POJ 1068 Parencodings（模拟）

Parencodings

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 15075		Accepted: 8989

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

 

 

 

/*

POJ 1068 Parencodings

简单模拟

用数组a[i]表示第i个右括号和第i+1个右括号间的左括号数

然后逐渐找和右括号匹配的左括号所处的位置

i-j



*/





#include<stdio.h>

#include<iostream>



const int MAXN=50;

int P[MAXN];

int W[MAXN];

int a[MAXN];



int main()

{

    int T;

    int n;

    scanf("%d",&T);

    while(T--)

    {

        scanf("%d",&n);

        for(int i=1;i<=n;i++)

           scanf("%d",&P[i]);

        a[0]=P[1];

        for(int i=1;i<n;i++) a[i]=P[i+1]-P[i];

        int j;

        for(int i=1;i<=n;i++)

        {

            for(j=i-1;j>=0;j--)//匹配该右括号

            {

                if(a[j]>0)

                {

                    a[j]--;

                    break;

                }

            }

            W[i]=i-j;

        }

        for(int i=1;i<n;i++)printf("%d ",W[i]);

        printf("%d\n",W[n]);

    }

    return 0;

}