Project Euler その10
Problem 11
In the 20×20 grid below, four numbers along a diagonal line have been marked in red. 08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08 49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00 81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65 52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91 22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80 24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50 32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70 67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21 24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72 21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95 78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92 16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57 86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58 19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40 04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66 88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69 04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36 20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16 20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54 01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48 The product of these numbers is 26 × 63 × 78 × 14 = 1788696. What is the greatest product of four adjacent numbers in the same direction (up, down, left, right, or diagonally) in the 20×20 grid?
訳
下の20x20のグリッドに、対角線にそって4つの赤く印のついた数がある。
08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08
49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00
81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65
52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91
22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80
24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50
32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70
67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21
24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72
21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95
78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92
16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57
86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58
19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40
04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66
88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69
04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36
20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16
20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54
01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48
それらの数の積は26 × 63 × 78 × 14 = 1788696である。
20x20のグリッドの中で同じ方向(上下左右、そして斜め)に隣接した4つの数の最も大きい積は何か?
コード
N ="""08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08 49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00 81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65 52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91 22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80 24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50 32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70 67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21 24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72 21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95 78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92 16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57 86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58 19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40 04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66 88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69 04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36 20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16 20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54 01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48""" N = N.split("\n") N = [map(int, i.split()) for i in N] imax = [0] for i in range(20): for j in range(0,16): imax.append(N[i][j]*N[i][j+1]*N[i][j+2]*N[i][j+3]) imax = [max(imax)] for i in range(16): for j in range(0,20): imax.append(N[i][j]*N[i+1][j]*N[i+2][j]*N[i+3][j]) for j in range(0,16): imax.append(N[i][j]*N[i+1][j+1]*N[i+2][j+2]*N[i+3][j+3]) for j in range(3,20): imax.append(N[i][j]*N[i+1][j-1]*N[i+2][j-2]*N[i+3][j-3]) imax = [max(imax)] print imax
Project Euler その9
Problem11がどうにもうまくいかないので先に進む。
Problem12
The sequence of triangle numbers is generated by adding the natural numbers. So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be: 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ... Let us list the factors of the first seven triangle numbers: 1: 1 3: 1,3 6: 1,2,3,6 10: 1,2,5,10 15: 1,3,5,15 21: 1,3,7,21 28: 1,2,4,7,14,28 We can see that 28 is the first triangle number to have over five divisors. What is the value of the first triangle number to have over five hundred divisors?
訳
三角数の数列は自然数を足していくことで表現される。なので7つ目の三角数は
1 + 2 + 3 + 4 + 5 + 6 + 7 = 28 となる。
三角数の最初の10個は 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...のようになる。
最初の7つの三角数の因数をリストアップしてみよう。
1: 1
3: 1,3
6: 1,2,3,6
10: 1,2,5,10
15: 1,3,5,15
21: 1,3,7,21
28: 1,2,4,7,14,28
28は5つ以上の除数を持つ最初の三角数だとわかる。
500個以上の除数を持つ最初の三角数の値はいくつか?
コード
from math import sqrt, ceil def count_divisors(N): fact = 1 j = 0 max = ceil(sqrt(N)) if N >= 3: while N!=1 and N%2==0: j += 1 N /= 2 i = 3 fact *=(j+1) j = 0 while N!=1 and i<=max: while N!=1 and N%i==0: j += 1 N /= i fact *=(j+1) i += 2 j = 0 if N > max: fact *=2 return fact i = 0 j = 1 while 1: i += j j += 1 if count_divisors(i) >= 500: print i break
素因数分解をまじめに書いたのでけっこう速い。
Project Euler その8
Problem 10
The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17. Find the sum of all the primes below two million.
コード
i = 3 prime = [2] flag = 0 while(1): for j in prime: if i%j == 0: flag = 1 break if flag == 0: prime.append(i) flag = 0 if i >= 2000000: break i += 1 print sum(prime)
素数判定がひどく重い。
Project Euler その7
Problem 9
A Pythagorean triplet is a set of three natural numbers, a < b < c, for which, a^2 + b^2 = c^2 For example, 32 + 42 = 9 + 16 = 25 = 52. There exists exactly one Pythagorean triplet for which a + b + c = 1000. Find the product abc.
訳
「ピタゴラスの三角数は3つの自然数の組から成り、a<b<cとしたとき、a^2+b^2=c^2となる。例えば3^2+4^2=9+16=25=5^2 a+b+c=1000となるピタゴラスの三角数がたしかに存在する。abcの積を求めよ。」
コード
canditate=[] for a in xrange(1,500): for b in xrange(1,500): for c in xrange(1,500): if a+b+c == 1000: canditate.append([a,b,c]) for i in canditate: if i[0]**2+i[1]**2 == i[2]**2: print i[0]*i[1]*i[2]
遅い(確信)
Problem 10
The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17. Find the sum of all the primes below two million.
コード
i = 3 prime = [2] flag = 0 while(1): for j in prime: if i%j == 0: flag = 1 break if flag == 0: prime.append(i) flag = 0 if i >= 2000000: break i += 1 print sum(prime)
これまた遅い。高速素数判定かprimetable使いたいところ。
Project Euler その6
Problem 4
A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 × 99. Find the largest palindrome made from the product of two 3-digit numbers.
コード
def isPalindrome(N): S = str(N) L = len(S) for i in xrange(L): if S[i] != S[L-1-i]: return False return True max = 0 for i in range(100,1000): for j in range(100,1000): if isPalindrome(i*j) and max < i*j: max = i*j print max
総当たりやるしか無い・・・よね?
Project Euler その5
飛ばしてたProblem3を解いた。
Problem 3
The prime factors of 13195 are 5, 7, 13 and 29. What is the largest prime factor of the number 600851475143 ?
訳
「13195の素因数は5と7,13と29である。600851475143の素因数の内最大の要素は何か? 」
コード
import time N = 600851475143 def factor(N): for i in range(2,1000000): if N%i == 0: print i break if N/i != 1: factor(N/i) factor(N)
forじゃなくてwhileで回せば素因数がすごく大きくても対応できるかも。(遅いだろうな・・・)
Project Euler その4
Problem8
Find the greatest product of five consecutive digits in the 1000-digit number. 73167176531330624919225119674426574742355349194934 96983520312774506326239578318016984801869478851843 85861560789112949495459501737958331952853208805511 12540698747158523863050715693290963295227443043557 66896648950445244523161731856403098711121722383113 62229893423380308135336276614282806444486645238749 30358907296290491560440772390713810515859307960866 70172427121883998797908792274921901699720888093776 65727333001053367881220235421809751254540594752243 52584907711670556013604839586446706324415722155397 53697817977846174064955149290862569321978468622482 83972241375657056057490261407972968652414535100474 82166370484403199890008895243450658541227588666881 16427171479924442928230863465674813919123162824586 17866458359124566529476545682848912883142607690042 24219022671055626321111109370544217506941658960408 07198403850962455444362981230987879927244284909188 84580156166097919133875499200524063689912560717606 05886116467109405077541002256983155200055935729725 71636269561882670428252483600823257530420752963450
訳
「この1000桁の数列の中から、連続した5桁の数字の最も大きい積を見つけよ」
コード
d="7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450" m=[] for i in range(0,len(d)-4): s=int(d[i])*int(d[i+1])*int(d[i+2])*int(d[i+3])*int(d[i+4]) m.append(s) print sorted(m)[::-1][0]
見た目が酷い・・・