Parencodings

Parencodings

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 13   Accepted Submission(s) : 7

Problem Description

Let S = s1 s2 … s2n be a well-formed string of parentheses. S can be encoded in two different ways: [ul] [li]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).[/li] [li]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).[/li] [/ul] Following is an example of the above encodings: [pre] S (((()()()))) P-sequence 4 5 6666 W-sequence 1 1 1456 [/pre] 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 file 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

2001 Asia Regional Teheran

 

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int T,num,i,j,sign,sum,K,k,z,tend,alto;
    scanf("%d",&T);
    while(T--)
    {
        int a[1000]={0},b[1000],c[1000]={0};
        scanf("%d %d",&num,&sign);
        a[sign]+=1;
        b[0]=sign;
        getchar();
        for(i=sign+1,j=1,k=1;j<num;j++,i++)
        {

            scanf("%d",&sum);
            if(sign==sum)
            {
                a[i]+=1;
                b[k]=i;
                k++;
            }
            else
            {
                K=sum-sign;
                a[i+K]+=1;
                b[k]=i+K;
                k++;
                i=i+K;
            }
            sign=sum;
        }

        for(j=0;j<k;j++)
            for(z=b[j]-1,tend=1;z>=0;z--)
            {
                if(a[z]==1)
                    continue;
                else
                {
                    if(a[z]==0)
                    {a[z]='N';c[j]+=1;break;}
                    else
                    {
                        if(a[z]=='N')
                        {
                            c[j]+=1;
                        }
                            continue;
                    }
                }
            }
        for(j=0;j<k;j++)
        {
                printf("%d",c[j]);
            if(j!=k-1)
            putchar(' ');
        }
        putchar('
');
    }
    return 0;

}