CodeForces 494A Treasure

A. Treasure

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

题目大意：有一个字符串包括字符“（”，“）”

#include <cstdio>#include <cstring>const int maxn=100010;char s[maxn];int main(){    int a=0,b=0,c=0,i;    scanf("%s",s);    int len=strlen(s);if(s[0]==')'||s[len-1]=='('||s[0]=='#')    {        printf("-1\n");        return 0;    }for(i=len-1;i>=0;i--){if(s[i]=='(')a++;if(s[i]==')')b++;if(a>b){printf("-1\n");return 0;}if(s[i]=='#')break;}a=0,b=0;    for(i=0;i<len;i++)    {        if(s[i]=='(')//记录“（”的个数            a++;        if(s[i]==')')//如果有合法的一对括号，把他们标记为“.”{if(s[i-1]=='('){s[i-1]='.';s[i]='.';b--;a--;}            b++;}        if(s[i]=='#'){b++;            c++;}if(b>a)//判断“#”后面的左右括号数是否相等，左括号多则不合法{printf("-1\n");return 0;}    }b-=c;    if(a-b<c)    {        printf("-1\n");        return 0;    }a=0;for(i=0;i<len;i++){if(s[i]=='(')a++;if(s[i]==')'){b--;a--;}if(s[i]=='#'){if(c>1){printf("1\n");a--;c--;}elseprintf("%d\n",a-b);}}    return 0;}

，“#”，遇到#时把它转变成“）”，求每个#转换成几个“）”能使所有左右括号合法。

解题思路：第一个字符不是“（”或者最后一个字符是“（”不合法，“#”后面左括号大于右括号不合法，“#”个数大于“（”个数-“）”个数的差值不合法。如果合法，前面的“#”转换为1个括号，最后的“#”转换为剩下的左括号的个数。

代码如下：