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Replace To Make Regular Bracket Sequence

Replace To Make Regular Bracket Sequence

 

You are given string s consists of opening and closing brackets of four kinds <>, {}, [], (). There are two types of brackets: opening and closing. You can replace any bracket by another of the same type. For example, you can replace < by the bracket {, but you can't replace it by ) or >.

The following definition of a regular bracket sequence is well-known, so you can be familiar with it.

Let's define a regular bracket sequence (RBS). Empty string is RBS. Let s1 and s2 be a RBS then the strings <s1>s2{s1}s2[s1]s2(s1)s2 are also RBS.

For example the string "[[(){}]<>]" is RBS, but the strings "[)()" and "][()()" are not.

Determine the least number of replaces to make the string s RBS.

 

Input

The only line contains a non empty string s, consisting of only opening and closing brackets of four kinds. The length of s does not exceed 106.

Output

If it's impossible to get RBS from s print Impossible.

Otherwise print the least number of replaces needed to get RBS from s.

Examples

Input
[<}){}
Output
2
Input
{()}[]
Output
0
Input
]]
Output
Impossible

题意:输入一行括号序列,左边括号可以和右括号抵消。右括号可以转换为其他类型右括号进行成对抵消。输出需要转换的个数。如果无法抵消输出"Impossible"
 1 #include <iostream>
 2 #include <stack>
 3 using namespace std;
 4 
 5 int main(){
 6     string input;
 7     while(cin >> input){
 8         int count = 0;
 9         stack<char> s;
10         s.push(input[0]); //[ 
11         for(int i = 1;i < input.length();i++){//<}){}
12             if(s.size() > 0 && (s.top() == '[' || s.top() == '<' || s.top() == '(' || s.top() == '{')){
13                 if(input[i] == ']' || input[i] == '>' || input[i] == ')' || input[i] == '}'){
14                     if(!((s.top() == '[' && input[i] == ']') || 
15                         (s.top() == '<' && input[i] == '>') || 
16                         (s.top() == '(' && input[i] == ')') ||
17                         (s.top() == '{' && input[i] == '}'))){
18                         count++;
19                     }
20                     s.pop();
21                 }else{
22                     s.push(input[i]);
23                 }
24             }else{
25                 s.push(input[i]);
26             }
27         }
28         if(s.size() != 0){
29             cout << "Impossible" << endl;
30         }else{
31             cout << count << endl;
32         }
33     }
34 }

 




posted on 2019-04-27 14:52  夏晴天  阅读(132)  评论(0编辑  收藏

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