中国剩余定理(一次同余问题) Java

/**
 * 中国剩余定理
 */
public class ChineseRemainderTheorem {
    
    // 扩展欧几里得算法，返回gcd(a, b)以及x, y使得ax + by = gcd(a, b)
    public static int extendedGCD(int a, int b, int[] xy) {
        if (b == 0) {
            xy[0] = 1;
            xy[1] = 0;
            return a;
        }
        int gcd = extendedGCD(b, a % b, xy);
        int temp = xy[0];
        xy[0] = xy[1];
        xy[1] = temp - (a / b) * xy[1];
        return gcd;
    }

    // 使用中国剩余定理求解同余方程组
    public static int chineseRemainder(int[] n, int[] a) {
        int prod = 1; // 模数乘积
        for (int i = 0; i < n.length; i++) {
            prod *= n[i];
        }

        int result = 0;
        for (int i = 0; i < n.length; i++) {
            int p = prod / n[i];
            int[] xy = new int[2];
            extendedGCD(p, n[i], xy); // 求解p * x ≡ 1 (mod n[i])
            int inverse = xy[0];
            result += a[i] * inverse * p;
        }

        return (result % prod + prod) % prod; // 保证结果为正数
    }

    public static void main(String[] args) {
        // 定义模数数组和余数数组
        int[] n = {3, 5, 7};
        int[] a = {2, 3, 2};

        // 使用中国剩余定理求解
        int result = chineseRemainder(n, a);
        System.out.println("结果是: " + result);
    }
}