AtCoder dwango Programming Contest B in Ruby, Perl and Java

Introduction

• AtCoder Problems * Use Recommendations to solve past problems. Thanks to AtCoder and AtCoder Problems.

This theme

• AtCoder dwango Programming Contest B --Nico Nico Character String * Difficulty: 832

The theme of this time is aggregation by reduce Ruby For example, the character string 1251251252525 is converted to 25-> ʻato make it easier to count characters. =>1a1a1aaa The number of consecutive ʻas is (1, 1, 3), so add up the combinations for each.

bit.rb

s = gets.chomp

s.gsub!(/25/, 'a')
p = []
c = 0
0.upto(s.size - 1) do |i|
  if s[i] == 'a'
    c += 1
  elsif c > 0
    p.push(c)
    c = 0
  end
  p.push(c) if i == s.size - 1 && c > 0
end
sum = 0
0.upto(p.size - 1) do |i|
  sum += (1..p[i]).reduce{|a, b| a + b}
end
puts sum

reduce.rb

  sum += (1..p[i]).reduce{|a, b| a + b}
  sum += (1..p[i]).reduce(&:+)
  sum += (1..p[i]).inject(:+)
  sum += (1..p[i]).sum

The reduce part can be either inject or sum (ruby 2.4 or later). ** Addition ** Added reduce (&: +) from the comments.

Perl

perl.pl

use v5.18; # strict say state
use List::Util qw/reduce/;

chomp (my $s = <STDIN>);
$s =~ s/25/a/g;
my $c = 0;
my @p;
for my $i (0..length($s)-1) {
  if (substr($s, $i, 1) eq 'a') {
    $c++;
  } elsif ($c > 0) {
    push @p, $c;
    $c = 0;
  }
  if ($i == length($s)-1 && $c > 0) {
    push @p, $c;
  }
}
my $sum = 0;
for my $i (0..$#p) {
  $sum += reduce {$a + $b} 1..$p[$i];
}
say $sum;

reduce.pl

use List::Util qw/reduce sum/;

  $sum += reduce {$a + $b} 1..$p[$i];
  $sum += sum(1..$p[$i]);

The reduce part can also be sum. Java

java.java

import java.util.*;
import java.util.stream.IntStream;

class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String s = sc.next();
        sc.close();
        s = s.replace("25", "a");
        List<Integer> p = new ArrayList<>();
        int c = 0;
        for (int i = 0; i < s.length(); i++) {
            if (s.substring(i, i + 1).equals("a")) {
                c++;
            } else if (c > 0) {
                p.add(c);
                c = 0;
            }
            if (i == s.length() - 1 && c > 0) {
                p.add(c);
            }
        }
        int sum = 0;
        for (int i = 0; i < p.size(); i++) {
            sum += IntStream.rangeClosed(1, p.get(i)).reduce(0, (a, b) -> a + b);
        }
        System.out.println(sum);
    }
}

reduce.java

            sum += IntStream.rangeClosed(1, p.get(i)).reduce(0, (a, b) -> a + b);
            sum += IntStream.rangeClosed(1, p.get(i)).sum();

The reduce part can also be sum.

Summary

• Solved dwango programming contest B
• Become familiar with Ruby
• Become familiar with Perl
• Become familiar with Java
• Became familiar with reduce

Referenced site

• The best way to sum array elements in Ruby *