POJ 1068 Parencodings

Parencodings

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 15654		Accepted: 9332

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

就是模拟题啦。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
using namespace std;
int main()
{
    int t;
    int n,a[50],v;
    bool vis[50];
    scanf("%d",&t);
    string c;
    while(t--)
    {
        c.clear();
        a[0]=0;
        scanf("%d",&n);
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&a[i]);
            for(int j=0; j<a[i]-a[i-1]; j++)
            {
                c+='(';
            }
            c+=')';
        }
        //cout<<c<<endl;
        memset(vis,0,sizeof(vis));
        for(int i=0; i<=2*n; i++)
        {
            if(c[i]==')')
            {
                int sum=0;
                for(int j=i-1; j>=0; j--)
                {
                    if(c[j]=='(')
                    {
                        sum++;
                        if(vis[j]==0)
                        {
                            vis[j]=1;
                            break;
                        }
                    }
                }
                cout<<sum;
                if(i==(2*n-1))cout<<endl;
                else cout<<" ";
            }
        }
    }
    return 0;
}