CodeForces 494A Treasure

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解决一个算法问题，通过替换特定字符使括号匹配正确。输入包含特殊字符的字符串，并确定每个特定字符需要替换成多少个闭合括号。

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
1
2

input
()((#((#(#()

output
2
2
1

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--;
			}
			else
			printf("%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：