# POJ1068 解法

Parencodings
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 27194 Accepted: 15982

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

``````#include <stdio.h>

int main(void)
{
int num;
int z;
scanf("%d",&z);
while(z--)
{
int data;
scanf("%d",&num);
int C[2*num];
int A[num];
int k=0;
for(int i=0;i<num;i++)
{
scanf("%d",&A[i]);
if(i==0)
data=A[i];
else
data=A[i]-A[i-1];
while(data>0)
{
C[k]=0;
k++;
data--;
}
C[k]=1;
k++;
}
int j;
for(int i=0;i<2*num;i++)
{
if(C[i]==1)
{
C[i]=-1;   //计算过的替换为-1
for(j=i-1;C[j]!=0;j--)
;
C[j]=-1;
int sum=(i-j+1)/2;
printf("%d ",sum);
}
}
printf("\n");
}

return 0;
}``````
Posted in ACM