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 题意：给出p序列，求w序列。
p[]表示：当出现匹配括号对时，每个右括号前面有多少左括号
w[]表示：当出现匹配括号对时，该括号对中包含多少个右括号，包含本身
思路：根据 p 算出 每两个 右括号之间 有多少个 左括号，用 w 数组记录，然后 对每一个 右括号，往前搜索左括号
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
using namespace std;
int p[],w[];
int main()
{
int t,n;
cin>>t;
while(t--)
{
memset(p,,sizeof(p));
memset(w,,sizeof(w));
cin>>n;
for(int i=;i<=n;i++)
{
cin>>p[i];
w[i]=p[i]-p[i-];
}
for(int i=;i<=n;i++)
{
int j=i;
while(!w[j]&&j>)
j--;
w[j]--;
if(i==)
printf("%d",i-j+);
else
printf(" %d",i-j+);
}
printf("\n");
}
return ;
}