leetcode650 2 Keys Keyboard
思路:
把给定的数分解质因子之后,对于每一个质因子x,都需要x次操作(一次copy all操作和x-1次paste),所以答案就是对分解后的所有质因子求和。
实现:
class Solution
{
public:
int minSteps(int n)
{
int sum = ;
for (int i = ; i <= n; i++)
{
while (n % i == )
{
sum += i;
n /= i;
}
}
return sum;
}
};
leetcode650 2 Keys Keyboard的更多相关文章
- leetcode650—2 Keys Keyboard
		
Initially on a notepad only one character 'A' is present. You can perform two operations on this not ...
 - [LeetCode] 4 Keys Keyboard 四键的键盘
		
Imagine you have a special keyboard with the following keys: Key 1: (A): Print one 'A' on screen. Ke ...
 - [LeetCode] 2 Keys Keyboard 两键的键盘
		
Initially on a notepad only one character 'A' is present. You can perform two operations on this not ...
 - LeetCode解题报告—— 2 Keys Keyboard & Longest Palindromic Substring & ZigZag Conversion
		
1. Longest Palindromic Substring Given a string s, find the longest palindromic substring in s. You ...
 - Leetcode 之 Keys Keyboard
		
1. 2 Keys Keyboard 先把dp的最小不走都设置为无穷大(Integer.MAX_VALUE),初始化条件:dp[0] = dp[1] = 0,状态转移方程为dp[i] = Math.m ...
 - LeetCode 4 Keys Keyboard
		
原题链接在这里:https://leetcode.com/problems/4-keys-keyboard/description/ 题目: Imagine you have a special ke ...
 - [leetcode] 650. 2 Keys Keyboard (Medium)
		
解法一: 暴力DFS搜索,对每一步进行复制还是粘贴的状态进行遍历. 注意剪枝的地方: 1.当前A数量大于目标数量,停止搜索 2.当前剪贴板数字大于等于A数量时,只搜索下一步为粘贴的状态. Runtim ...
 - [LeetCode] 650. 2 Keys Keyboard 两键的键盘
		
Initially on a notepad only one character 'A' is present. You can perform two operations on this not ...
 - [LeetCode] 651. 4 Keys Keyboard 四键的键盘
		
Imagine you have a special keyboard with the following keys: Key 1: (A): Print one 'A' on screen. Ke ...
 
随机推荐
- c++11中的线程、锁和条件变量
			
void func(int i, double d, const string& s) { cout << i << ", " << d ...
 - Hadoop 知识
			
Map Reduce & YARN 简介 Apache Hadoop 是一个开源软件框架,可安装在一个商用机器集群中,使机器可彼此通信并协同工作,以高度分布式的方式共同存储和处理大量数据.最初 ...
 - 实战恢复2950交换机的IOS
			
本来想用两台交换机做实验的,可是通过console口进入其中一台交换机后却发现这个台交换器的IOS文件丢失了 本来正常进入交换机后应该是首先进入到用户模式的,而且提示符应该是">&qu ...
 - VMWare9下基于Ubuntu12.10搭建Hadoop-1.2.1集群—整合Zookeeper和Hbase
			
VMWare9下基于Ubuntu12.10搭建Hadoop-1.2.1集群-整合Zookeeper和Hbase 这篇是接着上一篇hadoop集群搭建进行的.在hadoop-1.2.1基础之上安装zoo ...
 - AES加密算法的C++实现
			
摘要:作为新一代的加密标准,AES 旨在取代 DES(请看<DES加密算法的C++实现>),以适应当今分布式开放网络对数据加密安全性的要求.本文在分析了 AES 加密原理的基础上着重说明了 ...
 - Appium、selenium与Robot Framework
			
Robot Framework + Appium Appium-Python-Client: 与Appium Library和Appium Server的交互Appium Library通过Appii ...
 - Java 中 modifer 'public' is reduntant for interface methods
			
http://androidren.com/index.php?qa=322&qa_1=java-%E4%B8%AD-modifer-public-is-reduntant-for-inter ...
 - codeforces 552 C Vanya and Scales
			
这个题的意思就是给出一个数m.以及一个以1为首元素.w为比例常数的等比数列,数列长度为101,数列中每一个数字最多仅仅能用一次.问是否存在xa+wb+--=wc+wd+--+we+m. 非常显然,换句 ...
 - gitlab常见错误解决办法
			
1 fail: gitlab-workhorse: runsv not running systemctl start gitlab-runsvdir.service systemctl status ...
 - 每个sql结果返回值的字节大小的峰值统计
			
#5KB/per结合数据库配置mylimit = 50