Wednesday, December 2, 2009
Monday, October 26, 2009
Thursday, October 22, 2009
Project Euler 59
Äntligen lite kryptoknäckning!
ciphertext = open( "cipher1.txt", "r" ).readline().split(',')
frequencies = [[0]*255,[0]*255,[0]*255]
for char in range(0,len(ciphertext)):
frequencies[char%3][int(ciphertext[char])] += 1
print sum(int(ciphertext[char]) ^ 32 ^ frequencies[char%3].index(max(frequencies[char%3])) for char in range(0, len(ciphertext)))
Wednesday, October 21, 2009
Project Euler 148
to A <3
def f(x):
if x <= 6:
return (x)*(x+1)/2
else:
k = 1
r = 1
while (7*k <= x):
k *= 7
r *= 28
return f(x/k)*r + (x/k+1)*f(x-k*int(x/k))
Använde identiteten att x^{7} + y^{7} = (x+y)^{7} (mod 7). Det är vetvärt att lösningen är alla ickenoll rutor i en Sierpinskitriangel.
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