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 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

   题意: P表示的是当前位置前有几个左括号‘（’，当前位置有一个右括号‘）’，W表示的是当前位置下一对括号”（）“中有几对括号（包括自身），S为实际的括号情况。方法是将括号数组用‘1’‘0’表示出来1为左0为右，在输入P的时候构建括号模拟数组，构建一次就从当前位置开始向前遍历，设一个累加器x，遇到右括号累加器x+1，遇到左括号累加器x-1并且sum累加器+1表示出现一对括号，当x=0时停止遍历，将sum值存入答案数组中。最后输出答案数组。

#include <stdio.h>#include <string.h>#include <stdlib.h>#include <math.h>#include <algorithm>#include <iostream>using namespace std;int main(){    int arr[1000];    int ans[100];    int t,n,a,b,j,m;    int x,sum;    scanf("%d",&t);    while (t--)    {        int top=0;b=0;        int i=0;        scanf("%d",&n);        while (n--)        {            scanf("%d",&a);            m=a-b;            while(m--)            {                arr[i++]=1;            }            arr[i++]=0;            b=a;            x=0;sum=0;            for (j=i-1;j>=0;j--)            {                if(arr[j])                {                    x--;                    sum++;                }                else x++;                if(x==0)                {                    ans[top++]=sum;                    break;                }            }        }        for (i=0;i<top;i++)        {            printf("%d%c",ans[i],i==top-1?'\n':' ');        }    }    return 0;}