POJ 1068 - Parencodings

Parencodings

Time Limit: 1000MS Memory Limit: 10000K

Total Submissions: 24828 Accepted: 14627

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

给出一行括号，可以用数字量化，即从左到右遍历中，右括号左边有多少个左括号。

题目要求根据这一行括号，分别输出各个括号里面包含的左括号（包括其自身括号）

解题思路:

先是把这些存到char数组里，然后从左到右遍历，如果遇到右括号，那就从这个右括号向左开始遍历，遇到的第一个左括号停止，期间把走过的路径都标志掉，记录下走的步数，然后（step+1）2，就是左括号的个数.

AC代码

#include<stdio.h>
#include<string.h>
int main()
{
char s[100];
int n[100];
int r[100];
int loop;
scanf("%d",&loop);
while(loop--)
{
int sum;
int i;
scanf("%d",&sum);
for(i = 1;i <= sum;i++)
scanf("%d",&n[i]);
n[0] = 0;
int j;
int s_index = 0;
for(i = 1;i <= sum;i++)
{
for(j = 1;j <= n[i]-n[i-1];j++)
s[s_index++] = '(';
s[s_index++] = ')';
}
s[s_index] = '\0';
int r_index = 0;
int step = 0;
for(i = 0;i <= s_index-1;i++)
{
step = 0;
if(s[i] == ')')
{
j = i;
while(j>=0)
{
if(s[j] == '(')
{
s[j] = 0;
break;
}
s[j] = 0;
j--;
step++;
}
r[r_index++] = (step+1)/2;
}
}
for(i = 0;i < r_index;i++)
{
if(!i)
printf("%d",r[i]);
else
printf(" %d",r[i]);
}
printf("\n");
}
return 0;
}