poj 1068 Parencodings

Parencodings

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 22797		Accepted: 13363

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

2
6
4 5 6 6 6 6
9 
4 6 6 6 6 8 9 9 9

Sample Output

1 1 1 4 5 6
1 1 2 4 5 1 1 3 9

Source

Tehran 2001

//poj  1068
#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
#include <iostream>
using namespace std;
stack<int> s;
int main(){
    //freopen("D:\\INPUT.txt","r",stdin);
    int t,n;
    cin>>t;
    while(t--){
        while(!s.empty()){
            s.pop();
        }
        cin>>n;
        int i,j,k,num,pre=0;
        for(i=0;i<n;i++){
            if(i){
               cout<<" ";
            }
            scanf("%d",&num);
            //cout<<num<<endl;
            j=num-pre;//相邻两个右括号之间的左括号数目-1
            pre=num;//记录前一个右括号前面哪有几个左括号
            if(j==1){//多一个左括号
                cout<<1;
                if(!s.empty()){
                        s.top()++;
                }
            }
            else{
                if(!j){//相等
                    int next=++s.top();
                    cout<<s.top();
                    s.pop();
                    if(!s.empty()){
                        s.top()+=next;
                    }
                }
                else{//隔了不止一个左括号
                    cout<<1;
                    j--;
                    while(j--){
                        s.push(0);
                    }
                    s.top()++;
                }
            }
        }
        cout<<endl;
        //cout<<"aa"<<endl;
    }
    return 0;
}