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 9Sample Output
1 1 1 4 5 6 1 1 2 4 5 1 1 3 9 题意：一个括号表达式可以按照如下的规则表示，就是每个右括号之前的左括号数。 比如(((()()())))，每个右括号之前的左括号数序列为P=4 5 6 6 6 6，而每个右括号所在的括号内包含的括号数为W=1 1 1 4 5 6. 现在给定P，输出W。 我的思路：先根据P还原整个括号表达式，存在数组中，然后递归解出W.. #include<iostream> using namespace std; char y[10000]; int p[50],w[50],n,l,j; int f() { int s=1; while(1) if(y[j]=='(') { j++; s+=f(); } else { w[l++]=s; j++; return s; } } int main() { int t,i,k; cin>>t; while(t--) { cin>>n; for(i=0,l=0,k=0;i<n;i++) { cin>>p[i]; for(j=0;j<p[i]-k;j++) y[l++]='('; y[l++]=')'; k=p[i]; } l=j=0; f(); for(i=0;i<n;i++) cout<<w[i]<<' '; cout<<endl; } return 0; }