POJ1068——Parencodings

Parencodings

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

题目大意：
设序列 S 是一个完全匹配的括号序列。序列 S 能用下面两种不同方法加密：
通过一个整数序列 P = p1 p2...pn ，其中 pi 表示序列 S 中第 i 个右括号 ")" 之前的左括号 “(” 的数量。当然这些括号是从左到右数的。
通过一个整数序列 W = w1 w2...wn ，wi 表示从第 i 个右括号 ")" 相匹配的左括号 “(” 的位置开始数的右括号 “)” 的数目，包括第 i 个右括号 “)” 本身。 （摘自百度知道）
输入序列P输出序列W。

解题思路：模拟做的。根据序列P将字母串写出来，在根据W的规则写出序列W。（被String的用法坑了。。）
Code:

#include<string>
#include<iostream>
using namespace std;
int main()
{
    string tmp;
    int T,n,a[10000],i,j,k,t,cnt,sum;
    cin>>T;
    while (T--)
    {
        cin>>n;
        k=0;
        tmp="";
        for (i=1; i<=n; i++)
        {
            cin>>a[i];
            if (i!=1) t=a[i]-a[i-1];
            else t=a[i];
            for (j=1; j<=t; j++)
                tmp+='(';
            tmp+=')';
        }
        k=0;
        for (i=0; i<=tmp.length()-1; i++)
            if (tmp[i]==')')
            {
                cnt=0,sum=0;
                for (j=i; j>=0; j--)
                {
                    if (tmp[j]==')') cnt++,sum++;
                    else cnt--;
                    if (!cnt) break;
                }
                a[k++]=sum;
            }
        for (i=0; i<k; i++)
        {
            cout<<a[i];
            if (i!=k-1) cout<<' ';
            else cout<<endl;
        }
    }
    return 0;
}