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MIT OpenCourseWare 6.00 Introduction to Computer Science and Programming: Lesson 3, HW 1
#Problem Set 2 (Part I)
#Name: mattb
#Time: 00:10
#Collaborators: Curious Reef
for n in range (50,55):
for a in range (0, n/6 + 1):
for b in range (0, n/9 + 1):
for c in range (0,n/20 + 1):
if (6 * a) + (9 * b) + (20 * c) == n:
print a,'a', ' + ',b,'b', '+', c,'c = ', n
# Problem Set 2 (Part II)
# Name: mattb
# Collaborators: Curious Reef
# Time: 5 minutes
#All values greater than x + 5 can be attained
#by adding multiples of 6 to the appropriate value in
#the range x, x + 1,... x + 5
# Problem Set 2 (Part III)
# Name: mattb
# Collaborators: Curious Reef
# Time: 2:00
maxn = 6
ncount = 0
n = 7
while ncount < 6:
n += 1
ans = False
for a in range(0,n/6 + 1):
for b in range(0,n/9 + 1):
for c in range(0,n/20 + 1):
if (6*a)+(9*b)+(20*c) == n:
ans = True
if ans:
ncount += 1
else:
ncount = 0
maxn = n
print maxn
print 'Largest number of McNuggets that \
cannot be bought in exact quantity: ', maxn
# Problem Set 2 (Part IIII)
# Name: mattb
# Collaborators: Curious Reef
# Time: 20 mins
x = int(raw_input('x = '))
y = int(raw_input('y = '))
z = int(raw_input('z = '))
packages = (x,y,z)
maxn = 0
ncount = 0
n = 0
while ncount < 6 and n <200:
n += 1
ans = False
for a in range(0,n/packages[0] + 1):
for b in range(0,n/packages [1] + 1):
for c in range(0,n/packages [2] + 1):
if (x*a)+(y*b)+(z*c) == n:
ans = True
if ans:
ncount += 1
else:
ncount = 0
maxn = n
print maxn
print 'Given package sizes ', packages,' the largest \
number of McNuggets that cannot be bought in exact quantity is: ', maxn
mattb
9 months ago
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MIT OpenCourseWare 6.00 Introduction to Computer Science and Programming: Lesson 2, HW 1
PS1 - any suggestions/comments would be appreciated.
# Problem Set 1a
# Name: mattb
# Collaborators: Curious Reef
# 2 days
count = 1
candidate = 3
while count < 1000:
divisor = candidate/2
while divisor > 1:
if candidate % divisor == 0:
candidate += 2
divisor = candidate / 2
else:
divisor -= 1
count += 1
if count < 1000:
print count, ' : ', candidate
candidate += 2
print 'The 1000th prime is:', candidate
# Problem Set 1b
# Name: mattb
# Collaborators: Curious Reef
# Time: 6 hours
import math
candidate = 3
n = int(raw_input('n = '))
logsum = math.log(2)
while candidate < n - 1:
divisor = candidate/2
while divisor > 1:
if candidate % divisor == 0:
candidate += 2
divisor = candidate / 2
else: divisor -= 1
lastprime = candidate
logsum += math.log(candidate)
candidate += 2
print 'n = ', n
print 'Sum of logs = ', logsum
print 'Ratio = ', (logsum/(n))
mattb
9 months ago
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