POJ 1068：Parencodings：模拟水题

Parencodings

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 18885		Accepted: 11380

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

Source

Tehran 2001

找一下左括号和右括号数量之间的规律，很容易AC。

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
int t,n,a[25],b[25],c[25];
void input()
{
	int i,j;
	scanf("%d",&n);
	for(i=0;i<n;i++)
	{
		scanf("%d",&a[i]);
		if(i==0)
			b[i]=a[i]-1;
		else
			b[i]=a[i]-a[i-1]-1;
	}
}
int priot(int x)
{
	int i,j;
	for(i=x;i>=0;i--)
	{
		if(b[i]>0)
			break;
	}
	if(i>=0)
		return i;
	else
		return 0;
}
void chuli()
{
	int i,j;
	for(i=0;i<n;i++)
	{
		if(b[i]>=0)
			c[i]=1;
		else
		{
			j=priot(i);
			b[j]--;
			c[i]=i-j+1;
			b[i]=0;
		}
	}
}
void output()
{
	int i;
	for(i=0;i<n;i++)
	{
		if(i==0)
			printf("%d",c[i]);
		else
			printf(" %d",c[i]);
	}
	printf("\n");
}
int main()
{
	int i,j;
	while(scanf("%d",&t)!=EOF)
	{
		while(t--)
		{
			input();
			chuli();
			output();
		}
	}
	return 0;
}