检查两个给定的有理数是否相等

给定两个表示非负有理数的字符串S和T ，任务是检查S和T的值是否相等。如果发现是真的，则打印“是” 。否则，打印“否” 。

注意：任何有理数都可以用以下三种方式之一表示：

• (例如0、12、123 )
• <。> (例如0.5、1、2.12、2.00001)
• <。> <(> <)> (例如0.1(6)，0.9(9)，0.00(1212))

例子：

Input: S = “0.(52)”, T = “0.5(25)”
Output: YES
Explanation:
The rational number “0.(52)” can be represented as 0.52525252…
The rational number “0.5(25)” can be represented as 0.525252525….
Therefore, the required output is “YES”.

Input: S = “0.9(9)”, T = “1.”
Output: YES
Explanation:
The rational number “0.9(9)” can be represented as 0.999999999…, it is equal to 1.
The rational number “1.” can be represented as the number 1.
Therefore, the required output is “YES”.

方法：想法是将有理数转换为分数，然后检查两个有理数的分数是否相等。如果发现是真的，则打印“是” 。否则，打印“否” 。以下是观察结果：

The repeating portion of a decimal expansion is conventionally denoted within a pair of round brackets.
For example: 1 / 6 = 0.16666666… = 0.1(6) = 0.1666(6) = 0.166(66)
Both 0.1(6) or 0.1666(6) or 0.166(66) are correct representations of 1 / 6.

根据以下观察，可以将任何有理数转换为分数：

Let x = 0.5(25) —> (1)
Integer part = 0, Non-repeating part = 5, Repeating part = 25
Multiply both sides of equation (1) by 10 raised to the power of length of non-repeating part, i.e. 10 * x = 5.(25) —> (2)
Multiply both sides of equation (1) by 10 raised to the power of (length of non-repeating part + length of repeating part), 1000 * x = 525.(25) —> (3)
Subtract equation (2) from equation (3)
1000 * x – 10 * x = 525.(25)-5.(25)
990 * x = 520
⇒ x = 520 / 990

请按照以下步骤解决问题：

• 使用以上观察结果，将两个有理数均转换为分数。
• 检查数字的两个分数是否相等。如果发现是真的，则打印“是” 。
• 否则，打印“否” 。

下面是上述方法的实现：

Java

// Java program to implement
// the above approach
import java.io.*;
import java.util.*;
  
class GFG {
  
    // Function to check if the string S and T
    // are equal or not
    public static boolean isRationalEqual(String s,
                                          String t)
    {
  
        // Stores the fractional part of s
        Fraction f1 = Rational.parse(s).toFraction();
  
        // Stores the fractional part of t
        Fraction f2 = Rational.parse(t).toFraction();
  
        // If the condition satisfies, returns true
        // otherwise return false
        return f1.p * f2.q == f2.p * f1.q;
    }
  
    // Rational class having integer, non-repeating
    // and repeating part of the number
    public static class Rational {
        private final String integer, nonRepeating,
            repeating;
  
        // Constructor function to initialize
        // the object of the class
        private Rational(String integer,
                         String nonRepeating,
                         String repeating)
        {
  
            // Stores integer part
            this.integer = integer;
  
            // Stores non repeating part
            this.nonRepeating = nonRepeating;
  
            // Stores repeating part
            this.repeating = repeating;
        }
  
        // Function to split the string into
        // integer, repeating & non-repeating part
        public static Rational parse(String s)
        {
  
            // Split s into parts
            String[] parts = s.split("[.()]");
  
            return new Rational(
                parts.length >= 1 ? parts[0] : "",
                parts.length >= 2 ? parts[1] : "",
                parts.length >= 3 ? parts[2] : "");
        }
  
        // Function to convert the string
        // into fraction
        public Fraction toFraction()
        {
  
            long a = tenpow(nonRepeating.length());
            long i = Long.parseLong(integer + nonRepeating);
  
            // If there is no repeating part, then
            // form a new fraction of the form i/a
            if (repeating.length() == 0) {
                return new Fraction(i, a);
            }
  
            // Otherwise
            else {
                long b = tenpow(nonRepeating.length()
                                + repeating.length());
  
                long j = Long.parseLong(
                    integer + nonRepeating + repeating);
  
                // Form the new Fraction and return
                return new Fraction(j - i, b - a);
            }
        }
  
        public String toString()
        {
            return String.format("%s.%s(%s)", integer,
                                 nonRepeating, repeating);
        }
    }
  
    // Fraction class having numerator as p
    // and denominator as q
    public static class Fraction {
        private final long p, q;
  
        // Constructor function to initialize
        // the object of the class
        private Fraction(long p, long q)
        {
            this.p = p;
            this.q = q;
        }
  
        public String toString()
        {
            return String.format("%d/%d", p, q);
        }
    }
  
    // Function to find 10 raised
    // to power of x
    public static long tenpow(int x)
    {
        assert x >= 0;
        long r = 1;
        while (--x >= 0) {
            r *= 10;
        }
        return r;
    }
  
    // Driver Code
    public static void main(String[] args)
    {
  
        // Given S and T
        String S = "0.(52)", T = "0.5(25)";
  
        // Function Call
        if (isRationalEqual(S, T)) {
            System.out.println("YES");
        }
        else {
  
            System.out.println("NO");
        }
  
        // Print result
    }
}

输出：

YES

时间复杂度： O(N)，其中N是S和T的最大长度
辅助空间： O(1)