def main(argv):
    rna = fasta.read_one(argv[0])

    au_max = max(rna.count('A'), rna.count('U'))
    au_min = min(rna.count('A'), rna.count('U'))

    cg_max = max(rna.count('C'), rna.count('G'))
    cg_min = min(rna.count('C'), rna.count('G'))

    print combinatorics.permutations(
        au_max, au_min) * combinatorics.permutations(cg_max, cg_min)
示例#2
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def euler_41():
    "What is the largest n-digit pandigital prime that exists?"
    from combinatorics import permutations
    # Pan-digital primes are 4 or 7 digits. Others divisible by 3
    prime._refresh(2766)    # sqrt(7654321)
    x = list(range(7, 0, -1))
    for perm in permutations(x):
        num = 0
        for n in perm:
            num = num * 10 + n
        if prime._isprime(num):
            print("Lösung:", num)
            break
示例#3
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def euler_43():
    "Find the sum of all pandigital numbers with an unusual sub-string divisibility property."
    from combinatorics import permutations
    def num(l):
        s = 0
        for n in l:
            s = s * 10 + n
        return s
    def subdiv(l, n):
        return not num(l) % n
    total = 0
    for perm in permutations((0,1,2,3,4,6,7,8,9)):
        perm.insert(5, 5)               # d6 must be 5
        if (subdiv(perm[7:10], 17) and
            subdiv(perm[6:9],  13) and
            subdiv(perm[5:8],  11) and
            subdiv(perm[4:7],   7) and
            subdiv(perm[3:6],   5) and
            subdiv(perm[2:5],   3) and
            subdiv(perm[1:4],   2)):
                total += num(perm)
    print("Lösung:", total)    
示例#4
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def euler_49():
    "Find arithmetic sequences, made of prime terms, whose four digits are permutations of each other."
    from combinatorics import permutations
    prime._refresh(10000)
    for num in range(1000, 10000):
        if str(num).find('0') >= 0:
            continue
        if prime.isprime(num):
            prime_permutations = { num: 1 }
            for x in permutations(list(str(num))):
                next_num = int(''.join(x))
                if prime.isprime(next_num):
                    prime_permutations[next_num] = 1
            primes = sorted(prime_permutations.keys())
            for a in range(0, len(primes)):
                if primes[a] == 1487:
                    continue
                for b in range(a+1, len(primes)):
                    c = (primes[a] + primes[b]) // 2
                    if c in prime_permutations:
                        print("Lösung:", str(primes[a]) + str(c) + str(primes[b]))
                        return
示例#5
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文件: 041.py 项目: sanand0/euler
'''
We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once. For example, 2143 is a 4-digit pandigital and is also prime.

What is the largest n-digit pandigital prime that exists?
'''

import prime
from combinatorics import permutations

# Pan-digital primes are 4 or 7 digits. Others divisible by 3
prime._refresh(2766)    # sqrt(7654321)
for perm in permutations(range(7, 0, -1)):
    num = 0
    for n in perm: num = num * 10 + n
    if prime._isprime(num):
        print num
        break
示例#6
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from combinatorics import permutations

def num(l):
   s = 0
   for n in l: s = s * 10 + n
   return s

product = {}
for perm in permutations(range(1,10)):
    for cross in range(1,4):           
        for eq in range(cross+1, 6):    
            a = num(perm[0:cross])
            b = num(perm[cross:eq])
            c = num(perm[eq:9])
            if a * b == c: product[c] = 1

print sum(p for p in product)
示例#7
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文件: 043.py 项目: sanand0/euler
d5 d6 d7 = 357 is divisible by 7
d6 d7 d8 = 572 is divisible by 11
d7 d8 d9 = 728 is divisible by 13
d8 d9 d10= 289 is divisible by 17

Find the sum of all 0 to 9 pandigital numbers with this property.
'''

from combinatorics import permutations

def num(l):
    s = 0
    for n in l: s = s * 10 + n
    return s

def subdiv(l, n): return num(l) % n == 0

total = 0
for perm in permutations((0,1,2,3,4,6,7,8,9)):
    perm.insert(5, 5)               # d6 must be 5
    if (subdiv(perm[7:10], 17) and
        subdiv(perm[6:9],  13) and
        subdiv(perm[5:8],  11) and
        subdiv(perm[4:7],   7) and
        subdiv(perm[3:6],   5) and
        subdiv(perm[2:5],   3) and
        subdiv(perm[1:4],   2)):
            total += num(perm)

print total
示例#8
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def test_permutations_of_n_options():
    assert permutations(10) == 3628800
示例#9
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def test_permutations_of_non_numeric_throws_exception():
    with pytest.raises(Exception):
        permutations([])
示例#10
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文件: 049.py 项目: sanand0/euler
The arithmetic sequence, 1487, 4817, 8147, in which each of the terms increases by 3330, is unusual in two ways: (i) each of the three terms are prime, and, (ii) each of the 4-digit numbers are permutations of one another.

There are no arithmetic sequences made up of three 1-, 2-, or 3-digit primes, exhibiting this property, but there is one other 4-digit increasing sequence.

What 12-digit number do you form by concatenating the three terms in this sequence?

'''

import prime
from combinatorics import permutations

prime._refresh(10000)
for num in xrange(1000, 10000):
    if str(num).find('0') >= 0: continue

    if prime.isprime(num):
        prime_permutations = { num: 1 }
        for x in permutations(list(str(num))):
            next_num = int(''.join(x))
            if prime.isprime(next_num):
                prime_permutations[next_num] = 1

        primes = sorted(prime_permutations.keys())
        for a in xrange(0, len(primes)):
            if primes[a] == 1487: continue
            for b in xrange(a+1, len(primes)):
                c = (primes[a] + primes[b]) / 2
                if prime_permutations.has_key(c):
                    print str(primes[a]) + str(c) + str(primes[b])
                    exit()