POJ-1068-Parencodings

Parencodings

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 24991		Accepted: 14734

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

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <stack>
#include <queue>
using namespace std;

int main()
{
int n,i,T,x,j,t,k;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
stack<char>S;
queue<int>Q;
k=0;
for(i=0;i<n;i++)
{
scanf("%d",&x);
for(j=k;j<x;j++)
{
S.push('(');
}
t=1;
k=x;
while(!S.empty())
{
if(S.top()=='*')
{
t++;
S.pop();
}
else
break;
}
S.pop();
Q.push(t);
while(t--)
{
S.push('*');
}
}
printf("%d",Q.front());
Q.pop();
while(!Q.empty())
{
printf(" %d",Q.front());
Q.pop();
}
printf("\n");
}
return 0;
}