Project Euler: Difference between revisions
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==Problem 1== | ==Problem 1== | ||
< | <syntaxhighlight lang="python"> | ||
x=[] | x=[] | ||
for i in range(1000): | for i in range(1000): | ||
| Line 10: | Line 10: | ||
y=np.sum(x) | y=np.sum(x) | ||
print y | print y | ||
</ | </syntaxhighlight> | ||
==Problem 2== | ==Problem 2== | ||
< | <syntaxhighlight lang="python"> | ||
i=1 | i=1 | ||
j=1 | j=1 | ||
y=[] | y=[] | ||
while True: | while True: | ||
k=i+j | |||
if k>4e6:break | |||
i=j | |||
j=k | |||
if k%2==0:y.append(k) | |||
print sum(y) | print sum(y) | ||
</ | </syntaxhighlight> | ||
==Problem 3== | ==Problem 3== | ||
<!---syntaxhighlight lang="python"> | |||
</syntaxhighlight---> | |||
==Problem 4== | ==Problem 4== | ||
==Problem 5== | ==Problem 5== | ||
Latest revision as of 18:19, 7 August 2014
Problem 1
<syntaxhighlight lang="python">
x=[]
for i in range(1000):
if i%3==0 or i%5==0:
x.append(i)
x=np.array(x) #turns list into numpy array
y=np.sum(x)
print y
</syntaxhighlight>
Problem 2
<syntaxhighlight lang="python">
i=1
j=1
y=[]
while True:
k=i+j
if k>4e6:break
i=j
j=k
if k%2==0:y.append(k)
print sum(y)
</syntaxhighlight>