POJ 1068 Parencodings

Parencodings

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 15654 Accepted: 9332

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

就是模拟题啦。

#include<iostream>#include<cstdio>#include<cstring>#include<string>using namespace std;int main(){    int t;    int n,a[50],v;    bool vis[50];    scanf("%d",&t);    string c;    while(t--)    {        c.clear();        a[0]=0;        scanf("%d",&n);        for(int i=1; i<=n; i++)        {            scanf("%d",&a[i]);            for(int j=0; j<a[i]-a[i-1]; j++)            {                c+='(';            }            c+=')';        }        //cout<<c<<endl;        memset(vis,0,sizeof(vis));        for(int i=0; i<=2*n; i++)        {            if(c[i]==')')            {                int sum=0;                for(int j=i-1; j>=0; j--)                {                    if(c[j]=='(')                    {                        sum++;                        if(vis[j]==0)                        {                            vis[j]=1;                            break;                        }                    }                }                cout<<sum;                if(i==(2*n-1))cout<<endl;                else cout<<" ";            }        }    }    return 0;}