「CF494A」Treasure

2014年12月14日2,8842

Malek has recently found a treasure map. While he was looking for a treasure he found a locked door. There was a string swritten 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.

If there are several possible answers, you may output any of them.

Sample test(s)

input

(((#)((#)

(((#)((#)

output

1
2

input

()((#((#(#()

1	()((#((#(#()

output

2
2
1

input

#

output

-1

-1

input

(#)

(#)

output

-1

-1

Note

|s| denotes the length of the string s.

题解

除了最后一个#，其余的都取1，最后补全，然后判断合法性即可

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<cstring>
#define inf 2000000000
#define ll long long 
#define N 100005
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
char ch[100005];
int l,r,x,last;
int main()
{
    scanf("%s",ch+1);
    int L=strlen(ch+1);
    for(int i=1;i<=L;i++)
        if(ch[i]=='(')l++;
        else if(ch[i]==')')r++;
        else x++,last=i;
    if(r+x>l){puts("-1");return 0;}
    int now=0;
    for(int i=1;i<=L;i++)
    {
        if(ch[i]=='(')now++;
        else now--;
        if(i==last)now-=(l-r-x);
        if(now<0){puts("-1");return 0;}
    }
    for(int i=1;i<x;i++)puts("1");
    printf("%d\n",l-r-x+1);
    return 0;
}