poj 1068 -- Parencodings

Parencodings

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 19068		Accepted: 11503

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

简单模拟题，给出若干左右括号，能写出两种数列，其一:在第i个右括号前有多少个左括号pi就是几。 其二：找距离第i个右括号最近的左括号（没有配过对的）。
此题给出其一队列，求其二，模拟即可，不多说。

/*======================================================================
 *           Author :   kevin
 *         Filename :   Parencodings.cpp
 *       Creat time :   2014-05-23 13:41
 *      Description :
========================================================================*/
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#define clr(a,b) memset(a,b,sizeof(a))
#define M 50
using namespace std;
char str[M];
int vis[M];
int main(int argc,char *argv[])
{
    int t,n;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        clr(str,0);
        clr(vis,0);
        int a,cc = 0,cnt = 0;
        for(int i = 0; i < n; i++){
            scanf("%d",&a);
            for(int j = cc; j < a; j++){
                str[cnt++] = '(';
            }
            str[cnt++] = ')';
            cc = a;
        }
        int kong = 0;
        for(int i = 0; i < 2*n; i++){
            if(str[i] == ')'){
                int steps = 1;
                for(int j = i-1; j >= 0; j--){
                    if(str[j] == '(' && !vis[j]){
                        if(kong != n-1){
                            printf("%d ",steps);
                            kong++;
                        }
                        else{
                            printf("%d
",steps);
                        }
                        vis[j] = 1;
                        break;
                    }
                    if(str[j] == '(' && vis[j]){
                        steps++;
                    }
                }
            }
        }
    }
    return 0;
}