CF#282 Div 2 C Treasure

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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.

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

Sample test(s)
input
(((#)((#)
output
12
input
()((#((#(#()
output
221
input
#
output
-1
input
(#)
output
-1

Note

|s| denotes the length of the string s.

题意很好理解 #字符可以替换为若干个)字符问字符串中每个#字符应该替换为多少个)字符满足它设定的条件

条件有（1）前i个字符中 (字符的数量不能少于)字符的数量包括#替换之后！！！

（2）替换之后的字符串中 (字符的个数与)字符的个数相等

贪心的思路：

除了最后一个#字符之外每个#字符替换一个) 最后一个#字符来满足整个字符串中 (字符与)字符相等的条件

遇到一个(字符+1 遇到一个)字符 -1 #字符替换多少个)字符就减多少

每次循环判断值是否小于0 是否满足条件（1）

#include <cstdio>#include <iostream>#include <cstring>#include <cmath>#include <algorithm>#include <string.h>#include <string>#define eps 1e-8#define op operator#define MOD  10009#define MAXN  100100#define INF 0x7fffffff#define MEM(a,x)    memset(a,x,sizeof a)#define ll __int64using namespace std;char ch[MAXN];int num[MAXN];int main(){//freopen("ceshi.txt","r",stdin);    while(scanf("%s",ch)!=EOF)    {        int a=0,b=0,c=0;        int len=strlen(ch);        int last;        for(int i=0;i<len;i++)        {            if(ch[i]=='(') a++;            if(ch[i]==')') b++;            if(ch[i]=='#')            {                c++;                last=i;            }        }        if(a<b+c)        {            puts("-1");            continue;        }        int n=0;        int flag=1;        for(int i=0;i<len;i++)        {            if(ch[i]=='(') n++;            if(ch[i]==')') n--;            if(ch[i]=='#')            {                if(i==last)                {                    n-=a-(b+c-1);                }                else n--;            }//            cout<<i<<"   "<<n<<endl;            if(n<0)            {                flag=0;                puts("-1");                break;            }        }        if(flag)        {            for(int i=1;i<c;i++)                puts("1");            printf("%d\n",a-(b+c-1));        }    }    return 0;}

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