poj 1068 Parencodings

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最新推荐文章于 2018-04-22 15:15:24 发布

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分类专栏： ----------各大OJ---------- POJ ----------基础算法---------- 模拟题 ---Primary Algorithm(POJ)---

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Parencodings

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 24700		Accepted: 14533

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

Tehran 2001

提示

题意：

有一个括号序列它有两种表示方式：P-sequence，W-sequence。

P-sequence：序列是以从左到右，右括号的左边左括号的个数。

W-sequence：序列是以从左到右，右括号与之相匹配的左括号之间右括号个数（包括自身）。

思路：

模拟，将P-sequence序列转化为S序列（它原本的形式），再将S序列转化为W-sequence序列。

示例程序

Source Code

Problem: 1068		Code Length: 1118B
Memory: 392K		Time: 0MS
Language: GCC		Result: Accepted
#include <stdio.h>
int main()
{
    int t,i,n,x,i1,top,k,i2;
    char ch[300];
    scanf("%d",&t);
    for(i=1;t>=i;i++)
    {
        scanf("%d",&n);
        top=0;
        k=0;
        for(i1=1;n>=i1;i1++)
        {
            scanf("%d",&x);
            for(i2=x-k;i2>=1;i2--)
            {
                ch[top]='(';
                top++;
            }
            k=x;		//记录上一个左括号的个数
            ch[top]=')';
            top++;
        }
        x=0;
        for(i1=0;top>i1;i1++)
        {
            k=0;
            if(ch[i1]==')')
            {
                for(i2=i1;i2>=0&&ch[i2]!='(';i2--)
                {
                    if(ch[i2]!='0')
                    {
                        k++;
                    }
                }
                if(i2>=0)
                {
                    ch[i2]='0';		//匹配完成以后就不要了
                }
                if(x==0)
                {
                    printf("%d",k);
                    x=1;
                }
                else
                {
                    printf(" %d",k);
                }
            }
        }
        printf("\n");
    }
    return 0;
}