POJ 1068 Parencodings

Parencodings
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 15566 Accepted: 9275
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
考察点：栈
读懂题目就好做了
[cpp]
#include <iostream>
#include <cstring>
#include <stdio.h>
#include <math.h>
using namespace std;
int a[100],top,res[100],top2;
class str
{
    public:
    int pos;
    char c;
}s1[100];
class statck
{
    public:
    int pos;
    char c;
}s2[100];
int main()
{
    int i,j,n,m,s,t,pos1,pos2;
    cin>>t;
    while(t--)
    {
        cin>>n;
        for(i=0;i<=n-1;i++)
        {
            cin>>a[i];
        }
        top=0;
        for(i=0;i<=n-1;i++)
        {
            if(!i)
            {
                for(j=1;j<=a[i];j++)
                {
                    s1[top].c='(';
                    s1[top].pos=top;
                    top++;
                }
                s1[top].c=')';
                s1[top].pos=top;
                top++;
            }else
            {
                for(j=1;j<=(a[i]-a[i-1]);j++)
                {
                    s1[top].c='(';
                    s1[top].pos=top;
                    top++;
                }
                s1[top].c=')';
                s1[top].pos=top;
                top++;
            }
        }
        n=top;
        top=0;
        top2=0;
        memset(res,0,sizeof(res));
        for(i=0;i<=n-1;i++)
        {
            if(s1[i].c=='(')
            {
                s2[top].c='(';
                s2[top].pos=s1[i].pos;
                top++;
            }else
            {
                pos1=s2[top-1].pos;
                top-=1;
                pos2=s1[i].pos;
                for(j=pos1;j<=pos2;j++)
                {
                    if(s1[j].c==')')
                    {
                      &

补充：软件开发 , C++ ,