余数定理

by Anuj Pahade

由Anuj Pahade

如何用Java实现余数定理 (How to implement the Chinese Remainder Theorem in Java)

This post assumes that you know what Chinese Remainder Theorem (CRT) is and focuses on its implementation in Java. If you don’t, I’d recommend you read about it here.

这篇文章假定您知道什么是中文余数定理(CRT)，并将重点介绍其在Java中的实现。如果您不喜欢的话，建议您在这里阅读。

You can find the link to the complete code at the end of this post. So let’s get started.

您可以在本文末尾找到完整代码的链接。因此，让我们开始吧。

我们需要找到什么？ (What do we need to find?)

We need to find X. ?

我们需要找到X。

The statement goes as follows:

该语句如下：

There exists a minimum positive number X such that:

存在一个最小的正数 X，使得：

X % number[0]    =  remainder[0], X % number[1]    =  remainder[1], ...............X % number[k-1]  =  remainder[k-1]

So, here we have two arrays.

因此，这里有两个数组。

The array of numbers: All the numbers in this array are pairwise relatively prime. Which means, pick any two numbers from the array, you’ll find that their greatest common divisor is 1.

数字数组：此数组中的所有数字都是成对的相对质数。这意味着，从数组中选择任意两个数字，您会发现它们的最大公约数是1。
The array of remainders: As you can see in the expressions above, when X is divided by a number n from the number array it leaves a respective remainder from the remainder array.

余数数组：如您在上面的表达式中所看到的，当X用数字数组中的数字n进行除数时，它会从余数数组中留下相应的余数。

实施CRT的步骤 (Steps to implement the CRT)

These are the steps, or as we engineers say, the ‘algorithm’, to implement CRT.

这些是实施CRT的步骤，或者正如我们的工程师所说的“算法”。

步骤1：在第一个数组中找到所有数字的乘积。 (Step 1: Find the product of all the numbers in the first array.)

for(int i=0; i<number.length; i++ ){   product *= number[i];}

步骤2：找出每个数字的偏积。 (Step 2: Find the partial product of each number.)

Partial product of n= product/n

n的部分乘积= product / n

for(int i=0; i<num.length; i++){   partialProduct[i] = product/number[i];}

3.找到number [i]取模取取part [i]的模数乘法逆。 (3. Find the modular multiplicative inverse of number[i] modulo partialProduct[i].)

Here we find the inverse using the extended Euclidean algorithm. So, we call computeInverse(partialProduct[i],num[i])

在这里，我们使用扩展的欧几里得算法求逆。因此，我们将其称为computeInverse(partialProduct [i]，num [i])

public static int computeInverse(int a, int b){         int m = b, t, q;         int x = 0, y = 1;               if (b == 1)             return 0;               // Apply extended Euclid Algorithm         while (a > 1)         {             // q is quotient             q = a / b;             t = b;                   // now proceed same as Euclid's algorithm             b = a % b;a = t;             t = x;             x = y - q * x;             y = t;         }               // Make x1 positive         if (y < 0)          y += m;               return y;     }

步骤4：最终总和 (Step 4: Final Sum)

sum += partialProduct[i] * inverse[i] * rem[i];

步骤5：传回最小的X (Step 5: Return the smallest X)

In order to find the smallest of all solutions, we divide the sum from step 4 by the product from step 2.

为了找到所有解决方案中最小的一个，我们将步骤4的总和除以步骤2的乘积。

return sum % product;

Thus, we have found our X. I’d recommend you to try to implement the code on your own before looking at the code in the link below.

因此，我们找到了X。建议您在查看下面链接中的代码之前，先尝试自行实现代码。

Thanks for reading the post. I hope it helped you. Leave suggestions in the comments below or reach out to me with a better version of this code or queries on anujp5678[at]gmail[dot]com or connect with me on LinkedIn.

感谢您阅读这篇文章。希望对您有所帮助。在下面的评论中留下建议，或通过更好的代码版本或在anujp5678 [at] gmail [dot] com上与我联系，或在LinkedIn上与我联系。

Please clap. ? Claps motivate.

请拍手。？鼓掌激励。

Have fun and happy coding! :)

祝您编程愉快，愉快！ :)

翻译自: https://www.freecodecamp.org/news/how-to-implement-the-chinese-remainder-theorem-in-java-db88a3f1ffe0/