POJ1068--Parencodings--栈

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 6666W-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

264 5 6 6 6 69 4 6 6 6 6 8 9 9 9

Sample Output

1 1 1 4 5 61 1 2 4 5 1 1 3 9

#include <iostream>#include <cstdio>#include <stack>using namespace std;bool A[1008];int B[28];int C[28];int main(){int n,t;cin>>t;while(t--){cin>>n;for(int i=1;i<=n;i++){cin>>B[i];}B[0]=0;int k=0;for(int i=1;i<=n;i++){int len=B[i]-B[i-1];for(int j=1;j<=len;j++){A[++k]=0;//0代表左括号}A[++k]=1;//每一次匹配都以右括号结束，1代表右括号}stack <int> q;int m=0;for(int i=1;i<=k;i++){if(!A[i]){q.push(i);//推进的是i，仔细看else部分。}else{C[++m]=(i-q.top()+1)/2;q.pop();}}for(int i=1;i<=m;i++){cout<<C[i]<<" ";}cout<<endl;}return 0;}