poj1068 Parencodings【模拟】【刷题计划】

Parencodings

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 27375		Accepted: 16094

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

Source

题意：给你t组数据，每组数据输入一个n，再输入n个数，表示第i个右括号的左边有pi个左括号，比如题目上的 4 5 6 6 6 6，第1个右括号的左边有4个左括号，第2个右括号的左边有5个左括号,以此类推，要求输出的是，第i个右括号的左边包含多少个已经与左括号匹配好的完整括号，比如题目上的1 1 1 4 5 6 ，在第1个右括号的左边有1个左括号与之匹配，所以第1个右括号的左边有1个完整匹配的括号，后边两个1也是这样推出来的，直到第4个右括号，它和从左到右数的第3个左括号匹配，而它们中间包括了3个已经匹配好的完整括号，最后加上自身匹配好的完整括号，总共有4个已经完整匹配好的括号。

（为神魔感觉我的翻译这么辣鸡啊。。嘤嘤嘤）

思路：模拟大法好咯。看到数据量比较小，我就放心大胆的用数组模拟好了，左括号为0，右括号为1，完整括号为-1,读入的时候，模拟括号的排列顺序，最后进行统计，将统计的完整括号匹配的值存入一个新的数组，最后输出就好啦。

统计-1的总数除以2才是完整括号数。

#include<stdio.h>
#include<string.h>

int main()
{
    int str[50],num[50];
    int t,n,i,j,number,count,sum;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        memset(str,0,sizeof(str));
        for(i = 0; i < n; i ++)
        {
            scanf("%d",&number);
            str[number+i] = 1;//右括号处赋值为1 
        }
        sum = 0;//记录下标
        for(i = 0; i < 2*n; i ++)
        {
            count = 0;//统计完整括号数 
            if(str[i]== 1)//遇到右括号时 
            {
                for(j = i; j >= 0; j --)//从遇到的右括号处向左统计，共有多少个完整括号 
                {
                    if(str[j] == -1)//遇到完整括号，总数加1 
                        count ++;
                    if(str[j] == 0)//如果遇到与之匹配的左括号，结束循环 
                        break;
                }
                count /=2;//总数除以2才是完整括号数 
                str[i] = str[j] = -1;//将匹配好的左右括号标记为-1 
                num[sum++] = count +1; //最后加上自身 
            }
        }
        for(i = 0; i < n-1; i ++)
            printf("%d ",num[i]);
        printf("%d
",num[n-1]);
    }
    return 0;
}