National Contest for Private Universities (NCPU), 2019 F. Sequence Decoding

编辑代码

• 1000ms
• 262144K

Sequence Decoding

The amino acids in proteins are classified into two types of elements, hydrophobic (nonpolar) and hydrophilic (polar). Hydrophobic and hydrophilic are denoted by H and P respectively. A protein is represented by a sequence of H and P such as PPHHHPHPH. In order to reduce the representation length of a sequence, we use the notation $k[S]$ to denote the repeated sequence of $k$ times sequence $S$, where $2\le k\le 9$. A legal sequence is defined as following.

A sequence consists of only one character 'H' or 'P' is a legal sequence.

Let S1S_1S1​ and S2S_2S2​ be legal sequences. Then the sequence concatenated by S1S_1S1​ and S2S_2S2​ is also a legalsequence.

Let $S$ be a legal sequence. Then the sequence $k[S]$ is also a legal sequence, where $2\le k\le 9$. 

For example, PHPHPHPH is encoded as 4[PH]. Note that a repeated sequence may contain srepeated sequences recursively such as 2[PH4[P]4[H]].

Given a nonempty encoded protein sequence $S$, your job is to expand $S$ to its original sequence.That is, you should expand 2[PH4[P]4[H]] to PHPPPPHHHHPHPPPPHHHH.

Input Format

The first line is an integer $n$ indicating the number of test cases. Each of the next $n$ lines consistsof a legal sequence composed by number digits '2'~'9', '[', ']', 'P' and 'H'.

Output Format

For each test case, output the expanding sequence in one line.

Technical Specification

$1\le n\le 10$.

All the inputs are legal.

The length of each input sequence is less than $50$.

The length of each expanded sequence is less than $1000$.

样例输入 复制

3
PHPHP
2[3[P]H2[P]]
HH2[P3[H]]P

样例输出 复制

PHPHP
PPPHPPPPPHPP
HHPHHHPHHHP

题意:给你一段字符串只含 【，】，H，P，和数字，要把这些数字和中括号里的字母展开复制多少遍根据中括号外面的数字决定。这个深搜搜的我有点那啥，这个深搜的形式和以前的有些不同这个是更新的位置，一开始正为这个如何更新位置发愁对我来说也是一种新的深搜形式吧。

#include 
using namespace std;
char sv[10002];
int dfs(int w){
    int k=0;


    while(1){
    int p;
    k=0;
    while(sv[w]>='0'&&sv[w]<='9')
    {
        k=k*10+sv[w]-'0';
        w++;
    }
    k=k==0?1:k;
    if(sv[w]=='[')
    {
        for(int u=1;u<=k;u++)
            p=dfs(w+1);
        w=p;
    }
    else if(sv[w]=='P'||sv[w]=='H')
    {
        cout<>n;
    while(n--)
    {
        scanf("%s",sv);
        dfs(0);
        cout<