Hdu1361&&Poj1068 Parencodings 2017-01-18 17:17 45人阅读 评论(0) 收藏

Parencodings

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 5   Accepted Submission(s) : 5

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Problem Description

Let S = s1 s2 … s2n be a well-formed string of parentheses. S can be encoded in two different ways:

• 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).
• 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 file 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

2001 Asia Regional Teheran

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题意：

有两个规则：

1、p序列表示一个括号字符串的每一个右括号的前面有多少个左括号；

2、w序列表示一个括号字符串的每一个右括号在和自己匹配的左括号之间（包括自己这一对）有多少了完整的括号；

现在给出p序列求w序列；

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
#include<cmath>
using namespace std;
int P[50];
int W[50];
int a[50];
int main()
{
    int T;
    int n;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
           scanf("%d",&P[i]);
        a[0]=P[1];
        for(int i=1;i<n;i++)
            a[i]=P[i+1]-P[i];
        int j;
        for(int i=1;i<=n;i++)
        {
            for(j=i-1;j>=0;j--)
            {
                if(a[j]>0)
                {
                    a[j]--;
                    break;
                }
            }
            W[i]=i-j;
        }
        for(int i=1;i<n;i++)
            printf("%d ",W[i]);
        printf("%d\n",W[n]);
    }
    return 0;
}