Project Euler每帖一题（005）

题目：

引用

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?

这个题目解得很差，写了个函数，传入1到10，速度还行，传入1到20，竟然计算了XX秒...看来算法太烂了，自我笔试下@@@@

我的python代码：

def gcd(list):
    n = max(list)
    flag = True
    i = 1
    while True:
        for j in list:
            if (n * i) % j != 0:
                flag = False
                break
        if flag:
            return n * i
        else:
            flag = True
        i += 1
        
print gcd(range(1,21))

看了别人的解法，选了个比较大的基数，思路差不多：

i = 9699690 # Continuation.9699690=2*3*5*7*11*13*17*19
while 1:
	print "Testing: %s" % i
	for x in range(1, 21):
		print "%s:" % x,
		if i % x != 0:
			print "Fail"
			break
		else:
			print "Pass"
	else:
		print "Solution: %s" % i
		break
	i+=9699690

一种比较好的解决方案，核心思想是利用分解质因数来求解最大公倍数：

num = 20
'''
isSolved = False
i = 0
while not isSolved:
    for ii in range(2,num+1):
        if i % ii == 0 and i != 0:
            if ii == num: isSolved = True
        else:
            i = i + num
            break
print 'Lowest divisor: ', i
'''
 
def isPrime(a):
    prime=True
    for i in range(2,a):
        if (a % i == 0): prime=False
    if prime: return True
    else: return False
 
factorz = []
primeFactorz = []
def getPrimeFactors(a):
    gotFactor = False
    for ii in range(2,a+1):
        if a % ii == 0:
            gotFactor = True
            b = a/ii
            if isPrime(ii): factors.append(ii)
            else: getPrimeFactors(ii)
            if isPrime(b) and b != 1: factors.append(b)                
            else: getPrimeFactors(b)
        if gotFactor: break
 
for i in range(2,num+1):
    factors=[]
    getPrimeFactors(i)
    factorz.append(factors)
 
print 'Printing Factors of 2 through', num
for i in factorz:
    print i
 
new=[]
total=[]
def getHighestPow(a):
    counthigh = 0
    for i in factorz:
        count = 0
        for ii in i:
            if ii == a:
                count = count + 1
        if count > counthigh:
            counthigh = count
    if a not in new:
        new.append(a)
        total.append(a**counthigh)
 
for i in factorz:
    for ii in i:
        getHighestPow(ii)
 
sum=1
for i in total:
    sum = sum * i
 
print 'Lowest Common Multiple: ', sum