Educational Codeforces Round 4 612C Replace To Make Regular Bracket Sequence(脑洞)

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 exceed106.

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 test(s)

input

[<}){}

output

2

input

{()}[]

output

0

input

]]

output

Impossible

题目链接:点击打开链接

给出一个字符串, 要求左右括号相互匹配, 右括号间可以相互转换, 问最少需要转换的数量.

记录左括号的数量, 如果开始时就没有左括号, 那么无法转换. 否则如果右括号与左括号不匹配的话, ans+1. 最后如果还有剩余括号未匹

配, 则转换失败.

AC代码:

#include "iostream"#include "cstdio"#include "cstring"#include "algorithm"#include "queue"#include "stack"#include "cmath"#include "utility"#include "map"#include "set"#include "vector"#include "list"#include "string"#include "cstdlib"using namespace std;typedef long long ll;const int MOD = 1e9 + 7;const int INF = 0x3f3f3f3f;const int MAXN = 1e6 + 6;char s[MAXN], tmp[MAXN];int num, ans;int main(int argc, char const *argv[]){    scanf("%s", s);    int len = strlen(s);    for(int i = 0; i < len; ++i) {        if(s[i] == '<' || s[i] == '{' || s[i] == '[' || s[i] == '(') tmp[++num] = s[i];        else {            if(num == 0) {                printf("Impossible\n");                return 0;            }            if(abs(s[i] - tmp[num]) > 2) ans++;            num--;        }    }    if(num) {        printf("Impossible\n");        return 0;    }    printf("%d\n", ans);    return 0;}