POJ 1068 Parencodings（模拟）

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本文介绍了一种算法，用于将括号字符串的P序列转换为W序列。通过输入P序列，程序能够还原原始括号字符串，并进一步计算出相应的W序列。

Parencodings

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 22915		Accepted: 13434

Description

Let S = s1 s2...s2n be a well-formed string of parentheses. S can be encoded in two different ways:
q By an integer sequence P = p1 p2...pn where pi is the number of left parentheses before the ith right parenthesis in S (P-sequence).
q By an integer sequence W = w1 w2...wn where for each right parenthesis, say a in S, we associate an integer which is the number of right parentheses counting from the matched left parenthesis of a up to a. (W-sequence).

Following is an example of the above encodings:

	S		(((()()())))

	P-sequence	    4 5 6666

	W-sequence	    1 1 1456

Write a program to convert P-sequence of a well-formed string to the W-sequence of the same string.

Input

The first line of the input contains a single integer t (1 <= t <= 10), the number of test cases, followed by the input data for each test case. The first line of each test case is an integer n (1 <= n <= 20), and the second line is the P-sequence of a well-formed string. It contains n positive integers, separated with blanks, representing the P-sequence.

Output

The output file consists of exactly t lines corresponding to test cases. For each test case, the output line should contain n integers describing the W-sequence of the string corresponding to its given P-sequence.

Sample Input

2
6
4 5 6 6 6 6
9 
4 6 6 6 6 8 9 9 9

Sample Output

1 1 1 4 5 6
1 1 2 4 5 1 1 3 9

对于给出的原括号串，存在两种数字密码串：
1.p序列：当出现匹配括号对时，从该括号对的右括号开始往左数，直到最前面的左括号数，就是pi的值
2.w序列：当出现匹配括号对时，包含在该括号对中的所有右括号数（包括该括号对）,就是wi的值
题目的要求：对给出的p数字串，求出对应的s串

直接根据p  将括号串还原，直接根据括号串求出w
#include <iostream>
#include <cstdio>
#include <cstring>
#include <stdlib.h>
#define N 51
using namespace std;
char a[N];        
int main()
{
   int T;
   cin>>T;
   while(T--)
   {
       int n;
       scanf("%d",&n);
       int s=0;
       int x=0;
       for(int i=0;i<n;i++)
       {
           int r;
           cin>>r;
           for(int j=0;j<r-x;j++)
            a[s++]='(';
           a[s++]=')';
           x=r;
       }
       int k=0;
       int ans[N];
       memset(ans,0,sizeof(ans));
       for(int i=0;i<s;i++)
       {
           if(a[i]==')')
           {
               int sum=1;
               for(int j=i-1;j>=0;j--)
               {
                   if(a[j]=='(')
                        {
                            sum--;
                            ans[k]++;
                        }
                   else sum++;
                   if(sum==0) break;
               }
             k++;
           }
       }
       for(int i=0;i<k-1;i++)
        cout<<ans[i]<<" " ;
       cout<<ans[k-1]<<endl;
   }
    return 0;
}