poj1068 模拟

Parencodings

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 25010		Accepted: 14745

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

水

代码：

#include<iostream>
#include<string>
#include<cstdio>
using namespace std;
int main()
{
    int n,t;
    scanf("%d",&t);
    while(t--)
    {
        int q1=0,q2=0;
        string s;
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            scanf("%d",&q1);
            q2=q1-q2;
            while(q2--)
            s+="(";
            s+=")";
            q2=q1;
        }
        int num=s.size();
        int cnt=0;
        for(int i=0;i<num;i++)
        {
            if(s[i]==')')
            {
                cnt++;
                int flag=1;
                int sum=0;
                for(int j=i-1;j>=0;j--)
                {
                    if(flag==0)
                    break;
                    if(s[j]=='(')
                    {
                        sum++;
                        flag-=1;
                    }
                    else if(s[j]==')')
                    flag+=1;
                }
                if(cnt!=n)
                printf("%d ",sum);
                else printf("%d\n",sum);
            }
        }
    }
    return 0;
}