POJ 1068 (13.10.11)

Time Limit: 1000MS Memory Limit: 10000K

Total Submissions: 17713 Accepted: 10660

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

题意: 给出字符串的表达式P, 得到另一种表达式W

表达式P的规则是这样的: P1, P2, Pi...Pn, Pi表示的是, 第i个右括号左边有Pi个左括号

表达式W的规则是这样的: W1, W2, Wi...Wn, Wi表示的是, 第i个右括号与其对应的左括号形成的一对括号中, 有几对括号, 包括自己本身;

做法: 由P规则得出字符串, 再由该字符串从左至右逐个找右括号, 一对一对查, 查一对, 标记成"##", 以便下次查的时候不会重复, 以及可以得知这里已有一对括号, 便于统计

AC代码:

#include  
  
int main() {  
    int t;  
    scanf("%d", &t);  
    while(t--) {  
        char str[50];  
        int ans[50];  
        int n, formal, now;  
        scanf("%d", &n);  
        n--;  
        scanf("%d", &formal);  
        int i, pos;  
        for(pos = 0; pos < formal; pos++)  
            str[pos] = '(';  
        str[pos] = ')';  
        while(n--) {  
            scanf("%d", &now);  
            int num;  
            num = now - formal + pos + 1;  
            for(i = pos+1; i < num; i++)  
                str[i] = '(';  
            str[i] = ')';  
            pos = i;  
            formal = now;  
        }  
        int count;  
        int tpos = 0;  
        for(i = 0 ; i <= pos; i++) {  
            count = 0;  
            if(str[i] == ')') {  
                for(int j = i-1; j >= 0; j--) {  
                    if(str[j] == '#')  
                        count++;  
                    if(str[j] == '(') {  
                        str[j] = '#';  
                        str[i] = '#';  
                        break;  
                    }  
                }  
                ans[tpos++] = (count / 2) + 1;  
            }  
        }  
        for(int i = 0; i < tpos; i++) {  
            if(i != tpos-1)  
                printf("%d ", ans[i]);  
            else  
                printf("%d\n", ans[i]);  
        }  
    }  
    return 0;  
}