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Use timing decorator for problems 21-40

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This commit is contained in:
daniele 2023-06-07 18:17:42 +02:00
parent 634eb3f750
commit be5e97dfbb
Signed by: fuxino
GPG Key ID: 981A2B2A3BBF5514
20 changed files with 172 additions and 288 deletions
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24
Python/p021.py
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@ -9,34 +9,28 @@
# Evaluate the sum of all the amicable numbers under 10000.
from timeit import default_timer
from projecteuler import sum_of_divisors
from projecteuler import sum_of_divisors, timing
def main():
start = default_timer()
@timing
def p021():
sum_ = 0
for i in range(2, 10000):
# Calculate the sum of proper divisors with the function
# implemented in projecteuler.py.
# Calculate the sum of proper divisors with the function
# implemented in projecteuler.py.
n = sum_of_divisors(i)
# If i!=n and the sum of proper divisors of n=i,
# sum the pair of numbers and add it to the total.
# If i!=n and the sum of proper divisors of n=i,
# sum the pair of numbers and add it to the total.
if i != n and sum_of_divisors(n) == i:
sum_ = sum_ + i + n
sum_ = sum_ // 2
end = default_timer()
print('Project Euler, Problem 21')
print(f'Answer: {sum_}')
print(f'Elapsed time: {end - start:.9f} seconds')
if __name__ == '__main__':
main()
p021()

14
Python/p022.py
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@ -9,12 +9,12 @@
# What is the total of all the name scores in the file?
import sys
from timeit import default_timer
from projecteuler import timing
def main():
start = default_timer()
@timing
def p022():
try:
with open('p022_names.txt', 'r', encoding='utf-8') as fp:
names = list(fp.readline().replace('"', '').split(','))
@ -37,13 +37,9 @@ def main():
sum_ = sum_ + score
i = i + 1
end = default_timer()
print('Project Euler, Problem 22')
print(f'Answer: {sum_}')
print(f'Elapsed time: {end - start:.9f} seconds')
if __name__ == '__main__':
main()
p022()

25
Python/p023.py
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@ -12,29 +12,26 @@
#
# Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers.
from timeit import default_timer
from projecteuler import sum_of_divisors
from projecteuler import sum_of_divisors, timing
def is_abundant(n):
return sum_of_divisors(n) > n
def main():
start = default_timer()
@timing
def p023():
ab_nums = [False] * 28124
# Find all abundant numbers smaller than 28123.
# Find all abundant numbers smaller than 28123.
for i in range(12, 28124):
ab_nums[i] = is_abundant(i)
sums = [False] * 28124
# For every abundant number, sum every other abundant number greater
# than itself, until the sum exceeds 28123. Record that the resulting
# number is the sum of two abundant numbers.
# For every abundant number, sum every other abundant number greater
# than itself, until the sum exceeds 28123. Record that the resulting
# number is the sum of two abundant numbers.
for i in range(1, 28123):
if ab_nums[i]:
for j in range(i, 28123):
@ -45,18 +42,14 @@ def main():
sum_ = 0
# Sum every number that was not found as a sum of two abundant numbers.
# Sum every number that was not found as a sum of two abundant numbers.
for i in range(1, 28124):
if not sums[i]:
sum_ = sum_ + i
end = default_timer()
print('Project Euler, Problem 23')
print(f'Answer: {sum_}')
print(f'Elapsed time: {end - start:.9f} seconds')
if __name__ == '__main__':
main()
p023()

16
Python/p024.py
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@ -8,23 +8,19 @@
# What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?
from itertools import permutations
from timeit import default_timer
from projecteuler import timing
def main():
start = default_timer()
# Generate all the permutations in lexicographic order and get the millionth one.
@timing
def p024():
# Generate all the permutations in lexicographic order and get the millionth one.
perm = list(permutations('0123456789'))
res = int(''.join(map(str, perm[999999])))
end = default_timer()
print('Project Euler, Problem 24')
print(f'Answer: {res}')
print(f'Elapsed time: {end - start:.9f} seconds')
if __name__ == '__main__':
main()
p024()

15
Python/p025.py
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@ -21,32 +21,27 @@
#
# What is the index of the first term in the Fibonacci sequence to contain 1000 digits?
from timeit import default_timer
from projecteuler import timing
def main():
start = default_timer()
@timing
def p025():
fib1 = 1
fib2 = 1
fibn = fib1 + fib2
i = 3
# Calculate the Fibonacci numbers until one with 1000 digits is found.
# Calculate the Fibonacci numbers until one with 1000 digits is found.
while len(str(fibn)) < 1000:
fib1 = fib2
fib2 = fibn
fibn = fib1 + fib2
i = i + 1
end = default_timer()
print('Project Euler, Problem 25')
print(f'Answer: {i}')
print(f'Elapsed time: {end - start:.9f} seconds')
if __name__ == '__main__':
main()
p025()

27
Python/p026.py
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@ -16,27 +16,24 @@
#
# Find the value of d < 1000 for which 1/d contains the longest recurring cycle in its decimal fraction part.
from timeit import default_timer
from projecteuler import timing
def main():
start = default_timer()
@timing
def p026():
max_ = 0
for i in range(2, 1000):
j = i
# The repeating cycle of 1/(2^a*5^b*p^c*...) is equal to
# that of 1/p^c*..., so factors 2 and 5 can be eliminated.
# The repeating cycle of 1/(2^a*5^b*p^c*...) is equal to that of 1/p^c*..., so factors 2 and 5 can be eliminated.
while j % 2 == 0 and j > 1:
j = j // 2
while j % 5 == 0 and j > 1:
j = j // 5
# If the denominator had only factors 2 and 5, there is no
# repeating cycle.
# If the denominator had only factors 2 and 5, there is no repeating cycle.
if j == 1:
n = 0
else:
@ -44,10 +41,10 @@ def main():
k = 9
div = j
# After eliminating factors 2s and 5s, the length of the repeating cycle
# of 1/d is the smallest n for which k=10^n-1/d is an integer. So we start
# with k=9, then k=99, k=999 and so on until k is divisible by d.
# The number of digits of k is the length of the repeating cycle.
# After eliminating factors 2s and 5s, the length of the repeating cycle
# of 1/d is the smallest n for which k=10^n-1/d is an integer. So we start
# with k=9, then k=99, k=999 and so on until k is divisible by d.
# The number of digits of k is the length of the repeating cycle.
while k % div != 0:
n = n + 1
k = k * 10
@ -57,13 +54,9 @@ def main():
max_ = n
max_n = i
end = default_timer()
print('Project Euler, Problem 26')
print(f'Answer: {max_n}')
print(f'Elapsed time: {end - start:.9f} seconds')
if __name__ == '__main__':
main()
p026()

19
Python/p027.py
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@ -19,20 +19,17 @@
#
# Find the product of the coefficients, a and b, for the quadratic expression that produces the maximum number of primes for consecutive values of n, starting with n=0.
from timeit import default_timer
from projecteuler import is_prime
from projecteuler import is_prime, timing
def main():
start = default_timer()
@timing
def p027():
max_ = 0
# Brute force approach, optimized by checking only values of b where b is prime.
# Brute force approach, optimized by checking only values of b where b is prime.
for a in range(-999, 1000):
for b in range(2, 1001):
# For n=0, n^2+an+b=b, so b must be prime.
# For n=0, n^2+an+b=b, so b must be prime.
if is_prime(b):
n = 0
count = 0
@ -51,13 +48,9 @@ def main():
save_a = a
save_b = b
end = default_timer()
print('Project Euler, Problem 27')
print(f'Answer: {save_a * save_b}')
print(f'Elapsed time: {end - start:.9f} seconds')
if __name__ == '__main__':
main()
p027()

21
Python/p028.py
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@ -12,12 +12,11 @@
#
# What is the sum of the numbers on the diagonals in a 1001 by 1001 spiral formed in the same way?
from timeit import default_timer
from projecteuler import timing
def main():
start = default_timer()
@timing
def p028():
N = 1001
limit = N * N
@ -27,10 +26,10 @@ def main():
step = 0
sum_ = 1
# Starting with the central 1, it's easy to see that the next four numbers in the diagonal
# are 1+2, 1+2+2, 1+2+2+2 and 1+2+2+2+2, then for the next four number the step is increased
# by two, so from 9 to 9+4, 9+4+4 etc, for the next four number the step is again increased
# by two, and so on. We go on until the value is equal to N*N, with N=1001 for this problem.
# Starting with the central 1, it's easy to see that the next four numbers in the diagonal
# are 1+2, 1+2+2, 1+2+2+2 and 1+2+2+2+2, then for the next four number the step is increased
# by two, so from 9 to 9+4, 9+4+4 etc, for the next four number the step is again increased
# by two, and so on. We go on until the value is equal to N*N, with N=1001 for this problem.
while j < limit:
if i == 0:
step = step + 2
@ -38,13 +37,9 @@ def main():
sum_ = sum_ + j
i = (i + 1) % 4
end = default_timer()
print('Project Euler, Problem 28')
print(f'Answer: {sum_}')
print(f'Elapsed time: {end - start:.9f} seconds')
if __name__ == '__main__':
main()
p028()

17
Python/p029.py
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@ -13,23 +13,22 @@
#
# How many distinct terms are in the sequence generated by ab for 2 ≤ a ≤ 100 and 2 ≤ b ≤ 100?
from timeit import default_timer
from numpy import zeros
from projecteuler import timing
def main():
start = default_timer()
@timing
def p029():
powers = zeros(9801)
# Generate all the powers
# Generate all the powers
for i in range(2, 101):
a = i
for j in range(2, 101):
powers[(i-2)*99+j-2] = a ** j
# Sort the values and count the different values.
# Sort the values and count the different values.
powers = list(powers)
powers.sort()
@ -39,13 +38,9 @@ def main():
if powers[i] != powers[i-1]:
count = count + 1
end = default_timer()
print('Project Euler, Problem 29')
print(f'Answer: {count}')
print(f'Elapsed time: {end - start:.9f} seconds')
if __name__ == '__main__':
main()
p029()

19
Python/p030.py
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@ -12,17 +12,16 @@
#
# Find the sum of all the numbers that can be written as the sum of fifth powers of their digits.
from timeit import default_timer
from projecteuler import timing
def main():
start = default_timer()
@timing
def p030():
tot = 0
# Straightforward brute force approach. The limit is chosen considering that
# 6*9^5=354294, so no number larger than that can be expressed as sum
# of 5th power of its digits.
# Straightforward brute force approach. The limit is chosen considering that
# 6*9^5=354294, so no number larger than that can be expressed as sum
# of 5th power of its digits.
for i in range(10, 354295):
j = i
sum_ = 0
@ -35,13 +34,9 @@ def main():
if sum_ == i:
tot = tot + i
end = default_timer()
print('Project Euler, Problem 30')
print(f'Answer: {tot}')
print(f'Elapsed time: {end - start:.9f} seconds')
if __name__ == '__main__':
main()
p030()

13
Python/p031.py
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@ -10,7 +10,7 @@
#
# How many different ways can £2 be made using any number of coins?
from timeit import default_timer
from projecteuler import timing
# Simple recursive function that tries every combination.
@ -29,20 +29,15 @@ def count(coins, value, n, i):
return n
def main():
start = default_timer()
@timing
def p031():
coins = [1, 2, 5, 10, 20, 50, 100, 200]
n = count(coins, 0, 0, 0)
end = default_timer()
print('Project Euler, Problem 31')
print(f'Answer: {n}')
print(f'Elapsed time: {end - start:.9f} seconds')
if __name__ == '__main__':
main()
p031()

53
Python/p032.py
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@ -9,34 +9,31 @@
#
# HINT: Some products can be obtained in more than one way so be sure to only include it once in your sum.
from timeit import default_timer
import numpy as np
from numpy import zeros
from projecteuler import is_pandigital
from projecteuler import is_pandigital, timing
def main():
start = default_timer()
@timing
def p032():
n = 0
# Initially I used a bigger array, but printing the resulting products
# shows that 10 values are sufficient.
# Initially I used a bigger array, but printing the resulting products
# shows that 10 values are sufficient.
products = zeros(10, int)
# To get a 1 to 9 pandigital concatenation of the two factors and product,
# we need to multiply a 1 digit number times a 4 digit numbers (the biggest
# one digit number 9 times the biggest 3 digit number 999 multiplied give
# 8991 and the total digit count is 8, which is not enough), or a 2 digit
# number times a 3 digit number (the smallest two different 3 digits number,
# 100 and 101, multiplied give 10100, and the total digit count is 11, which
# is too many). The outer loop starts at 2 because 1 times any number gives
# the same number, so its digit will be repeated and the result can't be
# pandigital. The nested loop starts from 1234 because it's the smallest
# 4-digit number with no repeated digits, and it ends at 4987 because it's
# the biggest number without repeated digits that multiplied by 2 gives a
# 4 digit number.
# To get a 1 to 9 pandigital concatenation of the two factors and product,
# we need to multiply a 1 digit number times a 4 digit numbers (the biggest
# one digit number 9 times the biggest 3 digit number 999 multiplied give
# 8991 and the total digit count is 8, which is not enough), or a 2 digit
# number times a 3 digit number (the smallest two different 3 digits number,
# 100 and 101, multiplied give 10100, and the total digit count is 11, which
# is too many). The outer loop starts at 2 because 1 times any number gives
# the same number, so its digit will be repeated and the result can't be
# pandigital. The nested loop starts from 1234 because it's the smallest
# 4-digit number with no repeated digits, and it ends at 4987 because it's
# the biggest number without repeated digits that multiplied by 2 gives a
# 4 digit number.
for i in range(2, 9):
for j in range(1234, 4987):
p = i * j
@ -51,10 +48,10 @@ def main():
products[n] = p
n = n + 1
# The outer loop starts at 12 because 10 has a 0 and 11 has two 1s, so
# the result can't be pandigital. The nested loop starts at 123 because
# it's the smallest 3-digit number with no digit repetitions and ends at
# 833, because 834*12 has 5 digits.
# The outer loop starts at 12 because 10 has a 0 and 11 has two 1s, so
# the result can't be pandigital. The nested loop starts at 123 because
# it's the smallest 3-digit number with no digit repetitions and ends at
# 833, because 834*12 has 5 digits.
for i in range(12, 99):
for j in range(123, 834):
p = i * j
@ -70,7 +67,7 @@ def main():
products[n] = p
n = n + 1
# Sort the found products to easily see if there are duplicates.
# Sort the found products to easily see if there are duplicates.
products = np.sort(products[:n])
sum_ = products[0]
@ -79,13 +76,9 @@ def main():
if products[i] != products[i-1]:
sum_ = sum_ + products[i]
end = default_timer()
print('Project Euler, Problem 32')
print(f'Answer: {sum_}')
print(f'Elapsed time: {end - start:.9f} seconds')
if __name__ == '__main__':
main()
p032()

20
Python/p033.py
View File

@ -11,19 +11,19 @@
# If the product of these four fractions is given in its lowest common terms, find the value of the denominator.
from math import gcd
from timeit import default_timer
from projecteuler import timing
def main():
start = default_timer()
@timing
def p033():
prod_n = 1
prod_d = 1
for i in range(11, 100):
for j in range(11, 100):
# If the example is non-trivial, check if cancelling the digit that's equal
# in numerator and denominator gives the same fraction.
# If the example is non-trivial, check if cancelling the digit that's equal
# in numerator and denominator gives the same fraction.
if i % 10 != 0 and j % 10 != 0 and\
i != j and i % 10 == j // 10:
n = i // 10
@ -36,16 +36,12 @@ def main():
prod_n = prod_n * i
prod_d = prod_d * j
# Find the greater common divisor of the fraction found.
# Find the greater common divisor of the fraction found.
div = gcd(prod_n, prod_d)
end = default_timer()
print('Project Euler, Problem 33')
print(f'Answer: {prod_d // div}')
print(f'Elapsed time: {end - start:.9f} seconds')
if __name__ == '__main__':
main()
p033()

16
Python/p034.py
View File

@ -7,23 +7,23 @@
# Note: as 1! = 1 and 2! = 2 are not sums they are not included.
from math import factorial
from timeit import default_timer
from numpy import ones
from projecteuler import timing
def main():
start = default_timer()
@timing
def p034():
a = 10
sum_ = 0
factorials = ones(10, int)
# Pre-calculate factorials of each digit from 0 to 9.
# Pre-calculate factorials of each digit from 0 to 9.
for i in range(2, 10):
factorials[i] = factorial(i)
# 9!*7<9999999, so 9999999 is certainly un upper bound.
# 9!*7<9999999, so 9999999 is certainly un upper bound.
while a < 9999999:
b = a
sum_f = 0
@ -38,13 +38,9 @@ def main():
a = a + 1
end = default_timer()
print('Project Euler, Problem 34')
print(f'Answer: {sum_}')
print(f'Elapsed time: {end - start:.9f} seconds')
if __name__ == '__main__':
main()
p034()

29
Python/p035.py
View File

@ -6,9 +6,7 @@
#
# How many circular primes are there below one million?
from timeit import default_timer
from projecteuler import sieve
from projecteuler import sieve, timing
# Calculate all primes below one million, then check if they're circular.
@ -17,28 +15,28 @@ primes = sieve(N)
def is_circular_prime(n):
# If n is not prime, it's obviously not a circular prime.
# If n is not prime, it's obviously not a circular prime.
if primes[n] == 0:
return False
# The primes below 10 are circular primes.
# The primes below 10 are circular primes.
if primes[n] == 1 and n < 10:
return True
tmp = n
count = 0
# If the number has one or more even digits, it can't be a circular prime.
# because at least one of the rotations will be even.
# If the number has one or more even digits, it can't be a circular prime.
# because at least one of the rotations will be even.
while tmp > 0:
if tmp % 2 == 0:
return False
# Count the number of digits.
# Count the number of digits.
count = count + 1
tmp = tmp // 10
for _ in range(1, count):
# Generate rotations and check if they're prime.
# Generate rotations and check if they're prime.
n = n % (10 ** (count - 1)) * 10 + n // (10 ** (count - 1))
if primes[n] == 0:
@ -47,23 +45,18 @@ def is_circular_prime(n):
return True
def main():
start = default_timer()
@timing
def p035():
count = 13
# Calculate all primes below one million, then check if they're circular.
# Calculate all primes below one million, then check if they're circular.
for i in range(101, N, 2):
if is_circular_prime(i):
count = count + 1
end = default_timer()
print('Project Euler, Problem 35')
print(f'Answer: {count}')
print(f'Elapsed time: {end - start:.9f} seconds')
if __name__ == '__main__':
main()
p035()

21
Python/p036.py
View File

@ -6,32 +6,25 @@
#
# (Please note that the palindromic number, in either base, may not include leading zeros.)
from timeit import default_timer
from projecteuler import is_palindrome
from projecteuler import is_palindrome, timing
def main():
start = default_timer()
@timing
def p036():
N = 1000000
sum_ = 0
# Brute force approach. For every number below 1 million,
# check if they're palindrome in base 2 and 10 using the
# function implemented in projecteuler.c.
# Brute force approach. For every number below 1 million,
# check if they're palindrome in base 2 and 10 using the
# function implemented in projecteuler.c.
for i in range(1, N):
if is_palindrome(i, 10) and is_palindrome(i, 2):
sum_ = sum_ + i
end = default_timer()
print('Project Euler, Problem 36')
print(f'Answer: {sum_}')
print(f'Elapsed time: {end - start:.9f} seconds')
if __name__ == '__main__':
main()
p036()

32
Python/p037.py
View File

@ -7,29 +7,26 @@
#
# NOTE: 2, 3, 5, and 7 are not considered to be truncatable primes.
from timeit import default_timer
from projecteuler import is_prime
from projecteuler import is_prime, timing
def is_tr_prime(n):
# One-digit numbers and non-prime numbers are
# not truncatable primes.
# One-digit numbers and non-prime numbers are
# not truncatable primes.
if n < 11 or not is_prime(n):
return False
tmp = n // 10
# Remove one digit at a time from the right and check
# if the resulting number is prime. Return 0 if it isn't.
# Remove one digit at a time from the right and check
# if the resulting number is prime. Return 0 if it isn't.
while tmp > 0:
if not is_prime(tmp):
return False
tmp = tmp // 10
# Starting from the last digit, check if it's prime, then
# add back one digit at a time on the left and check if it
# is prime. Return 0 when it isn't.
# Starting from the last digit, check if it's prime, then add back one digit at a time on the left and check if it
# is prime. Return 0 when it isn't.
i = 10
tmp = n % i
@ -39,31 +36,26 @@ def is_tr_prime(n):
i = i * 10
tmp = n % i
# If it gets here, the number is truncatable prime.
# If it gets here, the number is truncatable prime.
return True
def main():
start = default_timer()
@timing
def p037():
i = 0
n = 1
sum_ = 0
# Check every number until 11 truncatable primes are found.
# Check every number until 11 truncatable primes are found.
while i < 11:
if is_tr_prime(n):
sum_ = sum_ + n
i = i + 1
n = n + 1
end = default_timer()
print('Project Euler, Problem 37')
print(f'Answer: {sum_}')
print(f'Elapsed time: {end - start:.9f} seconds')
if __name__ == '__main__':
main()
p037()

27
Python/p038.py
View File

@ -13,22 +13,19 @@
#
# What is the largest 1 to 9 pandigital 9-digit number that can be formed as the concatenated product of an integer with (1,2, ... , n) where n > 1?
from timeit import default_timer
from projecteuler import is_pandigital
from projecteuler import is_pandigital, timing
def main():
start = default_timer()
@timing
def p038():
max_ = 0
# A brute force approach is used, starting with 1 and multiplying
# the number by 1, 2 etc., concatenating the results, checking if
# it's 1 to 9 pandigital, and going to the next number when the
# concatenated result is greater than the greatest 9 digit pandigital
# value. The limit is set to 10000, since 1*10000=10000, 2*10000=20000 and
# concatenating this two numbers a 10-digit number is obtained.
# A brute force approach is used, starting with 1 and multiplying
# the number by 1, 2 etc., concatenating the results, checking if
# it's 1 to 9 pandigital, and going to the next number when the
# concatenated result is greater than the greatest 9 digit pandigital
# value. The limit is set to 10000, since 1*10000=10000, 2*10000=20000 and
# concatenating this two numbers a 10-digit number is obtained.
for i in range(1, 10000):
n = 0
j = 1
@ -54,13 +51,9 @@ def main():
if n > 987654321:
break
end = default_timer()
print('Project Euler, Problem 38')
print(f'Answer: {max_}')
print(f'Elapsed time: {end - start:.9f} seconds')
if __name__ == '__main__':
main()
p038()

31
Python/p039.py
View File

@ -6,18 +6,17 @@
#
# For which value of p ≤ 1000, is the number of solutions maximised?
from timeit import default_timer
from numpy import zeros
from projecteuler import timing
def main():
start = default_timer()
@timing
def p039():
max_ = 0
savedc = zeros(1000, int)
# Start with p=12 (the smallest pythagorean triplet is (3,4,5) and 3+4+5=12.
# Start with p=12 (the smallest pythagorean triplet is (3,4,5) and 3+4+5=12.
for p in range(12, 1001):
count = 0
a = 0
@ -25,16 +24,15 @@ def main():
c = 0
m = 2
# Generate pythagorean triplets.
# Generate pythagorean triplets.
while m * m < p:
for n in range(1, m):
a = m * m - n * n
b = 2 * m * n
c = m * m + n * n
# Increase counter if a+b+c=p and the triplet is new,
# then save the value of c to avoid counting the same
# triplet more than once.
# Increase counter if a+b+c=p and the triplet is new, then save the value of c to avoid counting the same
# triplet more than once.
if a + b + c == p and savedc[c] == 0:
savedc[c] = 1
count = count + 1
@ -44,14 +42,13 @@ def main():
tmpb = b
tmpc = c
# Check all the triplets obtained multiplying a, b and c
# for integer numbers, until the perimeters exceeds p.
# Check all the triplets obtained multiplying a, b and c for integer numbers, until the perimeters exceeds p.
while tmpa + tmpb + tmpc < p:
tmpa = a * i
tmpb = b * i
tmpc = c * i
# Increase counter if the new a, b and c give a perimeter=p.
# Increase counter if the new a, b and c give a perimeter=p.
if tmpa + tmpb + tmpc == p and savedc[tmpc] == 0:
savedc[tmpc] = 1
count = count + 1
@ -60,19 +57,15 @@ def main():
m = m + 1
# If the current value is greater than the maximum,
# save the new maximum and the value of p.
# If the current value is greater than the maximum,
# save the new maximum and the value of p.
if count > max_:
max_ = count
res = p
end = default_timer()
print('Project Euler, Problem 39')
print(f'Answer: {res}')
print(f'Elapsed time: {end - start:.9f} seconds')
if __name__ == '__main__':
main()
p039()

21
Python/p040.py
View File

@ -10,21 +10,20 @@
#
# d_1 × d_10 × d_100 × d_1000 × d_10000 × d_100000 × d_1000000
from timeit import default_timer
from numpy import zeros
from projecteuler import timing
def main():
start = default_timer()
@timing
def p040():
digits = zeros(1000005, int)
i = 1
value = 1
# Loop on all numbers and put the digits in the right place
# in an array. Use modulo and division to get the digits
# for numbers with more than one digit.
# Loop on all numbers and put the digits in the right place
# in an array. Use modulo and division to get the digits
# for numbers with more than one digit.
while i <= 1000000:
if value < 10:
digits[i-1] = value
@ -61,16 +60,12 @@ def main():
i = i + 6
value = value + 1
# Calculate the product.
# Calculate the product.
n = digits[0] * digits[9] * digits[99] * digits[999] * digits[9999] * digits[99999] * digits[999999]
end = default_timer()
print('Project Euler, Problem 40')
print(f'Answer: {n}')
print(f'Elapsed time: {end - start:.9f} seconds')
if __name__ == '__main__':
main()
p040()