poj 1068 Parencodings

Parencodings

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 24285 Accepted: 14257

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

Source

Tehran 2001

就是给出P序列，求出W序列；

只要根据P序列求出原始的括号序列，再统计右括号数，就能得到答案；

#include<iostream>#include<algorithm>#include<cmath>#include<string>#include<cstring>using namespace std;int p[30];int w[30];int str[50];int main(){    ios_base::sync_with_stdio(false);    cin.tie(0);    int t;    cin>>t;    while(t--)    {        int n;        cin>>n;        memset(str,0,sizeof(str));        for(int i=1;i<=n;++i)        {            cin>>p[i];        }        int cnt=0;        int pos=0;        for(int i=1;i<=n;++i)        {            while(cnt<p[i])            {                str[pos++]=0;                ++cnt;            }            str[pos++]=1;        }        for(int i=0;i<pos;++i)        {            if(str[i]==1)            {                cnt=2;                for(int j=i-1;j>=0;j--)                {                    if(str[j]==0)                    {                        str[i]=str[j]=-1;                        break;                    }                    else                        ++cnt;                }                cout<<cnt/2<<" ";            }        }        cout<<endl;    }    return 0;}