标签：
| 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
1 #include <iostream>
2 using namespace std;
3
4 int main() {
5 int num;
6 int n;
7 int p[21];
8 int flag[21];
9 cin>>num;
10 for(int i=0;i<num;i++){
11 cin>>n;
12 for(int j=0;j<n;j++){
13 cin>>p[j];
14 flag[j]=0;
15 }
16 for(int k=0;k<n;k++){
17 int count=0;
18 for(int m=p[k]-1;m>=0;m--){
19 if(flag[m]==0){
20 cout<<count+1<<" ";
21 flag[m]=1;
22 break;
23 }else{
24 count++;
25 }
26 }
27 }
28 cout<<endl;
29 }
30 return 0;
31 }
Parencodings - poj 1068
标签：
原文地址：http://www.cnblogs.com/sdxk/p/4617720.html
