Parencodings（POJ_1068）

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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

代码

第一次写的时候竟然用字符数组把括号存了起来= =简直傻。其实不用非要把括号写出来才能看出来结果，直接在输入的那串数字上操作就行= =

#include <iostream>#include <cstdio>#include <cstring>using namespace std;int main(){    int t, n;    int a[22];    scanf("%d", &t);    while(t--)    {        scanf("%d", &n);        for(int i = 0; i < n; i++)        {            scanf("%d", &a[i]);        }        for(int i = n - 1; i > 0; i--)        {            a[i] -= a[i-1];        }        for(int i = 0; i < n; i++)        {            if(a[i])            {                printf("1");                a[i]--;            }            else            {                int cnt = 0;                for(int j = i; j >= 0; j--)                {                    cnt++;                    if(a[j])                    {                        a[j]--;                        break;                    }                }                printf("%d", cnt);            }            if(i == n - 1)                printf("\n");            else                printf(" ");        }    }    return 0;}

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