【模拟】POJ-1068-Parencodings_poj parencodings-优快云博客

本文介绍了一种将括号序列的P-sequence转换为W-sequence的方法，并提供了一个具体的实现示例。通过输入由左括号数量定义的P-sequence，程序能够计算并输出相应的W-sequence，帮助理解括号配对的另一种表达方式。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Parencodings

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 27585		Accepted: 16217

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

#include <iostream>
#include<stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int a[20],w[20];
int vis[40];

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
         string s;
        memset(w,0,sizeof(w));

        int n;
        scanf("%d",&n);
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&a[i]);
        }
        a[0]=0;
        for(int i=1; i<=n; i++)
        {
            for(int j=1; j<=(a[i]-a[i-1]); j++)
            {
                s+='(';
            }
            s+=')';

        }

        int k=1;
        memset(vis,0,sizeof(vis));
        for(int i=0; i<2*n; i++)
        {
            int ans=1;
            if(s[i]==')')
            {
                for(int j=i-1; j>=0; j--)
                {
                    if(s[j]==')')ans++;
                    if(s[j]=='('&&!vis[j])
                    {
                        vis[j]=1;
                        break;
                    }
                }
                w[k++]=ans;
            }
        }
        for(int i=1; i<=n; i++)
        {
            printf("%d",w[i]);
            if(i!=n)printf(" ");
        }
        printf("\n");

    }
}