Parencodings

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

题目大意：
设序列 S 是一个完全匹配的括号序列。序列 S 能用下面两种不同方法加密：
通过一个整数序列 P = p1 p2...pn ，其中 pi 表示序列 S 中第 i 个右括号 ")" 之前的左括号 “(” 的数量。当然这些括号是从左到右数的。
通过一个整数序列 W = w1 w2...wn ，wi 表示从第 i 个右括号 ")" 相匹配的左括号 “(” 的位置开始数的右括号 “)” 的数目，包括第 i 个右括号 “)” 本身。 （摘自百度知道）
输入序列P输出序列W。

解题思路：模拟做的。根据序列P将字母串写出来，在根据W的规则写出序列W。（被String的用法坑了。。）
Code:

 #include<string>

 #include<iostream>

 using namespace std;

 int main()

 {

     string tmp;

     int T,n,a[],i,j,k,t,cnt,sum;

     cin>>T;

     while (T--)

     {

         cin>>n;

         k=;

         tmp="";

         for (i=; i<=n; i++)

         {

             cin>>a[i];

             if (i!=) t=a[i]-a[i-];

             else t=a[i];

             for (j=; j<=t; j++)

                 tmp+='(';

             tmp+=')';

         }

         k=;

         for (i=; i<=tmp.length()-; i++)

             if (tmp[i]==')')

             {

                 cnt=,sum=;

                 for (j=i; j>=; j--)

                 {

                     if (tmp[j]==')') cnt++,sum++;

                     else cnt--;

                     if (!cnt) break;

                 }

                 a[k++]=sum;

             }

         for (i=; i<k; i++)

         {

             cout<<a[i];

             if (i!=k-) cout<<' ';

             else cout<<endl;

         }

     }

     return ;

 }