poj 1068 Parencodings

Parencodings

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 24776 Accepted: 14582

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 6666W-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

264 5 6 6 6 69 4 6 6 6 6 8 9 9 9

Sample Output

1 1 1 4 5 61 1 2 4 5 1 1 3 9

开一个数组标记该位置是左括号还是右括号。扫描一遍所给序列，易得这个数组，这个题就简单了。

#include <iostream>#include <cstdio>#include <cstring>using namespace std;int main(){    int t;    scanf("%d", &t);    while(t--)    {        int n;        scanf("%d", &n);        int i, a[25];        for(i=0; i<n; i++)        {            scanf("%d", &a[i]);        }        int b[45];        int top=0;        int x=0;        for(i=0; i<n; i++)        {            int j=top;            for(; j<top+a[i]-x; j++)            {                b[j]=1;            }if(j==top+a[i]-x)            {                b[j]=0;                top=j+1;            }            x=a[i];        }        int flag=1;        for(i=0; i<top; i++)        {            int j;            if(b[i]==0)            {                int sum=0;                for(j=i; j>=0; j--)                {                    if(b[j]==-1)sum++;                    if(b[j]==1)                    {                        sum++;                        b[j]=-1;                        break;                    }                }              if(flag)  {printf("%d",sum);flag=0;}              else printf(" %d", sum);            }        }        printf("\n");    }    return 0;}