[CF]Codeforces Round #551 (Div. 2)

C. Serval and Parenthesis Sequence

Description

Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.

In his favorite math class, the teacher taught him the following interesting definitions.

A parenthesis sequence is a string, containing only characters "(" and ")".

A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.

We define that |𝑠||s| as the length of string 𝑠s. A strict prefix 𝑠[1…𝑙]s[1…l] (1≤𝑙<|𝑠|)(1≤l<|s|) of a string 𝑠=𝑠1𝑠2…𝑠|𝑠|s=s1s2…s|s| is string 𝑠1𝑠2…𝑠𝑙s1s2…sl. Note that the empty string and the whole string are not strict prefixes of any string by the definition.

Having learned these definitions, he comes up with a new problem. He writes down a string 𝑠scontaining only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in 𝑠s independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.

After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.

Input

The first line contains a single integer |𝑠||s| (1≤|𝑠|≤3⋅1051≤|s|≤3⋅105), the length of the string.

The second line contains a string 𝑠s, containing only "(", ")" and "?".

Output

A single line contains a string representing the answer.

If there are many solutions, any of them is acceptable.

If there is no answer, print a single line containing ":(" (without the quotes

 

正确解法：

首先n肯定是偶数

如果有 “（”，n1++　　“）”，n2++.

如果 n1>n/2 || n2 > n/2  的时候，肯定也不行

然后就用贪心的操作，先用 （，没了再用 ）

最后检查一遍就好啦！

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <set>
#include <map>
#include <vector>
using namespace std;
typedef long long ll;
const int inf=0x7fffffff;
const int N=100000+100;
char s[3*N];
int n,n1,n2,x;
void fc()
{
    cout<<":("<<endl;
    exit(0);
}
int main()
{
    scanf("%d",&n);
    scanf("%s",&s);
    if(n%2==1)
        fc();
    for(int i=0;i<n;i++)
    {
        if(s[i]=='(')
            n1++;
        else if(s[i]==')')
            n2++;
    }
    if(n1>n/2||n2>n/2)
        fc();
    n1=n/2-n1;
    for(int i=0;i<n;i++)
    {
        if(s[i]=='?')
        {
            if(n1)
            {
                s[i]='(';
                n1--;
            }
            else    s[i]=')';
        }
    }
    for(int i=0;i+1<n;i++)
    {
        if(s[i]=='(')   x++;
        else    x--;
        if(x<=0)    fc();
    }
    printf("%s",s);

    return 0;
}