ZOJ_1016

ZOJ_1016: Parencodings

Time Limit: 2000 MS  Memory Limit: 64 MB  64bit IO Format: %lld
Submitted: 0  Accepted: 0
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Description

Let S = s1 s2 ... s2n be a well-formed string of parentheses. S can be encoded in two different ways:

• 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).
• 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 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 6 694 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

Asia 2001, Tehran (Iran)

这道题目不要想的太复杂了，先把串给模拟出来比较好

#include<cstdio>

using namespace std;




int n;
int a[100];
char ss[1000];
int ans[1000];
int anslen;
int main()
{


    int T;
    scanf("%d",&T);
    while(T--)
    {
        anslen=0;
      scanf("%d",&n);


        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        int len=0;


        for(int i=0;i<a[0];i++)
            ss[len++]='(';
            ss[len++]=')';
        for(int i=1;i<n;i++)
        {


            for(int j=0;j<a[i]-a[i-1];j++)
                ss[len++]='(';


            ss[len++]=')';


        }
        ss[len]='\0';


        int num=0;
        anslen=0;
        for(int i=0;i<len;i++)
        {
            if(ss[i]=='(')
                continue;
            int counl=0,counr=1;
            for(int j=i-1;j>=0;j--)
            {


                if(ss[j]=='(')
                    counl++;
                    else
                        counr++;
                if(counl==counr)
                {
                    ans[anslen++]=counl;
                    break;
                }
            }
        }


        for(int i=0;i<anslen;i++)
        {
            if(i)
                printf(" ");


            printf("%d",ans[i]);
        }
        puts("");
      }
    return 0;


}