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
水题，因为没注意空格错了一会。
题意就是告诉第i个右括号前有几个左括号
打印第i个括号，前有几个匹配的括号
思路用的：STL队列，逆匹配
#include
#include
#include
#include
#include
#include
using namespace std; char s[1000]; int main() { std::ios::sync_with_stdio(false); int t,n,i,j,a[30],b[30],l; cin>>t; while(t--) { cin>>n; l = 0; for(i = 0;i
>a[i]; b[i] = a[i] + i; } for( i = 0;i
q; int co = 0; for(i = 0;i
=0;j--) { if(q.empty()) break; if(s[j]==')') q.push(s[j]); else if(s[j]=='(' && q.front()==')') { q.pop(); co++; } } if(i