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 ;
 }