POJ：1086-Parencodings

Parencodings

Time Limit: 1000MS Memory Limit: 10000K

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

---

• 题意就是给你一串括号，输入是第i个右括号左边有多少个左括号，要你输出第i个右括号左边有多少对匹配好的括号（包括自身）。

• 数据量这么小，肯定是模拟啊，直接用一个标记数组标记左括号的匹配情况，每次有一个右括号就从当前的右括号开始向左找，一直找到左边第一个没有被标记的左括号，并且记录到第一个没被标记的左括号为止有多少个标记的左括号就好。然后输出就行。

---

#include<stdio.h>
#include<cstring>
using namespace std;
const int maxn = 100;
int num[maxn];
bool l[maxn];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        memset(l,0,sizeof(l));
        int n;
        scanf("%d",&n);
        for(int i=1; i<=n; i++)
            scanf("%d",&num[i]);
        for(int i=1; i<=n; i++)
        {
            int sum = 0;
            int k = num[i];
            while(k>0 && l[k])
            {
                k--;
                sum++;
            }
            if(k != 0)//左边没有了左括号，右括号只能打光棍
            {
                l[k] = true;
                printf("%d",sum+1);
            }
            else
                printf("%d",sum);
            if(i !=  n)
                printf(" ");
        }
        printf("
");
    }
    return 0;
}