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B - Parencodings

Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 1068

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

#include<stdio.h>
int main()
{
    int t,n,i,j,m,k,h,f;
    int a[50];
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&f);
        int k=0;
        j=0;
        n=f;
        while(f--)
        {
            scanf("%d",&h);
            m=h-j;
            j=h;
            for(i=k;i<(k+m);i++)
                a[i]=1;
            a[i]=0;
            k=i+1;
        }
        for(i=0;i<2*n-1;i++)
        {
            if(a[i]==0){k=0;
                for(j=i;a[j]!=1;j--)
                {
                    if(a[j]==2)k++;
                }
                k++;
                a[j]=2;
                printf("%d ",k);
            }
        }
        k=0;
        for(j=2*n-1;a[j]!=1;j--)
            if(a[j]==2)k++;
        k++;
        a[j]=2;
        printf("%d\n",k);
    }
    return 0;
}