POJ - 1068 Parencodings

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

将输入的P序列再模拟回括号（这里用01表示），再读出W序列

#include<iostream>
#include<stdio.h>
#include<string.h>

using namespace std;

int c[45];
int p[25],w[25];

int main()
{
    int T,n,s1,s2;
    scanf("%d",&T);
    while(T--)
    {
        memset(c,0,sizeof(c));
        memset(p,0,sizeof(p));
        memset(w,0,sizeof(w));
        s1=s2=0;
        scanf("%d",&n);
        for(int i=0;i<n;i++)
            scanf("%d",&p[i]);
        for(int i=0;i<n;i++)
        {
            for(int j=s2;j<p[i]-s1+s2;j++)
                c[j]=0;
            s2+=p[i]-s1;
            s1=p[i];
            c[s2++]=1;
        }
        n=0;
    /*    for(int i=0;i<40;i++)
            cout<<c[i];*/
        for(int i=0;i<40;i++)
        {
            if(c[i])
            {
                //cout<<i<<" ";
                s1=s2=1;
                for(int j=i-1;j>=0;j--)    
                {
                    if(c[j])
                    {
                        s1++;
                        s2++;
                    }    
                        else
                            s1--;
                    if(!s1)
                    {
                        w[n++]=s2;
                        break;
                    }
                }
            }
        }
        for(int i=0;i<n;i++)
            printf("%d ",w[i]);
        printf("\n");
    }
    
    
    return 0;
}