杭电OJ题 1361 Parencodings解题报告

Parencodings

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 403    Accepted Submission(s): 231

Problem Description

Let S = s1 s2 … s2n be a well-formed string of parentheses. S can be encoded in two different ways:

• 
  
• 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).
  
• 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 file 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

 
————————————————————————————————————————————
我先通过P-sequence找出S表达式，然后转换为W-sequence

/****************************
 *Name:Parencodings.c
 *Tags:ACM water
 ****************************/

#include <stdio.h>

int main()
{
      int t, n, i, p, w[20], j, l, r;
      char sequence[100];
      scanf("%d", &t);
      while(t--) {
	    scanf("%d", &n);
	    j = 0;
	    l = 0;
	    for(i = 0; i < n; i++) {
		  scanf("%d", &p);
		  while(l < p) {
			sequence[j] = '(';
			l++;
			j++;
		  }
		  sequence[j] = ')';
		  j++;
	    }
	    sequence[j] = '\0';
	    /* printf("%s\n", sequence); */
	    for(i = 0; i < 20; i++) {
		  w[i] = 0;
	    }
	    i = 0;
	    j = 0;
	    while(sequence[i] != '\0') {
		  if(sequence[i] == ')') {
			if(sequence[i-1] == '(') {
			      w[j] = 1;
			      sequence[i-1] = '0';
			} else {
			      r = i;
			      while(sequence[r] != '(') {
				    if(sequence[r] == ')') {
					  w[j]++;
				    }
				    r--;
			      }
			      sequence[r] = '0';
			}
			j++;
		  }
		  i++;
	    }
	    printf("%d", w[0]);
	    for(i = 1; i < j; i++) {
		  printf(" %d", w[i]);
	    }
	    printf("\n");
      }
      return 0;
}