ACM--Parencodings

Parencodings

Time Limit:1000MS  Memory Limit:65536K
Total Submit:176 Accepted:113

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 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 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

Asia 2001, Tehran (Iran)

题目其实很简单，只是i前面有p[i]个“）”，求i前面有几个“（）”

代码：

#include<stdio.h>
main()
{
	int t,n,i,j,k,s[40],f,c,count;
	while(scanf("%d",&t)!=EOF)
	{
		for(i=1;i<=t;i++)
		{
			scanf("%d",&n);
			int num[n+1];
			f=1;
			num[0]=0;
			for(j=1;j<=n;j++)
			{
				scanf("%d",&num[j]);
				for(k=1;k<=num[j]-num[j-1];k++)
				{
					s[f]=0;
					f++;
				}
				s[f]=1;
				f++;
			}
			f=f-1;
			for(j=1;j<=n;j++)
			{
				for(k=1;k<=f;k++)
				{
					if(s[k]==1)
					{
						s[k]=2;
						for(c=k-1;c>=1;c--)
						{
							if(s[c]==0)
							{
								s[c]=2;
								count=(k-c+1)/2;
								printf("%d ",count);
								break;
							}
						}
					}
				}
			}
			printf("\n");
		}
	}
}