英文:
Trying to calculate a score using a math formula in Java
问题
我正在尝试在Java中使用一个公式计算分数。我在将其翻译为Java代码方面遇到了问题。
公式：𝑠𝑐𝑜𝑟𝑒 = 3 ^ ( ( log2( points/3) ) + 1)
其中 points 为输入值。
我目前有以下代码片段，但是我得不到期望的结果：
int score = 3 ^ (int)(Math.pow(Math.log(12 / 3), 2) + 1);
有人能告诉我我做错了什么，并可能给我一个正确的片段吗？以输入值 12 为例，结果应为 27。非常感谢。
编辑：
输入值 3 = 3
输入值 6 = 9
输入值 12 = 27
输入值 24 = 81
输入值 48 = 243
输入值 96 = 729
输入值 192 = 2187
等等。
英文:
I am trying to calculate a score in Java using a formula. I am having trouble with translating it to Java code.
Formula: 𝑠𝑐𝑜𝑟𝑒 = 3 ^ ( ( log2( points/3) ) + 1)
points = the input.
I currently have this piece of code, but I am not getting the result I would like:
int score = 3 ^ (int)(Math.pow(Math.log(12 / 3), 2) + 1);
Could anyone tell me what I am doing wrong and possibly give me a snippet of how it should be? The result should be 27 with 12 as input. Thank you very much in advance.
Edit:
Input 3 = 3
Input 6 = 9
Input 12 = 27
Input 24 = 81
Input 48 = 243
Input 96 = 729
Input 192 = 2187
etc.
答案1
得分: 2
公式为：
int score = (int) Math.pow(3, Math.log(points/3.0) / Math.log(2) + 1);
- 在Java中，^ 是按位异或操作符，您需要使用
Math.pow来进行幂运算。 - log2(x) 表示以 2 为底的对数，没有内置的函数，但您可以使用 log(x)/log(2) 来计算它。
如果输出稍有偏差，您可能希望使用 Math.round 而不是强制转换为 (int)。
英文:
The formula is:
int score = (int) Math.pow(3, Math.log(points/3.0) / Math.log(2) + 1);
- ^ in Java is the bit-wise XOR operator, you need to use
Math.powinstead for power - log2(x) refers to "base 2" logarithm, there's no builtin function for it but you can you can compute it with log(x)/log(2)
You may want to use Math.round instead of cast to (int) in case the output is slightly off.
答案2
得分: 1
这段代码中存在多个错误，主要的错误是您在幂运算中使用了^运算符，而^运算符是用于“按位异或”的。
另一个错误是您将Math.log()用作以e为底的对数，尽管我们需要以2为底的对数。
现在让我们重新编写这段代码：
首先，我们要获取以2为底的n的对数，为此，我们使用以下公式：
> logb(n) = loge(n) / loge(b)
因此，要获取12/3的以2为底的对数，我们可以写成：
Math.log(12/3)/Math.log(2)
现在，获取幂次的其余操作很容易：
(int) (Math.log(12/3) / Math.log(2) + 1)
最后一部分是正确使用Math.pow()函数，如果我们要计算3^n，我们应该输入Math.pow(3, n)。
因此，最终的代码将是：Math.pow(3, (int) (Math.log(12/3) / Math.log(2) + 1) )
英文:
There are multiple mistakes in this code, the main one though is that you are using the ^ operator for power, the ^ operator for "bitwise exclusive OR",
Another mistake is that you used Math.log() as if it's in base 2, even though it's in base e.
Now let's rewrite the code again:
First we want to get log in base 2 of n, to do so, we use the formula:
> logb(n) = loge(n) / loge(b)
So to get log in base 2 of 12/3 we'd write:
Math.log(12/3)/Math.log(2)
Now the rest of the operation to get the number in the power is easy:
(int) (Math.log(12/3) / Math.log(2) + 1)
The last bit is using the Math.pow() function properly, if we want 3^n we'd need to type in Math.pow(3, n).
So the final code would be: Math.pow(3, (int) (Math.log(12/3) / Math.log(2) + 1) )
答案3
得分: 0
我已经尝试为您逐步拆分。请查看下面的代码片段：
public class Score {
public static void main(String[] args) {
int points = 12;
int dividedPoints = points / 3;
int log2Value = log2(dividedPoints);
int k = log2Value + 1;
int score = (int) Math.pow(3, k);
System.out.println("final score is = " + score);
}
public static int log2(int N) {
// using log() method
int result = (int) (Math.log(N) / Math.log(2));
System.out.println("log2 Value is = " + result);
return result;
}
}
它将产生以下输出：
log2 Value is = 2
final score is = 27
英文:
I have tried to split it step by step for you. Please check the snippet below.
public class Score {
public static void main(String[] args) {
int points = 12;
int dividedPoints = points / 3;
int log2Value = log2(dividedPoints);
int k = log2Value + 1;
int score = (int) Math.pow(3, k);
System.out.println("final score is = " + score);
}
public static int log2(int N) {
// using log() method
int result = (int) (Math.log(N) / Math.log(2));
System.out.println("log2 Value is = " + result);
return result;
}
}
It would produce the following output
log2 Value is = 2
final score is = 27
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