Question

我试图在projecteuler中解决问题5.问题如下

2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.

What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?

我写了一个简单的python程序

primes = [2,3,5,7,11,13,17,19]
prod = 1

#function returns the list of prime factors of 'n'
def find_pf(n):
        pf = []
        for j in primes:
                if i % j == 0:
                        pf.append(j)

        return pf

#multiplies all the prime factors of all the numbers 
#from 1[1..20]
for i in range(1,21):
        lst = find_pf(i)
        for i in lst:
                prod *= i 


#verifies that 'n' is diviible evenly by [1..20]
count = 0
def test_n(n):
        global count
        for i in range(1,21):
                if n % i != 0:
                        count += 1


print ('smallest number divisible by [1..20] {}'.format(prod))
test_n(prod)
print('Total failures {}'.format(count))

上述程序获得的结果是

smallest number divisible by [1..20] 1055947052160000
Total failures 0

答案1055947052160000不正确。？有人可以指出上述程序有什么问题吗？或者建议正确的方法来解决这个问题？

Answer 1

错误在

#multiplies all the prime factors of all the numbers 
#from 1[1..20]
for i in range(1,21):
    lst = find_pf(i)
    for i in lst:
            prod *= i

你只对任何素数的最高必要力量感兴趣。例如，您要查找的值应该可被16整除。您的代码会查找可被2 * 4 * 8 * 16整除的数字。

Answer 2

您的代码正在寻找太多素数。寻找最大的就足够了：例如如果数字可以被16分割，则它已经可以被8分割。

primes = [2,3,5,7,11,13,17,19]
prod = 1 
for p in primes:
    n = 2
    prod *= p
    while (p**n < 21):
        prod *= p
        n += 1

print prod

你来了

232792560

Answer 3

def lcm(*values):
    values = [value for value in values]
    if values:
        n = max(values)
        m = n
        values.remove(n)
        while any(n % value for value in values):
            n += m
        return n
    return 0

    reduce(lcm, range(1, 20))
In [3]: reduce(lcm, range(1, 20))
Out[3]: 232792560

reduce将“两个参数的函数累加到从左到右的可迭代项中，以便将迭代减少到单个值。”

 def add_nums(a,b):
    print a,b # added print to show values
    return a+b
 reduce(add_nums, [1, 2, 3, 4, 5]) #calculates ((((1+2)+3)+4)+5)

欧拉5：最小的数字，可以除以1到20之间的数字

3 个答案: