Question

假设值为：

int p=9; // (any odd no)
int q=5; // (any odd no)
int e=3;
int n=p*q;

我通过以下代码读取文本文件并将内容转换为字节数组：

        while ((eof_line = br.readLine()) != null) {
        byte_array = eof_line.getBytes();
        String decoded = new String(byte_array, "UTF-8");
        // pow=byte_array.to;
        for (i = 0; i < byte_array.length; i++) {
            double input = byte_array[i];
            double t1 = Math.pow(e, -1);
            double t2 = n % 2;
            double encrypt = (Math.pow(input, e)) * t2;//formula
            double p_k = t1 * t2;
            double decrypt = (Math.pow(en, p_k)) * t2;//formula
            }}

这里我的输入是byte []，解密的值是double.So我想将decrypt（double）转换为原始文本文件内容。怎么办？

Answer 1

终于......我找到了解决方案..

     public String[] OpenFile() throws IOException {
    sp1.setVisible(true);
    String k = t2.getText();
    e = Integer.valueOf(k);
    d = (Math.pow(e, -1)) * (n % 2);
    System.out.println("d=" + d);
    FileReader fr = new FileReader(path);
    BufferedReader br = new BufferedReader(fr);
    String eof_line;
    int numberOfLines = 0;
    while ((eof_line = br.readLine()) != null) {
        arr = eof_line.toCharArray();
        byte_array = eof_line.getBytes();
        for (i = 0; i < byte_array.length; i++) {
            ta1.append(Integer.toString(byte_array[i]));
            double input = byte_array[i];
            double t1 = Math.pow(e, -1);
            double t2 = n % 2;
            double en = (Math.pow(input, e)) * t2;
            p_key = t1 * t2;
            double de = (Math.pow(en, p_key)) * t2;
            b = (char) java.lang.Math.round(de);
            ta2.append("" + b);
            }
            br.close();
            return null;
          }

如何进行加密和解密？

1 个答案: