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