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最小公倍数I

for i in range(0,len(L)-1): if L[i]&gt;L[i+1]:         num=L[i] else:         num=L[i+1] while(True): if(num % L[i] == 0) and (num % L[i+1] == 0):            L[i+1]=num break else:             num=num+1     L[i+1]=num print(L[i+1])

俺终于独自完成一个啦

m = max(L) L.pop(L.index(m)) L.sort() mul = 1 s = True while s: for i in L: if m * mul % i != 0: mul += 1 break else: s = False print(m * mul)

新手傻瓜办法，但好理解

m = max(L) while True: for i in L: if m%i != 0:             m+=1 break else: break print(m)

一般解法

L.sort(reverse = True) a = L[0] for b in L[1:]: for x in range(a, a * b +1 ): if x % a ==0 and x % b == 0:             a = x break print(a)

我的傻瓜办法

def min_multi(L):     lst=sorted(L)     n=1 while n&gt;=1:         num=lst[len(L)-1]*n         flag=True for i in range(len(lst)-1,-1,-1): if num%lst[i]!=0:                 flag=False break if flag: return num         n+=1 print(min_multi(L))

笨方法

def can(a, b):    # 输入一个数字a和一个数组b，如果a可以被数组中的任一元素整除，则返回True，反之返回False     i = 0 while i &lt; len(b): if a % b[i] != 0: return False         i += 1 return True L.sort() ans = L[-1] while True: if not can(ans, L):         ans += 1 else: print(ans) break

比较直接的方法

<pre><code>L.sort(reverse = True) #对由高到低列表排序 for i in range(L[-1]*L[-2],10000000000000): flag = [] for j in L: if i%j ==0: #判断i是否为L中的倍数，若i为L中每个元素的倍数，则flag中均为yes。 flag.append(&#x27;yes&#x27;) else: flag.append(&#x27;no&#x27;) if &#x27;no&#x27; not in flag: print(i) break</code></pre>

一种比较笨的办法

def fun(a, b): lcm = a*b if a &lt; b: a,b = b,a while a % b: remainder = a%b a = b b = remainder return lcm//b def solve_it(l): tmp = 1 for i in range(0,len(l)): tmp = fun(l[i],tmp) return tmp print(solve_it(L))

辗转相除

def gcd(a, b): if b == 0: return a else: return gcd(b, a % b) def lcm(a, b): return int(a * b / gcd(a, b)) def find_list_lcm(L): if len(L) == 2: return lcm(L[0], L[1]) return lcm(L[0], find_list_lcm(L[1:])) print(find_list_lcm([L]))

递归

len=len(L) p=[0 for i in range(len)] def GD(a,b): n=max(a,b) m=min(a,b) mod=n%m while mod!=0: n=m m=mod mod=n%m return m for i in range(len-1): p[i]=GD(L[i],L[i+1]) L[i+1]=L[i]*L[i+1]/p[i] print(int(L[len-1]))

一顿辗转相除法

def lcm(x, y): # 获取最大的数 if x &gt; y: greater = x else: greater = y while(True): if((greater % x == 0) and (greater % y == 0)): lcm = greater break greater += 1 return lcm for i in range(len(L)-1): L[i+1] = lcm(L[i], L[i+1]) print(L[-1])

通过了

def beis(num,minnum,maxnum): &#x27;&#x27;&#x27;返回minnum到maxnum的倍数(包括minnum和maxnum)&#x27;&#x27;&#x27; if minnum&gt;maxnum: raise ValueError(&quot;最小值不能大于最大值！&quot;) if num&lt;=0 or type(num)!=int or type(maxnum)!=int or type(minnum)!=int: raise ValueError(&quot;数值必须是非零自然数！&quot;)                num1=num                lists=[] while True: if num1&gt;=minnum: if num1&gt;maxnum:return lists                                              lists.append(num1)                               num1+=num def gbeis(num1,num2,num3): &#x27;&#x27;&#x27;返回三个数的公倍数&#x27;&#x27;&#x27; return min(list(set(beis(num1,max([num1,num2,num3]),num1*num2*num3))&amp;set(beis(num2,max([num1,num2,num3]),num1*num2*num3))&amp;set(beis(num3,max([num1,num2,num3]),num1*num2*num3)))) print(gbeis(L[0],L[1],L[2]))

为什么不能过？

for i in range(len(L)-1): 	for j in range(i,len(L)): 		if i!=j: 			if L[j]%L[i]==0: 				L[j]=L[j]//L[i] s=1 for i in range(len(L)): 	s*=L[i] print(s)

求解

<pre class="prettyprint lang-py linenums">def lcm(x, y):
 
 # 获取最大的数
 if x &gt; y:
 greater = x
 else:
 greater = y
 
 while(True):
 if((greater % x == 0) and (greater % y == 0)):
 lcm = greater
 break
 greater += 1
 
 return lcm

for i in range(len(L)-1):
 L[i+1] = lcm(L[i], L[i+1])

print(L[-1])</pre>

第60题：最小公倍数I

计算GPA

乘法运算

平分果子

路径总数