模方程解数的Java程序

给定 A 和 B，任务是找到 X 可以取的可能值的数量，使得给定的模方程(A mod X) = B成立。这里，X 也称为模方程的解。

例子：

Input : A = 26, B = 2
Output : 6
Explanation
X can be equal to any of {3, 4, 6, 8,
12, 24} as A modulus any of these values
equals 2 i. e., (26 mod 3) = (26 mod 4) 
= (26 mod 6) = (26 mod 8) = .... = 2 

Input : 21 5
Output: 2
Explanation
X can be equal to any of {8, 16} as A modulus 
any of these values equals 5 i.e. (21 mod 
8) = (21 mod 16) = 5

如果我们仔细分析方程 A mod X = B，很容易注意到，如果 (A = B)，那么 X 可以取的大于 A 的值是无限多的。在 (A < B) 的情况下，不可能有任何可能的 X 值符合模方程。所以我们唯一需要调查的情况是何时（A> B）。所以现在我们深入关注这个案例。

现在，在这种情况下，我们可以使用一个众所周知的关系，即

Dividend = Divisor * Quotient + Remainder

我们正在寻找给定 A 即股息和 B 即余数的所有可能的 X 即除数。所以，

We can say,
A = X * Quotient + B

Let Quotient be represented as Y
∴ A = X * Y + B
A - B = X * Y

∴ To get integral values of Y, 
we need to take all X such that X divides (A - B)

∴ X is a divisor of (A - B)

因此，问题归结为找到 (A – B) 的除数，而这些除数的数量是 X 可以取的可能值。
但是我们知道 A mod X 会产生 (0 到 X – 1) 的值，我们必须取所有这样的 X 使得 X > B。

因此，我们可以得出结论，(A – B) 大于 B 的除数数是 X 满足 A mod X = B 的所有可能值

// Java Program to print all cube free
// numbers smaller than or equal to n.
import java.lang.*;
  
class GFG {
    /* Returns the number of divisors of (A - B)
       greater than B */
    public static int calculateDivisors(int A, int B)
    {
        int N = (A - B);
        int noOfDivisors = 0;
  
        for (int i = 1; i <= Math.sqrt(N); i++) {
  
            // if N is divisible by i
            if ((N % i) == 0) {
  
                // count only the divisors greater than B
                if (i > B)
                    noOfDivisors++;
  
                // checking if a divisor isnt counted twice
                if ((N / i) != i && (N / i) > B)
                    noOfDivisors++;
            }
        }
        return noOfDivisors;
    }
  
    /* Utility function to calculate number of all
       possible values of X for which the modular
       equation holds true */
    public static int numberOfPossibleWaysUtil(int A, int B)
    {
        /* if A = B there are infinitely many solutions
           to equation  or we say X can take infinitely
           many values > A. We return -1 in this case */
        if (A == B)
            return -1;
  
        /* if A < B, there are no possible values of
           X satisfying the equation */
        if (A < B)
            return 0;
  
        /* the last case is when A > B, here we calculate
           the number of divisors of (A - B), which are
           greater than B */
        int noOfDivisors = 0;
        noOfDivisors = calculateDivisors(A, B);
        return noOfDivisors;
    }
  
    /* Wrapper function for numberOfPossibleWaysUtil() */
    public static void numberOfPossibleWays(int A, int B)
    {
        int noOfSolutions = numberOfPossibleWaysUtil(A, B);
  
        // if infinitely many solutions available
        if (noOfSolutions == -1) {
            System.out.print("For A = " + A + " and B = " + B
                             + ", X can take Infinitely many values"
                             + " greater than " + A + "\n");
        }
  
        else {
            System.out.print("For A = " + A + " and B = " + B
                             + ", X can take " + noOfSolutions
                             + " values\n");
        }
    }
    // Driver program
    public static void main(String[] args)
    {
        int A = 26, B = 2;
        numberOfPossibleWays(A, B);
        A = 21;
        B = 5;
        numberOfPossibleWays(A, B);
    }
}
// Contributed by _omg

输出：

For A = 26 and B = 2, X can take 6 values
For A = 21 and B = 5, X can take 2 values

上述方法的时间复杂度只不过是求 (A – B) 的除数个数的时间复杂度，即 O(√(A – B))

有关更多详细信息，请参阅有关模方程解数的完整文章！