Let S = s1 s2 ... s2n be a well-formed string of parentheses. S can be encoded in two different ways:
- 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).
- 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 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
先构建括号然后再查找w序列
AC代码：
#include<iostream>
#include<stdio.h>
#include<string>
using namespace std;
string p;
void add(int num)
{
int len=p.length();
for(int i=0;i<len;i++)
{
if(p[i]=='(')num--;
}
while(num--)
{
p+='(';
}
p+=')';
}
int main()
{
int cas;
cin>>cas;
while(cas--)
{
int n;
cin>>n;
int *num1=new int[n];
p="";
for(int i=0;i<n;i++)//构建括号
{
cin>>num1[i];
add(num1[i]);
}
// cout<<p<<endl;
//构建w序列
int len=p.length();
int start=0;
int k=0;
int w[100];
for(int i=start;i<len;i++)
{
int sum=1;
int count=0;
if(p[i]==')')
{
start=i+1;
for(int j=i-1;j>=0;j--)
{
if(p[j]=='(')
{
sum--;
count++;
}
if(p[j]==')')
{
sum++;
}
if(sum==0)
{
w[k++]=count;//找到匹配括号
break;//开始寻找下一个匹配括号
}
}
}
}
for(int i=0;i<k;i++)
{
cout<<w[i];
if(i==k-1)cout<<endl;
else cout<<' ';
}
}
}