[Codeforces494A] Treasure

2014-12-22
竞赛题解 / 模拟
Codeforces

描述 Description

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.

输入格式 InputFormat

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.

输出格式 OutputFormat

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.

样例输入 SampleInput

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

样例输出 SampleOutput

1
1
1
5

Codeforces 494A

代码 Code

因为可以把右括号全加到最后一个 #上去，所以只要保证每个 #都能有至少一个) 且符合条件即可。

#include <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int maxn=100005;
int i,j,t,n,m,l,r,k,z,y,x;
char a[maxn];
int main()
{
    scanf("%s",a);
    n=strlen(a);
    for (m=x=i=0;i<n;i++)
    {
        if (a[i]=='(') x++;
        else x--;
        if (a[i]=='#') m++;
        if (x<0)
        {
            printf("-1\n");
            return 0;
        }
    }
    z=x;
    for (x=0,i=n-1;i>=0 && a[i]!='#';i--)
    {
        if (a[i]=='(') x--;else x++;
        if (x<0) 
        {
            printf("-1\n");
            return 0;
        }
    }
    for (i=1;i<m;i++) printf("1\n");
    printf("%d\n",z+1);
}