feat: new dp problem formatted

This commit is contained in:
2025-12-19 07:49:39 -03:00
parent 409b0a324c
commit 1223e55c49
88 changed files with 6652 additions and 0 deletions

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# Normal directories
SRC_DIR := src
BIN_DIR := bin
DBG_DIR := bin/debug
# Grader directories
GRADER := $(wildcard $(SRC_DIR)/grader.cpp)
GRADER_DIR := $(SRC_DIR)/grader
HANDLER_DIR := $(SRC_DIR)/handler
GRADER_SRC := $(wildcard $(GRADER_DIR)/*.cpp)
GRADER_BIN := $(patsubst $(GRADER_DIR)/%.cpp, $(BIN_DIR)/%, $(GRADER_SRC))
GRADER_DBG := $(patsubst $(GRADER_DIR)/%.cpp, $(DBG_DIR)/%, $(GRADER_SRC))
# Change CPP source directories if grader is defined
ifdef GRADER
SRC := $(wildcard $(HANDLER_DIR)/*.cpp)
BIN := $(patsubst $(HANDLER_DIR)/%.cpp, $(BIN_DIR)/%, $(SRC))
DBG := $(patsubst $(HANDLER_DIR)/%.cpp, $(DBG_DIR)/%, $(SRC))
else
SRC := $(wildcard $(SRC_DIR)/*.cpp)
BIN := $(patsubst $(SRC_DIR)/%.cpp, $(BIN_DIR)/%, $(SRC))
DBG := $(patsubst $(SRC_DIR)/%.cpp, $(DBG_DIR)/%, $(SRC))
endif
SRC_C := $(wildcard $(SRC_DIR)/*.c)
BIN_C := $(patsubst $(SRC_DIR)/%.c, $(BIN_DIR)/%, $(SRC_C))
DBG_C := $(patsubst $(SRC_DIR)/%.c, $(DBG_DIR)/%, $(SRC_C))
SRC_JAVA := $(wildcard $(SRC_DIR)/*.java)
BIN_JAVA := $(patsubst $(SRC_DIR)/%.java, $(BIN_DIR)/%.class, $(SRC_JAVA))
DBG_JAVA := $(patsubst $(SRC_DIR)/%.java, $(DBG_DIR)/%.class, $(SRC_JAVA))
CHECKER := $(wildcard $(SRC_DIR)/checker.cpp)
C := gcc
CPP := g++
CXX_FLAGS := -Wall -O2
DEBUG_FLAGS := -Wall -g
BOCA_FLAGS := -static -DBOCA_SUPPORT
JV = javac
JV_DEBUG = -g
JV_DIR = -d bin
JV_DBG_DIR = -d bin/debug
.PHONY: all debug release checker clean
all: debug release checker
debug: $(DBG) $(DBG_C) $(DBG_JAVA) $(GRADER_DBG)
release: $(BIN) $(BIN_C) $(BIN_JAVA) $(GRADER_BIN)
ifdef CHECKER
checker: $(DBG_DIR)/checker-boca $(BIN_DIR)/checker-boca
endif
$(BIN): $(BIN_DIR)/% : $(SRC_DIR)/%.cpp | $(BIN_DIR)
$(CPP) $(CXX_FLAGS) $^ -o $@
$(DBG): $(DBG_DIR)/% : $(SRC_DIR)/%.cpp | $(DBG_DIR)
$(CPP) $(DEBUG_FLAGS) $^ -o $@
$(BIN_C): $(BIN_DIR)/% : $(SRC_DIR)/%.c | $(BIN_DIR)
$(C) $(CXX_FLAGS) $^ -o $@
$(DBG_C): $(DBG_DIR)/% : $(SRC_DIR)/%.c | $(DBG_DIR)
$(C) $(DEBUG_FLAGS) $^ -o $@
$(BIN_JAVA): $(BIN_DIR)/%.class : $(SRC_DIR)/%.java | $(BIN_DIR)
$(JV) $(JV_DIR) $^
$(DBG_JAVA): $(DBG_DIR)/%.class : $(SRC_DIR)/%.java | $(DBG_DIR)
$(JV) $(JV_DEBUG) $(JV_DBG_DIR) $^
$(GRADER_BIN): $(BIN_DIR)/% : $(GRADER_DIR)/%.cpp $(GRADER) $(GRADER_DIR)/*.h
$(CPP) $(CXX_FLAGS) $^ -o $@
$(GRADER_DBG): $(DBG_DIR)/% : $(GRADER_DIR)/%.cpp $(GRADER) $(GRADER_DIR)/*.h
$(CPP) $(DEBUG_FLAGS) $^ -o $@
$(BIN_DIR):
mkdir -p $@
$(DBG_DIR):
mkdir -p $@
$(BIN_DIR)/checker-boca: $(SRC_DIR)/checker.cpp
$(CPP) $(CXX_FLAGS) $(BOCA_FLAGS) $^ -o $@
$(DBG_DIR)/checker-boca: $(SRC_DIR)/checker.cpp
$(CPP) $(DEBUG_FLAGS) $(BOCA_FLAGS) $^ -o $@
clean:
@echo Cleaning problem files
rm -rf bin

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\documentclass{maratona}
\begin{document}
\begin{ProblemaAutor}{}{Maior Subsequência Palindrômica}{1}{256}{}
O problema consiste em determinar a maior subsequência palindrômica de uma sequência de caracteres.
Uma subsequência é formada ao remover zero ou mais elementos da sequência original, sem alterar a ordem relativa dos elementos restantes.
A subsequência procurada deve ser um palíndromo, ou seja, deve ser idêntica quando lida de frente para trás ou de trás para frente.
O objetivo é identificar essa subsequência de tamanho máximo e apresentar tanto o seu comprimento quanto os próprios elementos.
\Entrada
A entrada é composta por duas linhas.
A primeira linha contém um inteiro \( n \) (\( 1 \leq n \leq 1000 \)), representando o número de caracteres da sequência.
A segunda linha contém uma string \( s \) de comprimento \( n \), composta por letras minúsculas do alfabeto.
\Saida
A saída é composta por uma linha, um único inteiro, que representa representando o tamanho \( L \) da maior subsequência palindrômica.
\ExemploEntrada
\begin{Exemplo}
\texttt{5} & \texttt{4}\\
\texttt{bbbab} & \\
\rowcolor{gray!20}\texttt{4} & \texttt{2}\\
\rowcolor{gray!20}\texttt{cbbd} & \\
\texttt{1} & \texttt{1}\\
\texttt{a} & \\
\end{Exemplo}
\Notas
Para a sequência "bbbab", o tamanho da maior subsequência palindrômica é \( L = 4 \) e uma possível sequência é "bbbb".
Para a sequência "cbbd", a maior subsequência palindrômica é "bb", com tamanho \( L = 2 \).
Para a sequência "a", há apenas um elemento, então a maior subsequência palindrômica é o próprio caractere "a", com tamanho \( L = 1 \).\end{ProblemaAutor}
\end{document}

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\ProvidesPackage{maratona}
\LoadClass[11pt]{article}
% remove page numbers
\pagenumbering{gobble}
\RequirePackage{fancyhdr}
\RequirePackage{tabularx,colortbl}
%\RequirePackage{arial}
\RequirePackage{ifpdf}
\RequirePackage[T1]{fontenc}
\RequirePackage[utf8]{inputenc}
\RequirePackage[portuguese]{babel}
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\RequirePackage{amssymb,amsmath,wrapfig}
\RequirePackage{xcolor,colortbl}
\RequirePackage{xcolor}
\RequirePackage{ifthen}
\oddsidemargin 0cm
\evensidemargin -2cm
\topmargin -1cm
\textwidth 16cm
\textheight 23cm
\ifpdf
\RequirePackage[pdftex]{hyperref}
\else
\RequirePackage[hypertex]{hyperref}
\fi
\newcommand{\var}[1]{\ensuremath{{#1}}}
\hypersetup{
letterpaper,
colorlinks=true,
linkcolor=blue,
urlcolor=blue,
pdfpagemode=none,
pdftitle={IV Maratona de Programação do IFB \today},
pdfauthor={},
pdfsubject={Caderno de problemas da IV Maratona de Programação do IFB },
pdfkeywords={maratona, programação, IFB}
}
\DeclareGraphicsExtensions{png}
\lhead{DS Contest Tools}
\pagestyle{fancy}
% Capa
\newenvironment{Maratona}[3]
{
\begin{titlepage}
\begin{center}
\vspace{1cm}
\Large{\textbf{#1}} \\
\vspace{1cm}
{\textbf{Caderno de Problemas}} \\
\vspace{1cm}
\begin{small}
\textsl{#2}
\end{small} \\
\begin{figure}[htp]
\begin{center}
\includegraphics[scale=1]{logos/logo-maratona.png}
\end{center}
\end{figure}
{(Este caderno contém {#3} problemas)} \\
\vspace{1cm}
}
{
\vfill
\begin{small}
{QNM 40, Área Especial nº 01,
Taguatinga/DF, 72146-000 ,
Brasil } \\
{Telefone (61) 2103-2200 \\http://www.ifb.edu.br/taguatinga} \\
\end{small}
\end{center}
\end{titlepage}
}
\newcommand{\Organizacao}[2]{
{\small \vfill
\begin{center}
\textbf{Comissão Organizadora:} \\
{#1} \\
\bigskip
\textbf{Apoio:}\\
{#2}
\end{center}
}
\vfill
}
% Problema
\newcounter{problem}
\newenvironment{Problema}[4]{
\stepcounter{problem}
\newpage
\begin{center}
\Large{\ifthenelse{\equal{#1}{}}{\textbf{{#2}}}{\textbf{Problema {#1} -- {#2} }}}{\\\footnotesize \textbf{Limite de tempo: {#3}s}}{\\[-0.1cm]\footnotesize \textbf{Limite de memória: {#4}MB}}
\end{center}
}
\newcounter{problemAutor}
\newenvironment{ProblemaAutor}[5]{
\stepcounter{problemAutor}
\newpage
\begin{center}
\Large{\ifthenelse{\equal{#1}{}}{\textbf{{#2}}}{\textbf{Problema {#1} -- {#2} }}}{\\\footnotesize \textbf{Limite de tempo: {#3}s}}{\\[-0.1cm]\footnotesize \textbf{Limite de memória: {#4}MB\\}}{
\footnotesize Autor: {#5}
}
\end{center}
}
% Código-fonte
\newcommand{\codigofonte}[1]{Nome do arquivo fonte: {#1}}
% Entrada
\newcommand{\Entrada}{
\bigskip
\begin{large}
\textbf{Entrada} \\
\end{large}
}
% Saida
\newcommand{\Saida}{
\bigskip
\begin{large}
\textbf{Saída} \\
\end{large}
}
\newcommand{\Interacao}{
\bigskip
\begin{large}
\textbf{Interação} \\
\end{large}
}
\newcommand{\Notas}{
\bigskip
\begin{large}
\textbf{Notas} \\
\end{large}
}
% Exemplo
\newenvironment{Exemplo}
{
\tabularx{\textwidth}{XX}
% {@{\extracolsep{\fill}}|l|l|}
% {|l|l@{\extracolsep{\fill}|}}
\hline
Entrada & Saída \\\hline
}
{
\hline
\endtabularx
}
% Exemplo de Entrada
\newenvironment{ExemploEntrada}
{
\bigskip
\begin{large}
\textbf{Exemplo} \\
\end{large}
}
{
}
% Sample Output

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4

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3

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8

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4

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1

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84

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16

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18

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9

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68

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286

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2

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71

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199

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117

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223

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238

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35

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93

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299

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296

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216

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1

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195

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368

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131

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231

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1000

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327

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1

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4

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1

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8

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1

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8

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{
"version": "1.0",
"problem": {
"title": "Maior Subsequência Palindrômica",
"event": "",
"time_limit": 1.0,
"memory_limit_mb": 256,
"input_file": "stdin",
"output_file": "stdout",
"interactive": false,
"grader": false,
"subject": {
"en_us": [
"dynamic-programming", "DP", "longest-palindromic-subsequence", "LPS"
],
"pt_br": [
"programação-dinâmica", "maior-subsequência-palindrômica"
],
"es": [
""
]
}
},
"author": {
"name": "",
"affiliation": "",
"country": "",
"email": ""
},
"build": {
"run_generator": true,
"run_validator": true,
"produce_outputs": true,
"run_checker": true,
"run_all_solutions": true,
"run_specific_solution": "",
"generate_io_only": false,
"generate_pdf_only": false,
"cpu_count": 1,
"build_pdf": true,
"pdf_format": "ds",
"io_samples": 3
},
"solutions": {
"main-ac": "ac.cpp",
"alternative-ac": [],
"wrong-answer": [],
"time-limit": ["TLE.cpp"],
"time-limit-or-ac": [],
"time-limit-or-memory-limit": [],
"memory-limit": [],
"presentation-error": [],
"runtime-error": []
},
"polygon_config": {
"id": ""
},
"boca_config": {
"time_limit": 1,
"number_of_repetitions": 1,
"maximum_memory_mb": 512,
"maximum_output_size_kb": 24096
}
}

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#include <bits/stdc++.h>
using namespace std;
int solve(const string& s) {
if (s.length() == 0) return 0;
if (s.length() == 1) return 1;
if (s.front() == s.back()) {
return 2 + solve(s.substr(1, s.length() - 2));
} else {
return max(solve(s.substr(1)), solve(s.substr(0, s.length() - 1)));
}
}
int main() {
int n; cin >> n;
string s; cin >> s;
cout << solve(s) << endl;
return 0;
}

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#include <bits/stdc++.h>
using namespace std;
int main()
{
int n; cin >> n;
string s; cin >> s;
/*
dp[i][j] says if the substring from i to j is a palindrome
*/
vector<vector<int>> dp(n, vector<int>(n, 0));
for (int i = n - 1; i >= 0; i--)
{
dp[i][i] = 1;
for (int j = i + 1; j < n; j++)
{
if (s[i] == s[j])
{
dp[i][j] = dp[i + 1][j - 1] + 2;
}
else
{
dp[i][j] = max(dp[i + 1][j], dp[i][j - 1]);
}
}
}
cout << dp[0][n - 1] << endl;
return 0;
}

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#include "testlib.h"
#include <bits/stdc++.h>
using namespace std;
int main(int argc, char* argv[]) {
setName("compare two signed int%d's", 8 * int(sizeof(int)));
registerTestlibCmd(argc, argv);
int ja = ans.readInt();
int pa = ouf.readInt();
if (ja != pa)
quitf(_wa, "expected %d, found %d", ja, pa);
quitf(_ok, "all answers are correct");
}

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#include "testlib.h"
#include <bits/stdc++.h>
using namespace std;
const int MIN_N = 1;
const int MAX_N = 1000;
const int rnd_test_n = 30;
template <typename T> void append(vector<T> &dest, const vector<T> &orig) {
dest.insert(dest.end(), orig.begin(), orig.end());
}
string output_tc(const string &s) {
ostringstream oss;
oss << s.size() << endl;
oss << s << endl;
return oss.str();
}
vector<string> generate_sample_tests() {
vector<string> tests;
tests.push_back(output_tc("bbbab"));
tests.push_back(output_tc("cbbd"));
tests.push_back(output_tc("a"));
return tests;
}
vector<string> generate_manual_tests() {
vector<string> tests;
return tests;
}
string generate_structured_string(int n) {
string s = "";
if (rnd.next(0, 1) == 1) {
string half = rnd.next("[a-z]{" + to_string((n + 1) / 2) + "}");
s = half;
reverse(half.begin(), half.end());
s += half;
for (int i = 0; i < n / 10; i++) {
s[rnd.next(0, n - 1)] = rnd.next('a', 'z');
}
}
else {
s = rnd.next("[a-z]{" + to_string(n) + "}");
}
return s;
}
string rnd_test(int i) {
int min_n = MIN_N;
int max_n = MAX_N;
if (i < rnd_test_n / 3) {
max_n = 10;
} else if (i < rnd_test_n / 2) {
max_n = 100;
}
int n = rnd.next(min_n, max_n);
return output_tc(generate_structured_string(n));
}
vector<string> generate_random_tests() {
vector<string> tests;
for (int i = 0; i < rnd_test_n; i++) {
tests.push_back(rnd_test(i));
}
return tests;
}
vector<string> generate_extreme_tests() {
vector<string> tests;
tests.push_back(output_tc(string(1000, 'a')));
tests.push_back(output_tc(rnd.next("[a-z]{1000}")));
return tests;
}
int main(int argc, char *argv[]) {
registerGen(argc, argv, 1);
vector<string> tests;
size_t test = 0;
append(tests, generate_sample_tests());
append(tests, generate_manual_tests());
append(tests, generate_random_tests());
append(tests, generate_extreme_tests());
for (const auto &t : tests) {
startTest(++test);
cout << t;
}
return 0;
}

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generator

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#include "testlib.h"
#include <bits/stdc++.h>
using namespace std;
int main(int argc, char* argv[]) {
registerValidation(argc, argv);
int n = inf.readInt(1, 1000);
inf.readEoln();
string s = inf.readToken("[a-z]+");
ensuref(n == (int)s.size(), "string provided %s does not have %i characters", s, n);
inf.readEoln();
inf.readEof();
return 0;
}

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O problema consiste em determinar a maior subsequência palindrômica de uma sequência de caracteres.
Uma subsequência é formada ao remover zero ou mais elementos da sequência original, sem alterar a ordem relativa dos elementos restantes.
A subsequência procurada deve ser um palíndromo, ou seja, deve ser idêntica quando lida de frente para trás ou de trás para frente.
O objetivo é identificar essa subsequência de tamanho máximo e apresentar tanto o seu comprimento quanto os próprios elementos.

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A entrada é composta por duas linhas.
A primeira linha contém um inteiro \( n \) (\( 1 \leq n \leq 1000 \)), representando o número de caracteres da sequência.
A segunda linha contém uma string \( s \) de comprimento \( n \), composta por letras minúsculas do alfabeto.

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Para a sequência "bbbab", o tamanho da maior subsequência palindrômica é \( L = 4 \) e uma possível sequência é "bbbb".
Para a sequência "cbbd", a maior subsequência palindrômica é "bb", com tamanho \( L = 2 \).
Para a sequência "a", há apenas um elemento, então a maior subsequência palindrômica é o próprio caractere "a", com tamanho \( L = 1 \).

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A saída é composta por uma linha, um único inteiro, que representa representando o tamanho \( L \) da maior subsequência palindrômica.