单身狗

小Py的计算器

完美数II

在第一个题目里面是用估算的方式，这个题目呢，要精确方式就是每次都去找当前迭代次数的对应的大小，比如第2次循环就找找第2大的代码如下<pre class="prettyprint lang-py linenums">def perfect(n):
	if n == 1:
		return 1
	p2 = [2]
	p3 = [3]
	p5 = [5]
	for i in range(2, n+1):
		cur = min(p2[0], p3[0], p5[0])
		if cur == p2[0]:
			p2.pop(0)
		if cur == p3[0]:
			p3.pop(0)
		if cur == p5[0]:
			p5.pop(0)
		p2.append(cur * 2)
		p3.append(cur * 3)
		p5.append(cur * 5)
	return cur

#n = 3
print perfect(n)
</pre>

逐个获取

def isperfect(n): &nbsp; &nbsp; while n%2==0: &nbsp; &nbsp; &nbsp; &nbsp; n=n//2 &nbsp; &nbsp; while n%5==0: &nbsp; &nbsp; &nbsp; &nbsp; n=n//5 &nbsp; &nbsp; while n%3==0: &nbsp; &nbsp; &nbsp; &nbsp; n=n//3 &nbsp; &nbsp; if n==1: &nbsp; &nbsp; &nbsp; &nbsp; return True &nbsp; &nbsp; else: &nbsp; &nbsp; &nbsp; &nbsp; return False count=0 m=0 while(count&lt;=n): &nbsp; &nbsp; if(isperfect(m)): &nbsp; &nbsp; &nbsp; &nbsp; count=count+1 &nbsp; &nbsp; m=m+1 print(m-1)

复杂度过高

def perfect(n): &nbsp; &nbsp; if n==1:return 1 &nbsp; &nbsp; l2,l3,l5=[2],[3],[5] &nbsp; &nbsp; for i in range(2,n+1): &nbsp; &nbsp; &nbsp; &nbsp; i2,i3,i5 = l2[0],l3[0],l5[0] &nbsp; &nbsp; &nbsp; &nbsp; x=min(i2,i3,i5) &nbsp; &nbsp; &nbsp; &nbsp; if x==l2[0]:l2.pop(0) &nbsp; &nbsp; &nbsp; &nbsp; if x==l3[0]:l3.pop(0) &nbsp; &nbsp; &nbsp; &nbsp; if x==l5[0]:l5.pop(0) &nbsp; &nbsp; &nbsp; &nbsp; l2.append(x*2) &nbsp; &nbsp; &nbsp; &nbsp; l3.append(x*3) &nbsp; &nbsp; &nbsp; &nbsp; l5.append(x*5) &nbsp; &nbsp; return x print perfect(n)

第135题：完美数II

分解 n!

简单题之列表转换

正方形拼接

寻找最小的数