 = " + sum + "，需要减去" + remainder);
            } else {
                System.out.println("sum1 + sum2 = " + sum + "，需要减去" + remainder + "或者加上" + (10 - remainder));
            }

            System.out.println();
        } else {
            System.out.println("sum1 + sum2 = " + sum + "，合理");
        }
    }
}