<poj-1068>Parencodings

╰+哭是因爲堅強的太久メ 2022-07-15 07:20 114阅读 0赞

Parencodings

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;10000K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;25202</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;14863</td> 
  </tr> 
 </tbody> 
</table>

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 9

Sample Output

1 1 1 4 5 6
    1 1 2 4 5 1 1 3 9

解题思路

其实这个题也算的上是半个水题吧, 在我们的计划安排中是用模拟法做, 然后去百度了一下不知道什么是模拟法

就用数组做了,不知道开多大的数组就开的大一点,然后竟然A了

因为我比较渣,所以写的代码都是简单易懂但是很麻烦

//左0, 右1
    #include<stdio.h>
    #include<string.h>
    
    int s[10000];
    
    int main()
    {
    	int t, n;
    	int k, r;
    	scanf("%d",&t);
    	while(t--)
    	{
    		int flag = 0;
    		k = 0;
    		int left = 0;
    		memset(s, 0, sizeof(s));
    		scanf("%d",&n);
    		for(int i = 0; i < n; i++)
    		{
    			scanf("%d",&r);
    			while(left < r)
    			{
    				s[k++] = 0;//用一个数组来存储0和1
    				left++;
    			}
    			s[k++] = 1;
    		}
    		for(int i = 0; i < k; i++)
    		{
    			int count = 0;    //记前面位置出现了几个右括号
    			if(s[i] == 1)     //出现右括号,找离他最近的左括号
    				for(int j = i; j >= 0; j--)
    				{
    					if(s[j] == 1)
    						count++;
    					else if(s[j] == 0)
    					{
    						if(flag == 0) //第一个前面不输出空格
    						{
    							printf("%d",i-j-count+1);//计算出距离,然后减去右括号
    							flag = 1;
    						}
    						else
    							printf(" %d",i-j-count+1);
    						s[j] = 2;//将已经找完的右括号随便赋了个值
    						break;
    					}
    				}
    		}
    		printf("\n");
    	}
    	return 0;
    }