Codeforces Round #282 (Div. 1) A. Treasure 水题

A. Treasure

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/494/problem/A

Description

Malek has recently found a treasure map. While he was looking for a treasure he found a locked door. There was a string s written on the door consisting of characters '(', ')' and '#'. Below there was a manual on how to open the door. After spending a long time Malek managed to decode the manual and found out that the goal is to replace each '#' with one or more ')' characters so that the final string becomes beautiful.

Below there was also written that a string is called beautiful if for each i (1 ≤ i ≤ |s|) there are no more ')' characters than '(' characters among the first i characters of s and also the total number of '(' characters is equal to the total number of ')' characters.

Help Malek open the door by telling him for each '#' character how many ')' characters he must replace it with.

Input

The first line of the input contains a string s (1 ≤ |s| ≤ 105). Each character of this string is one of the characters '(', ')' or '#'. It is guaranteed that s contains at least one '#' character.

Output

If there is no way of replacing '#' characters which leads to a beautiful string print  - 1. Otherwise for each character '#' print a separate line containing a positive integer, the number of ')' characters this character must be replaced with.

Sample Input

(((#)((#)

Sample Output

1
2

HINT

题意

给你一个序列，其中#可以被替换成1个或多个)符号

你需要使得这个序列beautiful

beautiful的定义是，对于每一个i，使得(的个数大于或等于)，并且最后(的个数和)的个数相同

题解：

简单分析一下我就可以贪心的去搞一搞

前面的#我们都只替换一个，然后最后的#替换剩下的

然后我们再扫一遍进行check就好了

代码

#include<iostream>
#include<stdio.h>
#include<vector>
using namespace std;

string s;
vector<int> ans;
int main()
{
    cin>>s;
    int sum = 0;
    int flag = 0;
    for(int i=0;i<s.size();i++)
    {
        if(s[i]=='(')sum--;
        else if(s[i]==')')sum++;
        else
        {ans.push_back(i);flag++;}
        if(sum+flag>0)return puts("-1");
    }
    int k = ans.size();
    if(k+sum>0)return puts("-1");
    else if(k==0&&sum!=0)return puts("-1");
    else
    {
        int Ans = 0;
        for(int i=0;i<s.size();i++)
        {
            if(s[i]=='(')Ans--;
            else if(s[i]==')')Ans++;
            else if(i==ans[ans.size()-1])Ans+=0-(sum+k-1);
            else Ans++;
            if(Ans>0)return puts("-1");
        }
        for(int i=0;i<k;i++)
        {
            if(i==k-1)
                printf("%d
",0-(sum+k-1));
            else
                printf("1
");
        }
    }
}