Parencodings

Time Limit: 1000 MS Memory Limit: 10000 KB

64-bit integer IO format: %I64d , %I64u Java class name: Main

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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

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

#include <iostream>#include <string.h>#include <stdio.h>#include <stack>using namespace std;int main(){    int t,n,temp;    int a[1000];    int val[1000],w[1000];    cin>>t;    while(t--)    {        memset(val,0,sizeof(val));        cin>>n;        for(int j=0; j<n; j++)        {            cin>>a[j];            temp=a[j]+j;            val[temp]=1;  ///让）==1 （==0            /*for(int i=0;i<n*2;i++) ///输出1的地方            cout<<val[i]<<‘ ‘;            cout<<endl;*/            ///cout<<val[temp]<<endl;   ==1    从后往前数）的个数            for(int i=temp-1,sum=1,falg=0; i>=0; i--) ///  val[i]==) flag标记（是否与）匹配的            {                if(falg==0&&val[i]==0)   ///eg:)                {                    w[j]=sum;                    break;                }                else if(val[i]==0)   ///eg:((())                {                    falg--;                }                else if(val[i]==1) ///eg:)))                {                    sum++;                    falg++;                }            }        }        for(int i=0; i<n; i++)        {            cout<<w[i]<<‘ ‘;        }      cout<<endl;    }    return 0;}

注释：

题意：P:从左到右遇见第i个完整括号开始（的个数 q：从左到右遇见第一个）时与该）匹配的括号内）的个数（包含该））

1：因为不用输出括号所以将（标记为0 ）标记为1

2：对于q 要计算一个大括号内的全部）所以要从后向前遍历

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