Solving with Ruby, Perl, Java, and Python AtCoder ABC 065 C factorial

Introduction

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

This theme

AtCoder beginner Contets C - Reconciled? Difficulty: 647

This theme, factorial Ruby It is the factorial of the dog N and the factorial of the monkey M. If you make the factorial a function, you can use it for later contests.

ruby.rb

n, m = gets.split.map(&:to_i)
MOD = 1_000_000_007
def nPk(n, k)
  r = 1
  while k > 0
    r *= n
    r %= MOD
    n -= 1
    k -= 1
  end
  r
end
if (n - m).abs > 1
  puts 0
elsif n == m
  puts nPk(n, n) * nPk(m, m) * 2 % MOD
else
  puts nPk(n, n) * nPk(m, m) % MOD
end

nCk.rb

def nCk(n, k)
  r, j = 1, 1
  return 0 if k > n || k < 0
  k = n - k if n - k < k
  while j <= k
    r *= n
    n -= 1
    r /= j
    j += 1
  end
  r
end

This is also reusable ~~ copy ~~. Python

python.py

n, m = map(int, input().split())
MOD = 1000000007
def nPk(n, k):
    r = 1
    while k > 0:
        r *= n
        r %= MOD
        n -= 1
        k -= 1
    return r
if abs(n - m) > 1:
    print(0)
elif n == m:
    print(nPk(n, n) * nPk(m, m) * 2 % MOD)
else:
    print(nPk(n, n) * nPk(m, m) % MOD)

Perl

perl.pl

chomp (my ($n, $m) = split / /, <STDIN>);
my $MOD = 1_000_000_007;
sub nPk {
  my ($n, $k) = @_;
  my $r = 1;
  while ($k) {
    $r *= $n;
    $r %= $MOD;
    $n -= 1;
    $k -= 1;
  }
  $r;
}
if (abs($n - $m) > 1) {
  print "0\n";
} elsif ($n == $m) {
  print (nPk($n, $n) * nPk($m, $m) * 2 % $MOD), "\n";
} else {
  print (nPk($n, $n) * nPk($m, $m) % $MOD), "\n";
}

nCk.pl

sub nCk {
  my ($n, $k) = @_;
  my ($r, $j) = (1, 1);
  return 0 if $k > $n || $k < 0;
  $k = ($n - $k) if ($n - $k) < $k;
  while ($j <= $k) {
    $r *= $n--;
    $r /= $j++;
  }
  $r;
}

Java

java.java

import java.util.*;

class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = Integer.parseInt(sc.next());
        int m = Integer.parseInt(sc.next());
        sc.close();
        int MOD = 1_000_000_007;
        if (Math.abs(n - m) > 1) {
            System.out.println(0);
        } else if (n == m) {
            System.out.println((((nPk(n, n, MOD) * nPk(m, m, MOD)) % MOD) * 2) % MOD);
        } else {
            System.out.println((nPk(n, n, MOD) * nPk(m, m, MOD)) % MOD);
        }
    }

    static long nPk(int n, int k, int MOD) {
        long r = 1;
        while (k > 0) {
            r *= n;
            r %= MOD;
            n -= 1;
            k -= 1;
        }
        return r;
    }
}

• Java * will eat integer overflow if you allow it.

Summary

• Solved ABC 065 C
• Become familiar with Ruby
• Become familiar with Python
• Become familiar with Perl
• Become familiar with Java