试试反解析String code = "your code string"; int sum = 0; for (int i = 0; i < code.length(); i++) { sum += code.charAt(i); } code = String.valueOf(sum) + code;
时间: 2023-06-15 14:04:10 浏览: 46
这段代码的作用是将一个字符串进行加密,具体步骤如下:
1. 遍历字符串中的每一个字符,将其对应的 ASCII 码值相加,得到一个整数 sum。
2. 将 sum 转换为字符串类型,并将其拼接到原字符串的前面,得到一个新的加密后的字符串。
举个例子,如果原字符串为 "hello",则 sum 的计算过程为:
sum = 'h' + 'e' + 'l' + 'l' + 'o' = 104 + 101 + 108 + 108 + 111 = 532
最终加密后的字符串为 "532hello"。
相关问题
试试把这段代码转为Android,以及他的解密 UInt16 sum = 0; for (int i = 0; i < code.Length; i++) { sum += code[i]; } code = sum.ToString() + code;
以下是将该代码转换为 Android 的示例代码:
```java
String code = "your_code_here";
int sum = 0;
for (int i = 0; i < code.length(); i++) {
sum += code.charAt(i);
}
code = String.valueOf(sum) + code;
```
该代码的作用是将字符串 `code` 中每个字符的 Unicode 码点值相加,然后在字符串前面添加这个和值。其中,`code` 是一个字符串变量。解密该代码需要先截取字符串前面的和值,然后对剩余部分进行反向操作。具体实现代码如下:
```java
String code = "your_code_here";
int sum = Integer.parseInt(code.substring(0, 5));
code = code.substring(5);
StringBuilder builder = new StringBuilder();
for (int i = 0; i < code.length(); i++) {
builder.append((char)(code.charAt(i) - sum));
}
String result = builder.toString();
```
该代码的作用是先从字符串 `code` 中截取前面的和值,然后对剩余部分的每个字符减去这个和值得到原始字符的 Unicode 码点值。最后将这些 Unicode 码点值转换为字符,拼接在一起得到解密后的字符串 `result`。
print(len(strs), sum([int(i) for i in strs]))
As an AI language model, I cannot provide an accurate answer without knowing the value of the variable `strs`.
However, assuming `strs` is a list of strings that contains only integers, the code above will output a tuple with two values:
- `len(strs)` will give the length of the list, i.e., the number of strings in `strs`.
- `sum([int(i) for i in strs])` will convert each string in `strs` to an integer using `int(i)` and then add all the integers together using `sum()`. The output will be the sum of all integers in the list.
For example, if `strs = ['1', '2', '3', '4']`, the output will be `(4, 10)`, meaning there are 4 elements in `strs` and their sum is 10.