[模拟题] 利用STL中string解pku 1068

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

264 5 6 6 6 69 4 6 6 6 6 8 9 9 9

Sample Output

1 1 1 4 5 61 1 2 4 5 1 1 3 9

solution：

#include<iostream>#include<string>using namespace std;int main(){  int i,j,k,m,n,totel,right,an;  cin>>m;  while(m--)  {  string s;  cin>>n;  totel=0;  //通过读入的整数将原来的字符串复原出来   for(i=0;i<n;i++)  {  cin>>k;  for(j=0;j<k-totel;j++) s+='(';  s+=')';  totel=k;            }          //对每个')'做一次向前扫描    for(i=0;i<s.size();i++)  {  if(s[i]==')')  {  right=1; an=1;  for(j=i-1;j>=0;j--)  {  if(s[j]=='(') right--;  else {right++; an++;}     if(right==0) break;           }  cout<<an<<" ";  }                       }  cout<<endl;  }    }/*2064 5 6 6 6 69 4 6 6 6 6 8 9 9 9*/