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 #include<stdio.h> #include<stack> #include<queue> #include<string.h> using namespace std; int main() { char s[1000000+10]; while(scanf("%s",&s)!=EOF) { stack<char>sta; int i,j,k,l,ans=1; l=strlen(s); for(k=0,i=0;i<l;i++) { if(sta.empty()&&(s[i]==']'||s[i]=='}'||s[i]=='>'||s[i]==')')) { ans=0; break; } else if(s[i]=='<'||s[i]=='{'||s[i]=='['||s[i]=='(') { sta.push(s[i]); continue; } else { if(s[i]-sta.top()==2||s[i]-sta.top()==1) sta.pop(); else { sta.pop(); k++; } } } if(ans==0||!sta.empty()) printf("Impossible\n"); else printf("%d\n",k); } return 0; }