POJ 1068 Parencodings

Parencodings

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 19185 Accepted: 11570

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 6666W-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

Source

Tehran 2001

#include<stdio.h>
#include<string.h>
int main()
{
    int T,i,j;
    int a[10020],b[10010];
    int c[10010];
    scanf("%d",&T);
    while(T--)
    {
        int n,k = 0;
        scanf("%d",&n);
        a[0] = 0;
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        for(i=1;i<=n;i++)
        {
            for(j=a[i-1];j<a[i];j++)
            {
                b[k++] = 1;
            }
             b[k++] = 2;
        }
        int t = 0,p,pp;
        for(i=1;i<k;i++)
        {
            if(b[i] == 2)
            {
                b[i] = 1000;
                p = 0,pp = 0;
                for(j=i;j>=0;j--)
                {
                    p++;
                    if(b[j] == 1)
                    {
                       b[j] = 1000;
                       c[t++] = p/2;
                       break;
                    }
                }
            }
        }
        for(i=0;i<t;i++)
        {
            if(i<t-1)
            printf("%d ",c[i]);
            else
                printf("%d\n",c[i]);
        }
    }
    return 0;
}