【Codeforces】-612C-Replace To Make Regular Bracket Sequence（括号配对，栈）

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C. Replace To Make Regular Bracket Sequence

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

做题首先应该分清楚是栈还是队列，然后理清来做。

#include<cstdio>#include<cstring>#include<stack>using namespace std;char a[1e6+11];int main(){stack<char> stk;scanf("%s",a);int left=0,right=0;//统计左括号和右括号的个数，不同也不可能 int num=0,flag=1,l;l=strlen(a);for(int i=0;i<l;i++){if(a[i]=='<'||a[i]=='('||a[i]=='{'||a[i]=='['){left++;stk.push(a[i]);}else if(a[i]=='>'){right++;if(stk.empty()){flag=0;break;}else if(stk.top()=='<')//配对弹出 stk.pop();else{num++;//不配对，更换，num+1 stk.pop();}}else if(a[i]==')'){right++;if(stk.empty()){flag=0;break;}else if(stk.top()=='(')stk.pop();else{num++;stk.pop();}}else if(a[i]==']'){right++;if(stk.empty()){flag=0;break;}else if(stk.top()=='[')stk.pop();else{num++;stk.pop();}}else if(a[i]=='}'){right++;if(stk.empty()){flag=0;break;}else if(stk.top()=='{')stk.pop();else{num++;stk.pop();}}}if(left!=right) flag=0;if(flag==1)printf("%d\n",num);elseprintf("Impossible\n");return 0;}