Parencodings - poj 1068

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 22764		Accepted: 13344

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
这道题很简单，一遍AC，用flag数组记录左括号的状态，当是1时已经匹配，当是0时未匹配，向前查找第一个未匹配的左括号，记录中间匹配的左括号的个数

#include <iostream>
using namespace std;

int main() {
int num;
int n;
int p[21];
int flag[21];
cin>>num;
for(int i=0;i<num;i++){
cin>>n;
for(int j=0;j<n;j++){
cin>>p[j];
flag[j]=0;
}
for(int k=0;k<n;k++){
int count=0;
for(int m=p[k]-1;m>=0;m--){
if(flag[m]==0){
cout<<count+1<<" ";
flag[m]=1;
break;
}else{
count++;
}
}
}
cout<<endl;
}
return 0;
}