Alex Li arekusuri
arekusuri
/ best_route.py
Created
August 4, 2014 03:45
Find the maximum total from top to bottom of the triangle below:
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| mport numpy as np | |
| data = """ | |
| 75 | |
| 95 64 | |
| 17 47 82 | |
| 18 35 87 10 | |
| 20 04 82 47 65 | |
| 19 01 23 75 03 34 | |
| 88 02 77 73 07 63 67 | |
| 99 65 04 28 06 16 70 92 |
arekusuri
/ letter_counts.py
Created
August 3, 2014 23:12
If the numbers 1 to 5 are written out in words: one, two, three, four, five, then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total. If all the numbers from 1 to 1000 (one thousand) inclusive were written out in words, how many letters would be used? NOTE: Do not count spaces or hyphens. For example, 342 (three hundred and forty-two) contai…
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| info_less_than20 = { 0:"", 1:"one", 2:"two", 3:"three", 4:"four", 5:"five", | |
| 6:"six", 7:"seven", 8:"eight", 9:"nine", 10:"ten", | |
| 11:"eleven", 12:"twelve", 13:"thirteen", 14:"forteen", 15:"fifteen", | |
| 16:"sixteen", 17:"seventeen", 18:"eighteen", 19:"nineteen", | |
| } | |
| info_biger_than_twenty = { 20:"twenty", 30:"thirty", 40:"forty", 50:"fifty", | |
| 60:"sixty", 70:"seventy", 80:"eighty", 90:"ninety", | |
| } | |
| arr = [list(str(x)) for x in range(1000)] | |
| arr = [x[:-2]+["".join(x[-2:])] for x in arr] |
arekusuri
/ big_number.py
Created
August 3, 2014 20:19
2^15 = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26. What is the sum of the digits of the number 2^1000?
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| def multiply2(arr): | |
| arr = [x*2 for x in arr] | |
| arr.append(0) | |
| for i in range(len(arr)-1): | |
| arr[i+1] += arr[i] / 10 | |
| arr[i] = arr[i] % 10 | |
| if arr[-1] == 0: | |
| arr = arr[:-1] | |
| return arr |
arekusuri
/ find_route.py
Created
August 3, 2014 19:37
How many such routes are there through a 20×20 grid?
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| def f(n): | |
| arr = [1] | |
| for i in range(n): | |
| a = [1] * (len(arr) + 1) | |
| for j in range(1, len(arr)): | |
| a[j] = arr[j] + arr[j-1] | |
| arr = a | |
| return sum([x*x for x in arr]) | |
| print f(20) |
arekusuri
/ longest_chain.py
Created
August 3, 2014 18:21
Which starting number, under one million, produces the longest chain?
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| under = 1000000 + 1 | |
| info = [0] * (3 * under) | |
| def func(n): | |
| if n == 1: | |
| return 1 | |
| if n < len(info): | |
| if info[n]: | |
| return info[n] | |
| t = n/2 if n%2 == 0 else n*3 +1 | |
| info[n] = 1 + func(t) |
arekusuri
/ sum_100_row.py
Last active
August 29, 2015 14:04
Work out the first ten digits of the sum of the following one-hundred 50-digit numbers.
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| import numpy as np | |
| data = """ | |
| 37107287533902102798797998220837590246510135740250 | |
| 46376937677490009712648124896970078050417018260538 | |
| 74324986199524741059474233309513058123726617309629 | |
| 91942213363574161572522430563301811072406154908250 | |
| 23067588207539346171171980310421047513778063246676 | |
| 89261670696623633820136378418383684178734361726757 | |
| 28112879812849979408065481931592621691275889832738 | |
| 44274228917432520321923589422876796487670272189318 |
arekusuri
/ triangle_500.py
Created
August 3, 2014 03:50
What is the value of the first triangle number to have over five hundred divisors?
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| import math | |
| def func(target=500): | |
| n = 1 | |
| while True: | |
| n = n+1 | |
| total = n * (n+1)/2 | |
| count = 2 | |
| for i in range(2, int(math.sqrt(total))): | |
| if total%i == 0: |
arekusuri
/ max_grid_productor.py
Created
August 3, 2014 00:11
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 8…
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| import numpy as np | |
| from itertools import chain | |
| data = """ | |
| 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 |
arekusuri
/ sum_prime.py
Last active
August 29, 2015 14:04
The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17. Find the sum of all the primes below two million.
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| def find_prime_under(n=100): | |
| arr = range(n) | |
| nn = n - 1 | |
| for num in arr: | |
| if num > 1: | |
| i = 2 | |
| while True: | |
| mul = i*num | |
| i += 1 | |
| if mul > nn: |
arekusuri
/ find_pythagorean.py
Created
August 2, 2014 18:33
A Pythagorean triplet is a set of three natural numbers, a < b < c, for which, a^2 + b^2 = c^2 For example, 3^2 + 4^2 = 9 + 16 = 25 = 5^2. There exists exactly one Pythagorean triplet for which a + b + c = 1000. Find the product abc.
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| t = 1000. | |
| def calc_a(b): | |
| a = (t*t - 2*t*b)/2/(t-b) | |
| return a | |
| ab = [(calc_a(b), b) for b in range(1, int(t))] | |
| ab = filter(lambda x: float(int(x[0])) == x[0] and x[0] > 0, ab) | |
| abc = [list(x) + [t-x[0]-x[1]] for x in ab] | |
| productor = [int(x[0] * x[1] * x[2]) for x in abc] |
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