テスト用に使ったコード。 REQUEST_TOKEN_URI = 'https://api.twitter.com/oauth/request_token' CALLBACK_URI = 'http://localhost/callback' CONSUMER_KEY = '**********************' CONSUMER_SECRET = '**********************' や場合によってはいろい…

検証用に使っていた物です。 len = ARGV[0].to_i B = 10 ** len B2 = B << 1 pi = (len * 8 + 1).step(3, -2).inject(B) {|a, i| (i >> 1) * (a + B2) / i} - B puts "3.#{pi}" # time ruby pi.rb 100 3.14159265358979323846264338327950288419716939937510…

ちょっと必要だったので、小数も指定基数でto_sできるようにしてみた。 ただ、指定基数によっては無限小数になる場合があるので、第２引数でto_sの最大長を指定できるようにしてみた。 class Float alias :to_s10 :to_s def to_s(base=10, max=20) raise Arg…

http://rubyist.g.hatena.ne.jp/moira/20090729/1248837435 を見て別のアプローチで再帰使わずにやってみた。 （階乗の計算では再帰使っているけどw かなり速い！！ class Numeric def fact return 1 if self.zero? return @cache until @cache.nil? @cache …

問題 The prime 41, can be written as the sum of six consecutive primes: 41 = 2 + 3 + 5 + 7 + 11 + 13 This is the longest sum of consecutive primes that adds to a prime below one-hundred. The longest sum of consecutive primes below one-thou…

問題 The arithmetic sequence, 1487, 4817, 8147, in which each of the terms increases by 3330, is unusual in two ways: (i) each of the three terms are prime, and, (ii) each of the 4-digit numbers are permutations of one another. There are n…

問題 The series, 1^1 + 2^2 + 3^3 + ... + 10^10 = 10405071317. Find the last ten digits of the series, 1^1 + 2^2 + 3^3 + ... + 1000^1000. ソース puts (1..1000).inject(0){|s, i| s + i**i} % (10 ** 10) 解答 9110846700 感想 ここまで来て１行プ…

問題 The first two consecutive numbers to have two distinct prime factors are: 14 = 2 × 7 15 = 3 × 5 The first three consecutive numbers to have three distinct prime factors are: 644 = 2^2 × 7 × 23 645 = 3 × 5 × 43 646 = 2 × 17 × 19. Find …

問題 It was proposed by Christian Goldbach that every odd composite number can be written as the sum of a prime and twice a square. 9 = 7 + 2×1^2 15 = 7 + 2×2^2 21 = 3 + 2×3^2 25 = 7 + 2×3^2 27 = 19 + 2×2^2 33 = 31 + 2×1^2 It turns out tha…

問題 Triangle, pentagonal, and hexagonal numbers are generated by the following formulae: Triangle T(n)=n(n+1)/2 1, 3, 6, 10, 15, ... Pentagonal P(n)=n(3n−1)/2 1, 5, 12, 22, 35, ... Hexagonal H(n)=n(2n−1) 1, 6, 15, 28, 45, ... It can be ve…

問題 Pentagonal numbers are generated by the formula, P_(n)=n(3n−1)/2. The first ten pentagonal numbers are: 1, 5, 12, 22, 35, 51, 70, 92, 117, 145, ... It can be seen that P(4) + P(7) = 22 + 70 = 92 = P(8). However, their difference, 70 −…

問題 The number, 1406357289, is a 0 to 9 pandigital number because it is made up of each of the digits 0 to 9 in some order, but it also has a rather interesting sub-string divisibility property. Let d(1) be the 1st digit, d(2) be the 2nd …

問題 The n'th term of the sequence of triangle numbers is given by, t(n) = 1/2 n(n+1); so the first ten triangle numbers are: 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ... By converting each letter in a word to a number corresponding to its alp…

問題 We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once. For example, 2143 is a 4-digit pandigital and is also prime. What is the largest n-digit pandigital prime that exists? ソース cla…

問題 An irrational decimal fraction is created by concatenating the positive integers: 0.123456789101112131415161718192021... It can be seen that the 12th digit of the fractional part is 1. If d(n) represents the n'th digit of the fraction…

問題 If p is the perimeter of a right angle triangle with integral length sides, {a,b,c}, there are exactly three solutions for p = 120. {20,48,52}, {24,45,51}, {30,40,50} For which value of p &#8804; 1000, is the number of solutions maxim…

問題 Take the number 192 and multiply it by each of 1, 2, and 3: 192 × 1 = 192 192 × 2 = 384 192 × 3 = 576 By concatenating each product we get the 1 to 9 pandigital, 192384576. We will call 192384576 the concatenated product of 192 and (1…

問題 The number 3797 has an interesting property. Being prime itself, it is possible to continuously remove digits from left to right, and remain prime at each stage: 3797, 797, 97, and 7. Similarly we can work from right to left: 3797, 37…

問題 The decimal number, 585 = 1001001001_(2) (binary), is palindromic in both bases. Find the sum of all numbers, less than one million, which are palindromic in base 10 and base 2. (Please note that the palindromic number, in either base…

問題 The number, 197, is called a circular prime because all rotations of the digits: 197, 971, and 719, are themselves prime. There are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, and 97. How many circular …

問題 145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145. Find the sum of all numbers which are equal to the sum of the factorial of their digits. Note: as 1! = 1 and 2! = 2 are not sums they are not included. ソース class Intege…

問題 The fraction 49/98 is a curious fraction, as an inexperienced mathematician in attempting to simplify it may incorrectly believe that 49/98 = 4/8, which is correct, is obtained by cancelling the 9s. We shall consider fractions like, 3…

問題 We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once; for example, the 5-digit number, 15234, is 1 through 5 pandigital. The product 7254 is unusual, as the identity, 39 × 186 = 7254,…

問題 In England the currency is made up of pound, £, and pence, p, and there are eight coins in general circulation: 1p, 2p, 5p, 10p, 20p, 50p, £1 (100p) and £2 (200p). It is possible to make £2 in the following way: 1×£1 + 1×50p + 2×20p +…

問題 Surprisingly there are only three numbers that can be written as the sum of fourth powers of their digits: 1634 = 1^4 + 6^4 + 3^4 + 4^4 8208 = 8^4 + 2^4 + 0^4 + 8^4 9474 = 9^4 + 4^4 + 7^4 + 4^4 As 1 = 1^4 is not a sum it is not includ…

問題 Consider all integer combinations of a^b for 2 <= a <= 5 and 2 <= b <= 5: 2^2=4, 2^3=8, 2^4=16, 2^5=32 3^2=9, 3^3=27, 3^4=81, 3^5=243 4^2=16, 4^3=64, 4^4=256, 4^5=1024 5^2=25, 5^3=125, 5^4=625, 5^5=3125 If they are then placed in nume…

問題 Starting with the number 1 and moving to the right in a clockwise direction a 5 by 5 spiral is formed as follows: [21]22 23 24[25] 20 [7] 8 [9]10 19 6 [1] 2 11 18 [5] 4 [3]12 [17]16 15 14[13] It can be verified that the sum of both di…

問題 Euler published the remarkable quadratic formula: n^2 + n + 41 It turns out that the formula will produce 40 primes for the consecutive values n = 0 to 39. However, when n = 40, 40^2 + 40 + 41 = 40(40 + 1) + 41 is divisible by 41, and…

問題 A unit fraction contains 1 in the numerator. The decimal representation of the unit fractions with denominators 2 to 10 are given: 1/2 = 0.5 1/3 = 0.(3) 1/4 = 0.25 1/5 = 0.2 1/6 = 0.1(6) 1/7 = 0.(142857) 1/8 = 0.125 1/9 = 0.(1) 1/10 =…

問題 Using names.txt (right click and 'Save Link/Target As...'), a 46K text file containing over five-thousand first names, begin by sorting it into alphabetical order. Then working out the alphabetical value for each name, multiply this v…