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
[<}){}
2
{()}[]
0
]]
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"
#include <iostream>
#include <stack>
using namespace std; int main(){
string input;
while(cin >> input){
int count = ;
stack<char> s;
s.push(input[]); //[
for(int i = ;i < input.length();i++){//<}){}
if(s.size() > && (s.top() == '[' || s.top() == '<' || s.top() == '(' || s.top() == '{')){
if(input[i] == ']' || input[i] == '>' || input[i] == ')' || input[i] == '}'){
if(!((s.top() == '[' && input[i] == ']') ||
(s.top() == '<' && input[i] == '>') ||
(s.top() == '(' && input[i] == ')') ||
(s.top() == '{' && input[i] == '}'))){
count++;
}
s.pop();
}else{
s.push(input[i]);
}
}else{
s.push(input[i]);
}
}
if(s.size() != ){
cout << "Impossible" << endl;
}else{
cout << count << endl;
}
}
}