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
#include<stdio.h>
#include<string.h>
int main()
{
int m,n,a[50],i,j,k,s,flag,b[50],book[50];
scanf("%d",&m);
while(m--)
{
scanf("%d",&n);
memset(book,0,sizeof(book));
j = 1,s = flag = 0;
for(i = 1; i <= n; i++)
scanf("%d",&a[i]); //输入的a[i]数值表示前面有几个左括号
a[0] = 0;
for(i = 1;i <= n; i++) //表示输入的n组数据
{
if(a[i] == a[i-1])
b[j++] = 0; //将右括号标记为0
else
{
for(k = 1; k <= a[i]-a[i-1];k++) //k表示前一个右括号和本身的距离相差几
b[j++] = 1; //将左括号标记为1
b[j++] = 0; //将自己本身标记为0
}
}
for(i = a[1] + 1; i <= 2*n;i++) //从第一个右括号开始找
{
if(b[i] == 0) //找到右括号
{
s = 1; //表示这组括号带本身总共包含几个括号
for(k = i - 1; k >= 1; k--)
{
if(!book[k] && b[k] == 1) //找到一个没有用过的左括号
{
book[k] = 1; //标记，表示此左括号以用过
if(!flag)
{
printf("%d",s); //控制输出格式
flag = 1;
}
else
printf(" %d",s);
break;
}
if(book[k]) //表示组成一个括号，经过几组成对的括号
s++;
}
}
}
printf("\n");
}
return 0;
}