Parencodings
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
Source
题目类型:简单模拟
算法分析:从左到右同时遍历两个数组模拟计算即可
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#include <cstdio> #include <iostream> #include <fstream> using namespace std; int main() //important!!! { int cases, p[25], pairs[25]; p[0] = 0; char s[45]; //ifstream cin ("aaa.txt"); cin >> cases; while (cases--) { int n; cin >> n; int i; for (i = 1; i <= n; i++) cin >> p[i]; int j = 1, k; int cp; for (i = 1; i <= n; i++) { cp = p[i] - p[i-1]; for (k = 1; k <= cp; k++) s[j++] = '('; s[j++] = ')'; } s[j] = '\0'; n *= 2; j = 1; for (i = 1; i <= n; i++) { if (s[i] == '(') pairs[j++] = i; else { printf ("%d", (i - pairs[j-1] + 1) / 2); j--; if (i != n) printf (" "); } } printf ("\n"); } return 0; } |
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