poj1068 模拟

Parencodings

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 25010		Accepted: 14745

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

Source

Tehran 2001

水

代码：

 #include<iostream>

 #include<string>

 #include<cstdio>

 using namespace std;

 int main()

 {

     int n,t;

     scanf("%d",&t);

     while(t--)

     {

         int q1=,q2=;

         string s;

         scanf("%d",&n);

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

         {

             scanf("%d",&q1);

             q2=q1-q2;

             while(q2--)

             s+="(";

             s+=")";

             q2=q1;

         }

         int num=s.size();

         int cnt=;

         for(int i=;i<num;i++)

         {

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

             {

                 cnt++;

                 int flag=;

                 int sum=;

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

                 {

                     if(flag==)

                     break;

                     if(s[j]=='(')

                     {

                         sum++;

                         flag-=;

                     }

                     else if(s[j]==')')

                     flag+=;

                 }

                 if(cnt!=n)

                 printf("%d ",sum);

                 else printf("%d\n",sum);

             }

         }

     }

     return ;

 }

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