POJ 1068 Parencodings

標籤：poj   c++   

Parencodings

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 19409		Accepted: 11718

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

#include<iostream>#include<cstring>using namespace std;int main(){int T;cin>>T;while(T--){int n;int i,j;int a[22],b[45],c[22];memset(b,0,sizeof(b));cin>>n;int d,k=1;a[0]=0;for(i=1;i<=n;i++){cin>>a[i];d=a[i]-a[i-1];b[d+a[i-1]+k++]=1;}int ans,p=0;for(i=1;i<=2*n;i++){if(b[i]==1){b[i]=3;for(j=i-1;j>0;j--){ans=0;if(b[j]==0)  {b[j]=2;for(k=j;k<=i;k++)if(b[k]==3)  ans+=1;c[++p]=ans;break;}}}}for(i=1;i<p;i++)cout<<c[i]<<" ";cout<<c[p]<<endl;}return 0;}