ZJU 1016 Parencodings

ZOJ Problem Set - 1016

Parencodings

---

Time Limit:1 Second      Memory Limit: 32768KB

---

Let S = s1 s2 ... s2n be a well-formed string of parentheses. S can beencoded 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-sequenceof 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. Thefirst line of each test case is an integer n (1 <= n <= 20), and thesecond line is the P-sequence of a well-formed string. It contains n positiveintegers, separated with blanks, representing the P-sequence.

Output

The output consists of exactly t lines corresponding to test cases. For eachtest case, the output line should contain n integers describing the W-sequenceof 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:Asia 2001, Tehran(Iran)

 

思路：

此题关键在于记录相邻右括号间剩余的括号数。

当读取一个新的P序列的数nP后，计算它与前一个P序列的数nLastP的差（nLastP初始为0，则第一个差为第一个nP），并保存为对应两个右括号间剩余的括号数nLParenLeft[i]。当要计算W序列的数Wi时，把Wi初始为1，从nLParenLeft[i]开始考察，如果不等于0，表示在这个位置有左括号剩余，则对其减一，并输出Wi；如果等于0，对Wi加一，并考察前一位的nLParenLeft[]。

 

 

代码：

#include <iostream>

using namespace std;

int main()
{
    int nTestCaseNum;

    cin>>nTestCaseNum;
    cin.get();

    while (nTestCaseNum--)
    {
        int nRParenNum;

        cin>>nRParenNum;
        cin.get();

        int nLastP = 0;
        int nLParenLeft[20];


        for (int i = 0; i < nRParenNum; i ++)
        {
            int nP;
            cin>>nP;

            nLParenLeft[i] = nP - nLastP;
            nLastP = nP;


            int k = i;
            int nW = 1;
            while (k >= 0)
            {

                if(nLParenLeft[k]> 0)
               {
                   nLParenLeft[k]--;

                   break;
               }
               else
               {
                   k --;

                   nW ++;

               }
            }

 

            if (i > 0)
            {
               cout<<" ";
            }

            cout<<nW;           
        }
        cout<<endl;
    }

    return 0;

}