26、
#encoding=utf-8
”’
而怎么计算一个数的循环长度呢
只需要知道它能被多少长度的9整除就行了
3能被9整除，所以它的循环长度是1
7能被999999整除，商正好是循环体142857，所以它的循环长度是6
”’
from euler import is_prime
n = 997
for p in range(n, 1, -2):
if not is_prime(p):
continue
c = 1
while (pow(10, c) – 1) % p != 0:
c += 1
if p-c == 1:
break
print p
27、
#encoding=utf-8
from euler import is_prime
def get_primes_num(a, b):
n = 0
while 1:
value = n*n + a*n +b
if value <= 0:
break
if not is_prime(value):
break
n += 1
return n
#print get_primes_num(-79, 1601)
”’
b一定是正素数
因为当n为0时，表达式的值为b
”’
primes_b = []
for i in range(3, 1000):
if is_prime(i):
primes_b.append(i)
solution = ()
snum = 0
for a in range(-999, 1000):
for b in primes_b:
gpnum = get_primes_num(a, b)
if gpnum > snum:
solution = (a, b)
snum = gpnum
print solution, gpnum, solution[0] * solution[1]
28、
#encoding=utf-8
numbers_sum = 1
corder_sum = 1
circle_num = (1001 + 1) / 2 #2n-1等于边长，n为圈数
for circle in range(2, circle_num+1):
numbers_sum = (2 * (circle-1) – 1)**2
first_num = numbers_sum + 1 #这一圈的第一个数
step = 2*circle – 2 #边长-1
first_corder = first_num + step – 1
corder_sum += first_corder*4 + step*6
print corder_sum
29、
result = set()
for a in range(2, 101):
for b in range(2, 101):
result.add(a**b)
print len(result)
30、
resultsum = []
for num in range(2, 9**5*7):
tmpsum = 0
for i in str(num):
tmpsum += int(i)**5
if tmpsum == num:
resultsum.append(tmpsum)
print resultsum
print sum(resultsum)
31、
#encodin=utf-8
target = 200
coins = [1,2,5,10,20,50,100,200]
ways = [1]+[0]*target
”’
动态归划问题
”’
for coin in coins:
for i in range(coin, target+1):
ways[i] += ways[i-coin]
print ways[target]
32、
pandigitals = set()
for multiplicard in range(1, 5001):
for multiplier in range(1, 101):
product = multiplicard * multiplier
ss = str(product)+str(multiplicard)+str(multiplier)
ss = ”.join(sorted(ss))
if ss == ’123456789′:
print product
pandigitals.add(product)
print sum(pandigitals)
33、
d = 1
for i in range(1, 10):
for j in range(1, i):
for k in range(1, j):
ki = k*10 + i
ij = float(i)*10 + j
if (k*ij == ki*j):
d *= ij/ki
print d
34、
fact = (1,1,2,6,24,120,720,5040,40320,362880)
s = 0
for n in range(10, 50000):
x = sum( fact[int(d)] for d in str(n) )
if n == x:
s += n
print s
35、
import math
total=1
def prime(n):
if(n==2):
return True
if(not n%2):
return False
else:
for i in range(3,int(math.sqrt(n))+1,2):
if(not n%i):
return False
return True
def rotate(l):
l.insert(0,l.pop())
return (check(l))
def check(l):
n=0
for k in l:
n=n*10+int(k)
if(prime(n)):
return True
return False
for i in range(3,1000000,2):
flag=True
evenflag=False
even=[2,4,6,8,0]
l=list(str(i))
for k in even:
if(str(k) in l):
evenflag=True
break;
if(evenflag):
continue
for k in range(len(l)):
if(not rotate(l)):
flag=False
break
if(flag):
total+=1
print total
36、
result = 0
for i in range(1, 1000001):
if str(i) == str(i)[::-1]:
b = str(bin(i))[2:]
if b == b[::-1]:
#print i, b
result += i
print result
37、
from euler import is_prime
from math import log10
primes = [23,37,53,73,313,317,373,797,3137,3797,7937,31397,31973,37313,37397,71317,
71713,71917,73973,79397,313717,317197,319313,371737,371797,373717,373937,
379397,713737,713917,717317,717397,717917,719197,719713,719717,731713,
731737,739373,739397,791317,791797,793717,797917]
def trunc(n):
c = n
while c>10:
c = c % ( 10**(int(log10(c))) )
n = n//10
if not is_prime(c) or not is_prime(n): return False
return True
c = 0
s = 0
for p in primes:
if trunc(p):
print p
c += 1
s += p
print c,"primes found for a sum of",s
38、
def f1(n):
strNum = str(n)
for i in range(1,5):
tmpStr = strNum
num = int(strNum[:i])
index = 1
flag = False
while True:
str1 = str(num*index)
len1 = len(str1)
if str1 == tmpStr[:len1]:
if len(tmpStr)>len1:
tmpStr=tmpStr[len1:]
elif len(tmpStr) == len1:
flag = True
break
else:
break
index += 1
if flag:
return True
return False
numList=[1,['','12345678']]
while True:
if numList[0]==40320:
#8!=362880
break
for i in range(1,numList[0]+1):
left = numList[i][0]
right= numList[i][1]
l = left+right[0]
r = right.replace(right[0],”)
numList[i] = [l,r]
for j in right[1:]:
l = left+j
r = right.replace(j,”)
numList.append([l,r])
numList[0] += 1
numStrList = []
for i in numList[1:]:
numStrList.append(’9′+i[0]+i[1])
del numList
# print numStrList
# print len(numStrList)
result = 0
for i in numStrList:
if f1(int(i)):
if int(i) > result:
result = int(i)
print result
39、
t_max = 0
p_limit = 1000
for p in range(p_limit//2, p_limit+1, 2):
t = 0;
for a in range(2, p/4+1):
if p*(p – 2*a) % (2*(p-a)) == 0: t += 1
if t > t_max: (t_max, p_max) = (t, p)
print p_max
40、
num = 0
strs = ”
count = 0
while count < 1000001:
l = len(str(num))
count += l
strs += str(num)
num += 1
ss = 1
for i in range(7):
ss *= int( strs[10**i] )
print ss
41、
from euler import is_prime, is_pandigital
”’
any integer is divisible by 3 or 9 whose sum of digits is divisible by 3 or 9; therefore composite and not prime.
9+8+7+6+5+4+3+2+1 = 45,
8+7+6+5+4+3+2+1 = 36,
6+5+4+3+2+1 = 21, and
5+4+3+2+1 = 15,
all of which are divisible by 3 and therefore could not yield a 1 to {5, 6, 8, 9} pandigital prime. So that leaves 4 and 7 digit prime numbers less than 4321 and 7654321 respectively.
”’
n = 7654321
while not(is_prime(n) and is_pandigital(str(n), 7)):
n -= 2
print n
42、
import urllib
words = urllib.urlopen(‘http://projecteuler.net/project/words.txt’).read()
wordslist = [ w[1:-1] for w in words.split(‘,’) ]
tri_numbers = []
n = 1
while n*(n+1) / 2 <= 260:
tri_numbers.append(n*(n+1)/2)
n += 1
result = 0
for word in wordslist:
v = 0
for w in word:
v += ord(w)-ord(‘A’)+1
if v in tri_numbers:
result += 1
print result
43、
import itertools
#here is a better and faster way: http://blog.dreamshire.com/2009/04/28/project-euler-problem-43-solution/
iter_list = itertools.permutations(range(10))
result_sum = 0
divs = [2,3,5,7,11,13,17]
for number in iter_list:
if number[0] == 0: continue
number_str = ”.join([str(n) for n in number])
isok = True
for i in range(len(divs)):
if int(number_str[i+1: i+4]) % divs[i] != 0:
isok = False
break
if isok:
result_sum += int(number_str)
print result_sum
44、
pent = set( n * (3*n – 1) / 2 for n in range(2,2400) )
def pe44():
for pj in pent:
for pk in pent:
if pj – pk in pent and pj + pk in pent:
return pj – pk
print pe44()
45、
#encoding=utf8
from math import sqrt
”’
Note that Hexagonal numbers are a subset of Triangle numbers so we only determine the first occurrence of Hi = Pj to find our answer. We can safely start our search from 144 as alluded from the problem’s description.
”’
def is_pentagonal(n):
p = (sqrt(1 + 24*n) + 1) / 6
return p==int(p)
h = lambda n: n*(2*n – 1) #calculate the nth hexagonal number
n = 144
while not(is_pentagonal(h(n))): n += 1
print h(n)
46、
n = 5
f = 1
primes = set()
while (1):
if all( n % p for p in primes ):
primes.add(n)
else:
if not any( (n-2*i*i) in primes for i in range(1, n) ):
break
n += 3-f
f = -f
print n
47、
from euler import factor # the euler module can find here : http://blog.dreamshire.com/2009/03/26/94/ ci = 1 nf = 4 #number of distinct factors ns = 4 #number of consecutive integers n = 2*3*5*7 #starting candidate for search while ci != ns: n += 1 if len(factor(n)) == nf: ci += 1 else: ci = 0 print "Answer to PE47 = ", n-nf+1
48、
result = 0
for i in range(1, 1001):
result += i**i
print str(result)[-10:]
49、
from euler import is_prime, is_perm
n = 1489 # must be odd
while True:
b, c = n+3330, n+6660
if is_prime(n) and is_prime(b) and is_prime(c) \
and is_perm(n,b) and is_perm(b,c): break
n += 2
print "Answer to PE49 = ", str(n)+str(b)+str(c)
50、
from euler import prime_sieve, is_prime
max = 1000000
primes = prime_sieve(max)
prime_sum = [0]
sum = 0
count = 0
while(sum < max):
sum += primes[count]
prime_sum.append(sum)
count += 1
terms = 1
for i in range(count):
for j in range(i+terms, count):
n = prime_sum[j] – prime_sum[i]
if j-i > terms and is_prime(n):
terms, max_prime = j-i, n
print max_prime