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从别人的博客上学的：https://www.cnblogs.com/--ZHIYUAN/p/5910981.html #include<iostream> #include<string> #include<cstdio> using namespace std; int main(){ int n,T; scanf("%d",&T); while(T--){ int q1,q2 = 0; string s; scanf("%d",&n); for(int i = 0; i < n; i++){ scanf("%d",&q1); q2 = q1 - q2; while(q2--) s += "("; s += ")"; q2 = q1; } int num = s.length(); int cnt = 0; int sum; for(int i = 0; i < num; i++){ if(s[i] == ')'){ cnt++; int flag = 1; sum = 0; for(int j = i - 1; j >= 0; j--){ if(flag == 0) break; if(s[j] == ')'){ flag += 1; } else if(s[j] == '('){ flag -= 1; sum++; } } if(cnt != n) printf("%d ",sum); else printf("%d\n",sum); } } } return 0; } View Code
转载于:https://www.cnblogs.com/Weixu-Liu/p/10422650.html
