<poj-1068>Parencodings

Parencodings

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 25202		Accepted: 14863

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;
}