poj 1068 Parencodings 模拟题

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

题意：给出p序列，求w序列。
    p[]表示：当出现匹配括号对时，每个右括号前面有多少左括号
    w[]表示：当出现匹配括号对时，该括号对中包含多少个右括号，包含本身

思路：根据 p 算出 每两个 右括号之间 有多少个 左括号，用 w 数组记录，然后 对每一个 右括号，往前搜索左括号

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
using namespace std;
int p[25],w[25];
int main()
{
    int t,n;
    cin>>t;
    while(t--)
    {
        memset(p,0,sizeof(p));
        memset(w,0,sizeof(w));
        cin>>n;
        for(int i=1;i<=n;i++)
        {
             cin>>p[i];
             w[i]=p[i]-p[i-1];
        }
        for(int i=1;i<=n;i++)
        {
            int j=i;
            while(!w[j]&&j>1)
                j--;
            w[j]--;
          if(i==1)
                printf("%d",i-j+1);
          else
            printf(" %d",i-j+1);
        }
        printf("
");
    }
    return 0;
}