CodeForces 494A Treasure 【greedy】

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

Note

|s| denotes the length of the string s.

题意：给你一个字符串，里面包含‘（’， ‘）’， ‘#’三种字符，字符‘#’可以换成一个或多个‘）’， 问通过将‘#’换成‘)’能不能形成一个完美字符串（前i个字符中要满足‘（’个数大于‘）’，并且交换完成（所有的‘#’都交换完成）后的字符串满足的‘（’和‘）’的个数要相等）

分析：因为题目给的很宽松，如果有多组答案数据，输出任意一组就行。那么，我们可以将除了最后一个‘#’特殊处理之外，其他的都只与一个‘）’交换，最后一个特殊判断就行了。 

代码：

#include <stdio.h>#include <string.h>#include <algorithm>#define M 100050char s[M];int c[M];int main(){while(~scanf("%s", s)){memset(c, 0, sizeof(c));int i, a, c1, c2;int len = strlen(s);a = c1 = c2 = 0;for(i = 0; i < len; ++ i){if(s[i] == ')') ++c1;else if(s[i] == '(') ++c2;else ++a;}int tot = a;if(c1+a > c2){printf("-1\n"); continue;}int temp1 = 0, temp2 = 0, j;for(i = 0; i < len; ++ i){if(s[i] == ')') ++temp1;if(s[i] == '(') ++temp2;if(s[i] == '#'){if((--a) == 0){int cou = 0;j = len-1;while(j > i){if(s[j] == ')') ++cou;-- j;}int temp = c2-temp1-cou;temp1 += temp;c[a] = temp;}else{++temp1;c[a] = 1;}}if(temp1 > temp2) break;}if(temp1 > temp2) printf("-1\n");else{for(i = tot-1; i >= 0; i --)printf("%d\n", c[i]);}}return 0;}