A string consisting of lowercase English letters, (, and ) is said to be a good string if you can make it an empty string by the following procedure: First, remove all lowercase English letters. Then, repeatedly remove consecutive () while possible. For example, ((a)ba) is a good string, because removing all lowercase English letters yields (()), from which we can remove consecutive () at the 2-nd and 3-rd characters to obtain (), which in turn ends up in an empty string. You are given a good string S. We denote by S i ​ the i-th character of S. For each lowercase English letter a, b, …, and z, we have a ball with the letter written on it. Additionally, we have an empty box. For each i=1,2, … ,∣S∣ in this order, Takahashi performs the following operation unless he faints. If S i ​ is a lowercase English letter, put the ball with the letter written on it into the box. If the ball is already in the box, he faints. If S i ​ is (, do nothing. If S i ​ is ), take the maximum integer j less than i such that the j-th through i-th characters of S form a good string. (We can prove that such an integer j always exists.) Take out from the box all the balls that he has put in the j-th through i-th operations. Determine if Takahashi can complete the sequence of operations without fainting.用cpp解决，直接写代码

题意简述：给定一个字符串 S，它由小写英文字母、左括号 ( 和右括号 ) 组成，如果可以通过以下步骤将 S 变为空字符串，则 S 被称为好字符串：

1. 首先，删除所有小写英文字母。
2. 然后，当可能时，连续删除 ()。例如，((a)ba) 是一个好字符串，因为删除所有小写英文字母后得到 (()), 然后可以在第 2 到 3 个字符处连续删除 () 得到 (), 进而得到一个空字符串。

你现在已经得到了一个好字符串 S。我们用 S i 表示 S 的第 i 个字符。Takahashi 要进行如下操作（除非他晕倒）：

- 如果 S i 是小写英文字母，就将带有该字母的球放入盒子中。如果球已经在盒子中，他就晕倒。
- 如果 S i 是左括号 (，就什么也不做。
- 如果 S i 是右括号 )，就找到小于 i 的最大整数 j，使得 S 的第 j 到第 i 个字符组成的子串是好字符串。（我们可以证明这样的整数 j 总是存在）。从盒子中取出从第 j 次操作到第 i 次操作之间放入盒子的所有

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A string consisting of lowercase English letters, (, and ) is said to be a good string if you can make it an empty string by the following procedure: First, remove all lowercase English letters. Then, repeatedly remove consecutive () while possible. For example, ((a)ba) is a good string, because removing all lowercase English letters yields (()), from which we can remove consecutive () at the 2-nd and 3-rd characters to obtain (), which in turn ends up in an empty string. You are given a good string S. We denote by S i ​ the i-th character of S. For each lowercase English letter a, b, …, and z, we have a ball with the letter written on it. Additionally, we have an empty box. For each i=1,2, … ,∣S∣ in this order, Takahashi performs the following operation unless he faints. If S i ​ is a lowercase English letter, put the ball with the letter written on it into the box. If the ball is already in the box, he faints. If S i ​ is (, do nothing. If S i ​ is ), take the maximum integer j less than i such that the j-th through i-th characters of S form a good string. (We can prove that such an integer j always exists.) Take out from the box all the balls that he has put in the j-th through i-th operations. Determine if Takahashi can complete the sequence of operations without fainting.用cpp解决

这道题的意思是，给定一个字符串S，S是一个好的字符串，如果您可以通过以下过程将其转换为空字符串：首先，删除所有小写英文字母。然后，反复删除连续的()，直到不可能为止。例如，((a)ba)是一个好的字符串，因为删除所有小写英文字母后得到(())，从中我们可以在第2和第3个字符处删除连续的()以获得()，而后者最终变为空字符串。您将得到一个好的字符串S。我们用S i 表示S的第i个字符。对于小写英文字母a，b，…和z，我们都有一个写有该字母的小球。另外，我们有一个空盒子。对于按此顺序的i=1,2,…,|S|，高桥除非他昏倒，否则执行以下操作。如果S i 是小写英文字母，将写有该字母的球放入盒子中。如果球已经在盒子中，他就会昏倒。如果S i 是(，则什么也不做。如果S i 是)，则取出最大的整数j，使得S的第j到第i个字符形成一个好的字符串。 （我们可以证明这样的整数j总是存在的。）从第j次到第i次

There is a string a (unknown to you), consisting of lowercase Latin letters, encrypted according to the following rule into string s : after each character of string a , an arbitrary (possibly zero) number of any lowercase Latin letters, different from the character itself, is added; after each such addition, the character that we supplemented is added. You are given string s , and you need to output the initial string a . In other words, you need to decrypt string s .

To decrypt string s, you can iterate through each character of s and keep track of the last character seen. Whenever a new character is encountered, it means that all the characters before it (including the new character) belong to the same character in the original string a. So you can add the last character seen to the decrypted string a, and update the last character seen to the new character.

Here's the Python code to implement this:

def decrypt_string(s):
    a = ''
    last_char = ''
    for c in s:
        if c != last_char:
            a += c
            last_char = c
    return a

For example, if s = 'aabbccdd', the function will return 'abcd'.