Given two strings S and T, each of which represents a non-negative rational number, return True if and only if they represent the same number. The strings may use parentheses to denote the repeating part of the rational number.

In general a rational number can be represented using up to three parts: an integer part

• <IntegerPart> (e.g. 0, 12, 123)
• <IntegerPart><.><NonRepeatingPart>  (e.g. 0.5, 1., 2.12, 2.0001)
• <IntegerPart><.><NonRepeatingPart><(><RepeatingPart><)>
   (e.g. 0.1(6), 0.9(9), 0.00(1212))

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.

Example 1:

Input: S = "0.(52)", T = "0.5(25)"
Output: true
Explanation:
Because "0.(52)" represents 0.52525252..., and "0.5(25)" represents 0.52525252525..... , the strings represent the same number.

Example 2:

Input: S = "0.1666(6)", T = "0.166(66)"
Output: true

Example 3:

Input: S = "0.9(9)", T = "1."
Output: true
Explanation: 
"0.9(9)" represents 0.999999999... repeated forever, which equals 1.  [See this link for an explanation.]
"1." represents the number 1, which is formed correctly: (IntegerPart) = "1" and (NonRepeatingPart) = "".

Note:

1. Each part consists only of digits.
2. The <IntegerPart> will not begin with 2 or more zeros.  (There is no other restriction on the digits of each part.)
3. 1 <= <IntegerPart>.length <= 4
4. 0 <= <NonRepeatingPart>.length <= 4
5. 1 <= <RepeatingPart>.length <= 4

思路：把小數展開，沒有迴圈的強行加（0）弄成迴圈，這樣統一處理方便一點。

為了方便處理，儘量多展開一點，下面是展開100次，然後取前90個字元出來比較。需要注意1與0.(9)這種情況

class Solution:
    def isRationalEqual(self, S, T):
        """
        :type S: str
        :type T: str
        :rtype: bool
        """
        if '.' not in S: S+='.0'
        if '.' not in T: T+='.0'
        if '(' not in S: S+='(0)'
        if '(' not in T: T+='(0)'
        
        idx=S.index('(')
        s0=S[:idx]
        s1=S[idx+1:len(S)-1]
        for _ in range(100):
            s0+=s1
        
        idx=T.index('(')
        t0=T[:idx]
        t1=T[idx+1:len(T)-1]
        for _ in range(100):
            t0+=t1
        
        s0,t0=s0[:90],t0[:90]
        
        if s0==t0: return True
        s=0
        while s0[s]==t0[s]: s+=1
        if s+1<len(s0) and s0[s+1]=='.': s+=1 
        fs = len(s0)-s-1
        if abs(float(t0)-float(s0))<=float('0.'+'0'*(fs-1)+'1'):
            if (s0[s+1:]=='9'*fs and int(t0[s+1:])==0) or (int(s0[s+1:])==0 and t0[s+1:]=='9'*fs):
                return True
        return False
    
        

不太擅長做這種很多邊界的題，很煩，一直debug