HDU 1361 Parencodings （模拟）

Parencodings

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 913    Accepted Submission(s): 541

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

264 5 6 6 6 69 4 6 6 6 6 8 9 9 9

Sample Output

1 1 1 4 5 61 1 2 4 5 1 1 3 9

Source

2001 Asia Regional Teheran

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题意：

由括号序列S可经P规则和W规则变形为P序列和W序列。
p规则是：pi是第i个右括号左边的左括号的数；
w规则是：wi是第i右括号与它匹配的左括号之间右括号的数（其中包括它本身）。
输入p，输出w。
题解：
 设1是左括号，2是右括号。模拟一下咯。

AC代码：

#include<bits/stdc++.h>using namespace std;int n;vector<int > V;int main(){    int T;    scanf("%d",&T);    while(T--)    {        scanf("%d",&n);        int k=1;        for(int i=0;i<n;i++)        {            int x;            scanf("%d",&x);            for (int j=k;j<=x;j++)            {                V.push_back(1); //左括号             }            V.push_back(2); //右括号             k=x+1;        }                vector<int > v;        int d=V.size();        for (int i=d-1;i>=0;i--)        {            if (V[i]==2) //右括号             {                int ans=0;                int cnt1=0,cnt2=0;                for (int j=i;j>=0;j--)                {                    if (V[j]==2)//右括号                     {                        cnt2++;                        ans++;                    }                    else//左括号                     {                        cnt1++;                    }                    if (cnt1==cnt2) break;                }                                v.push_back(ans);            }        }        for (int i=v.size()-1;i>=0;i--) printf("%d%c",v[i],i==0?'\n':' ');        V.clear();    }    return 0;}