poj 1068 模拟水题

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 6
1 1 2 4 5 1 1 3 9
题意：P ‘）’前‘（’ 的数目
     Q “（）”之间‘（）’的数目（算上自己）
思路：注意一下 题目上说的是n<20 其实不只 我n开到1000就过了 还有 输出最后一行有空格 垃圾题
可以吧‘（’看成1  ‘）’看成0 然后构造01字符串
计算Q的时候可以从0入手 向前读 读到0就0加一 读到1就 0减一 sum++ 0的数目为0是时 跳出循环就好
代码：
#include<iostream>#include<cstdio>#include<cstring>int main(){    int t;    int a[1000] ;    scanf("%d",&t);    while(t--)    {        int n;        scanf("%d",&n);        int dis = 0;          int dis1 = 0;        for(int i = 0; i < n; i++)//构建0 1字符串        {            int s;            scanf("%d",&s);            while(dis1 < s)            {                a[dis++] = 1;                dis1++;            }            a[dis++] = 0;        }        for(int i = 0;i < dis; i++)//计算Q        {            int sum = 0;            int dis2 = 0;            if(a[i] == 0)            {                for(int k = i; k >= 0; k--)                {                    if(a[k] == 1)                    {                        dis2--;                        sum++;                    }                    if(a[k] == 0)                        dis2++;                    if(dis2 == 0)                        break;                }            }            if(sum != 0)             {                 printf("%d ",sum);             }        }        printf("\n");    }    return 0;}