java - 迭代整数中的每个数字并做 mod 11

Question

我正在尝试做一些程序并需要一些帮助。例如，我在文本框中输入了这个数字：00669253，然后对于每个数字，我想将它与数字 87654321 相乘，例如

(0x8) + (0x7) + (6x6).... 等等，然后得到所有数字的总和并做 11 的 mod。我该怎么做？

感谢您的提前帮助。

Answer 1

您可以先将数字转换为String，然后您可以这样做：

String no1="00669253";
String no2="87654321";
int sum=0;
for(int i=0;i<no1.length();i++){
        sum+=Integer.parseInt(""+no1.charAt(i))*Integer.parseInt(""+no2.charAt(i));
}

之后就可以进行任何操作了sum

Answer 2

尝试这个，

public class Dummy {
    public static void main(String[] args) {
        String s = "112211";
        String s2 = "010101";
        char[] sDummy = s.toCharArray();
        char[] s2Dummy = s2.toCharArray();
        int sum = 0;
        for (int i = 0; i < sDummy.length; i++) {
            System.out.println("multiplying..." + sDummy[i] + "with " + s2Dummy[i] + "and adding prev sum " + sum + " to it");
            sum = sum + (Integer.parseInt(sDummy[i] + "") * Integer.parseInt(s2Dummy[i] + ""));
        }
        sum = sum % 11;
        System.out.println("sum with mod is = " + sum);
    }
}

如果两个字符串的长度不同，您需要添加一些检查

Answer 3

你可以尝试这样的事情：

public class TestMult {

    public static void main(String[] args) {
        String first = "00669253";
        String second = "87654321";

        long sum = 0L;
        if (first.length() == second.length()) {
            for (int i = 0; i < first.length(); i++) {
                int firstInt = Character.getNumericValue(first.charAt(i));
                int secondInt = Character.getNumericValue(second.charAt(i));
                sum += firstInt * secondInt;
            }
            System.out.println("result: " + sum % 11);
        } else {
            System.err.println("lengths are not equal.");
        }

    }

}

Answer 4

尝试这个，

    String inputValue = "00669253";
    String multipleValue="87654321";

    int result = 0;
    for (int i = 0; i < inputValue.length(); i++)
            {
                result += Integer.parseInt(Character.toString(inputValue.charAt(i))) * 
                Integer.parseInt(Character.toString(multipleValue.charAt(i)));
            }