小Py的计算器

变态超级楼梯

完美数

import sysn=1500cnt=1if n==1: print(1) sys.exit()for i in range(2,10000000): j=i while j: if j%2==0: j=j//2 elif j%3==0: j=j//3 elif j%5==0: j=j//5 else: break if j==1: cnt+=1 if cnt==n: print(i) break

不知道为什么不可以

<pre class="prettyprint lang-py linenums"># n = 26


# 改变a, b, c各数字的倍数值
def change_multi(temp_x, min_x, multi_x):
 d = multi_x
 # 如果乘积等于最小乘积（可能有两个），则从列表中找到下一个值作为倍数值
 if temp_x == min_x:
 d = min_lst[min_lst.index(d) + 1]
 return d


num_a, num_b, num_c = 2, 3, 5
# 最小值的列表，第1到第5
min_lst = [1, 2, 3, 4, 5]
# 相应a, b, c的倍数值
multi_a, multi_b, multi_c = 3, 2, 2
temp_a, temp_b, temp_c = 0, 0, 0

# 前5个已经有了，从第6个开始
for i in range(6, n + 1):
 temp_a, temp_b, temp_c = num_a * multi_a, num_b * multi_b, num_c * multi_c
 min_temp = min(temp_a, temp_b, temp_c)
 min_lst.append(min_temp)
 multi_a = change_multi(temp_a, min_temp, multi_a)
 multi_b = change_multi(temp_b, min_temp, multi_b)
 multi_c = change_multi(temp_c, min_temp, multi_c)
 # print('#', i, ':', min_lst[-1], '\t(', multi_a, multi_b, multi_c, ')', '\t(', temp_a, temp_b, temp_c, ')')
print(min_lst[n - 1])</pre>

我的方式，循环找到2,3,5的倍数中最小的值

lst=[1] sst=set(lst) tmp = set([]) while(len(sst)&lt;=n*3): &nbsp; &nbsp; tmp = set([]) &nbsp; &nbsp; for i in sst: &nbsp; &nbsp; &nbsp; &nbsp; tmp.add(i * 2) &nbsp; &nbsp; &nbsp; &nbsp; tmp.add(i * 3) &nbsp; &nbsp; &nbsp; &nbsp; tmp.add(i * 5) &nbsp; &nbsp; sst.update(tmp) lst=list(sst) lst.sort() print lst[n-1]

set复习

直接使用集合进行不断更新，比方说一开始是1,之后用1去乘以2,3,5,得到[1,2,3,5]，再用这里面的每个数字分别去乘以2,3,，5,更新集合；这样得到我们的集合，但问题是如何确定停止的条件呢：直观上来说呢，就是要多遍历几次，直到新一轮得到的一个不存在过的数字比第n个大就可以了，但是这个条件不好判断，直接去测试得到对于目前的测试用例，只要让大小比n大3倍就可以了<pre class="prettyprint lang-py linenums">#n = 3
perfects = set([1])
while True:
	tmp = set([])
	for i in perfects:
		tmp.add(i * 2)
		tmp.add(i * 3)
		tmp.add(i * 5)
	perfects.update(tmp)
	if len(perfects) &gt; n * 3:
		break
perfects = list(perfects)
perfects.sort()
print perfects[n-1]</pre>

穷举

s=set([1]) while len(s)&lt;n*5: &nbsp; &nbsp; tmp=set([]) &nbsp; &nbsp; for i in s: &nbsp; &nbsp; &nbsp; &nbsp; tmp.add(i*2) &nbsp; &nbsp; &nbsp; &nbsp; tmp.add(i*3) &nbsp; &nbsp; &nbsp; &nbsp; tmp.add(i*5) &nbsp; &nbsp; s.update(tmp) &nbsp; &nbsp;&nbsp; s=list(s) s.sort() print s[n-1]

还好位数不多~穷举下吧~

复杂度过高

<pre><code>i=1 l = [] while 1&lt;=i&lt;=1500: if i == 1 or i%2 == 0 or i%3 == 0 or i%5 == 0:         l.append(i)         l.sort()     i+=1 print(l[n-1])</code></pre>

应该对的呀

from math import log #n=int(input()) class number(object): &nbsp; &nbsp; def __init__(self,i,j,k): &nbsp; &nbsp; &nbsp; &nbsp; self.i=i &nbsp; &nbsp; &nbsp; &nbsp; self.j=j &nbsp; &nbsp; &nbsp; &nbsp; self.k=k &nbsp; &nbsp; &nbsp; &nbsp; self.number_log=self.i * log(2,10) + self.j * log(3,10) + self.k * log(5,10) &nbsp; &nbsp; def __le__(self, other): &nbsp; &nbsp; &nbsp; &nbsp; return True if self.number_log&lt;=other.number_log else False l=[] for i in range(0,60): &nbsp; &nbsp; for j in range(0,60): &nbsp; &nbsp; &nbsp; &nbsp; for k in range(0,60): &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; l.append(number(i,j,k)) def number_compare(a): &nbsp; &nbsp; return a.number_log l.sort(key=number_compare) print(round(10**l[n-1].number_log))

大家看下代码，我在自己机子上通过了，在网站上就说有错误

ans=[] for i in range(0,60): 	for j in range(0,60): 		for k in range(0,60): 			ans.append(2**i*3**j*5**k) ans.sort() print ans[n-1]

第133题：完美数

球迷购票问题

吃糖果

新年祝福

又是字符串