poj 1068 Parencodings

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


继续熟悉下数值范围

vc++

int  4字节 

long 4字节  long long 8字节

采用位运算，注意此处使用long long 不然n>=16 只能得到32位结果

#include <iostream>
using namespace std;

int main()
{
    int a,b,c;
    cin>>a;
    for(int i=0;i<a;i++)
    {
        long long  pare=0;
        cin>>b;
        for(int i=0;i<b;i++)
        {
            cin>>c;
            pare^=long long (1)<<(c+i);
        }

        long long  l,r;
        for(int i=0;i<2*b;i++)
        {
            if((pare>>i)&1) 
            {
                l=0,r=1;
                for(int j=i-1;j>=0;j--)
                {
                    if((pare>>j)&1) r++;
                    else l++;
                    if(l==r) 
                    {
                        cout<<l<<" ";
                        break;
                    }
                }
            }
        }
        cout<<endl;
    }
    return 0;
}