[一起来刷leetcode吧][19]--No.692 2 Keys Keyboard

这篇文章是程序自动发表的,详情可以见 这里
href="http://ounix1xcw.bkt.clouddn.com/github.markdown.css" rel="stylesheet">

这是leetcode的第692题--2 Keys Keyboard

  题目 Initially on a notepad only one character 'A' is present. You can perform two operations on this notepad for each step:

Copy All: You can copy all the characters present on the notepad (partial copy is not allowed). Paste: You can paste the characters which are copied last time. Given a number n. You have to get exactly n 'A' on the notepad by performing the minimum number of steps permitted. Output the minimum number of steps to get n 'A'. Note:The n will be in the range [1, 1000]   思路 分析可知，可以复制一个，然后粘贴n个，胆这样次数太多。要想减少次数，就要复制个数较多（注意复制个数只能越来越多），但复制多了可能会超过。可以发现，若需2n个，可复制n个，需3n个，可复制nge，粘贴2次。而且越早复制越好。这其实就是将n质因数分解。代码7--15行求的li，是用素数筛求小于一个数的所有素数，后面只需将n的所有非1因数加起来即可   

show me the code

class Solution(object):
    def minSteps(self, n):
        """
        :type n: int
        :rtype: int
        """
        li = [i for i in range(2,1001)]
        i = 1
        while  i <len(li):
            prod =2 * li[i] 
            while prod <=li[-1]:
                if prod i ''''''n li:
                    li.remove(prod)
                prod =li[i]
            i =1
        rst = 0
        for i in li:
            if i > n:break
            ct = 0
            while n%i == 0:
                n/=i
                ct  =1
            rst  =ct*i
        return rst