CodeForces 612C Replace To Make Regular Bracket Sequence

题意：
给你一个只含有括号的字符串，你可以将一种类型的左括号改成另外一种类型，右括号改成另外一种右括号
问你最少修改多少次，才能使得这个字符串匹配，输出次数
思路：
用stack，每次将左括号压进stack里面，遇到右括号就判断一下就好了
非法就很简单，看看栈最后是否还有，看看右括号的时候，左括号的栈是否为空

#include<bits/stdc++.h>
using namespace std;

string s;
stack<char> S;
int main()
{
cin>>s;
int ans = 0;
for(int i=0;i<s.size();i++)
{

if(s[i]==']')
{
if(S.size()==0)return puts("Impossible");
if(S.top()=='[')
S.pop();
else
{
ans++;
S.pop();
}
}
else if(s[i]==')')
{
if(S.size()==0)return puts("Impossible");
if(S.top()=='(')
S.pop();
else
{
ans++;
S.pop();
}
}

else if(s[i]=='>')
{
if(S.size()==0)return puts("Impossible");
if(S.top()=='<')
S.pop();
else
{
ans++;
S.pop();
}
}
else if(s[i]=='}')
{
if(S.size()==0)return puts("Impossible");
if(S.top()=='{')
S.pop();
else
{
ans++;
S.pop();
}
}
else S.push(s[i]);
}
if(S.size()!=0)return puts("Impossible");
cout<<ans<<endl;
}

Description

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.

Sample Input

Input

[<}){}

Output

2

Input

{()}[]

Output

0

Input

]]

Output

Impossible