Find the first prime number of 10 consecutive digits of e (base of natural logarithm) with ruby. However, e is up to 200 digits
2.7182818284590452353602874713526624977572470936999595749669676277240766303535475945713821785251664274274663919320030599218174135966290435729003342952605956307381323286279434907632338298807531952510190
Copy this into text and read it.
By the way, this is the issue here (https://qiita.com/daddygongon/items/ba94b0f2a73990fc6a07).
For the time being, I will paste the code I wrote this time before the explanation.
def prime?(num)
warden = 0
i_max = Integer.sqrt(num)
[*2..i_max].each do |i|
warden = i
break if num % i == 0
end
warden == i_max #When I called rubocop, it became like this.It seems that you can bother to write return
end
def read_exp
exp1 = gets.to_s.chomp
end
def google_recruit
# e (The bottom of the natural logarithm)Read
exp1 = read_exp
first = 2
last = exp1.size - 1 - 10
[*first..last].each do |i|
#In the value of e,Generate consecutive 10-digit integers
target = exp1[i..i + 10 - 1].to_i
#Primality test
if prime?(target)
puts target
break #Break when you find the first consecutive 10-digit prime number
end
end
end
google_recruit if $PROGRAM_NAME == __FILE__
The explanation is on the link I posted earlier, so I will omit it. (I wrote the code according to the explanation, so I have nothing to say)
By the way,
warden == i_max
Originally, that part
if warden == i_max
return true
else
return false
end
Was written.
When I ran this on rubocop, I got warden == i_max.
Congratulations. You've made it to level 2. Go to www.Linux.org and enter Bobsyouruncle as the login and the answer to this equation as the password.
f(1)=7182818284
f(2)=8182845904
f(3)=8747135266
f(4)=7427466391
f(5)=__________
This problem is also an issue here (https://qiita.com/daddygongon/items/ba94b0f2a73990fc6a07).
The problem is to find a number of 10 consecutive digits of e (base of natural logarithm) with a sum of the numbers of each digit being 49. (Here (<https://q.hatena.ne) .jp/1169050304>) Seen.)
So, you can modify the code you wrote earlier. The modified code looks like this.
def sum_is49?(nums)
sum = 0
[*0..nums.size - 1].each do |k|
sum += nums[k]
end
sum == 49
end
def read_exp
exp1 = gets.to_s.chomp
end
######
# e (The bottom of the natural logarithm)Of 10 consecutive digits in the value of,Find the sum of the numbers in each digit is 49
#The reason why the sum wants 49,Found at the following URL
# https://q.hatena.ne.jp/1169050304
######
def google_recruit2
# e (The bottom of the natural logarithm)Read
exp1 = read_exp
first = 2
last = exp1.size - 1 - 10
[*first..last].each do |i|
nums = []
#In the value of e,Generate consecutive 10-digit integer group
[*i..i + 10 - 1].each do |j|
nums.push(exp1[j].to_i)
end
#Determine if the sum is 49
print "#{nums.join}\n" if sum_is49?(nums)
end
end
google_recruit2 if $PROGRAM_NAME == __FILE__
The corrected part is roughly like this.
--The number of consecutive 10 digits is made into an array with the numerical value of each digit in each element.
--Instead of the function prime?, We created and used a new function sum_is49? That determines whether the sum of each element of the array is 49.
I don't write any more commentary.
By the way, nums.join returns a string that joins each element of the array nums. For details, see Here in the reference manual.
Refactored as pointed out by @scivola. The code after refactoring looks like this.
# e (The bottom of the natural logarithm)Of 10 consecutive digits in the value of,Find the sum of the numbers in each digit is 49
def google_recruit2
# e (The bottom of the natural logarithm)Read
exp1 = gets.to_s.chomp
#Generate an array of integers for each digit of e
char_nums = exp1.chars
integer_nums = char_nums.map { |n| n.to_i }
#Generate consecutive 10-digit integer groups in the value of e.And judge whether the sum is 49,output.
integer_nums.each_cons(10) do |nums|
puts nums.join if nums.sum == 49
end
end
google_recruit2 if $PROGRAM_NAME == __FILE__
Very beautiful. Good feeling.