惯性聚合 高效追踪和阅读你感兴趣的博客、新闻、科技资讯
阅读原文 在惯性聚合中打开

推荐订阅源

Hacker News - Newest:
Hacker News - Newest: "LLM"
AI
AI
T
Troy Hunt's Blog
GbyAI
GbyAI
H
Hacker News: Front Page
SecWiki News
SecWiki News
V2EX - 技术
V2EX - 技术
A
About on SuperTechFans
人人都是产品经理
人人都是产品经理
Hacker News: Ask HN
Hacker News: Ask HN
S
Secure Thoughts
Last Week in AI
Last Week in AI
MyScale Blog
MyScale Blog
L
LINUX DO - 最新话题
C
CERT Recently Published Vulnerability Notes
C
Cyber Attacks, Cyber Crime and Cyber Security
O
OpenAI News
S
SegmentFault 最新的问题
Y
Y Combinator Blog
C
CXSECURITY Database RSS Feed - CXSecurity.com
The Cloudflare Blog
P
Proofpoint News Feed
Engineering at Meta
Engineering at Meta
奇客Solidot–传递最新科技情报
奇客Solidot–传递最新科技情报
美团技术团队
Google DeepMind News
Google DeepMind News
Simon Willison's Weblog
Simon Willison's Weblog
Know Your Adversary
Know Your Adversary
K
Kaspersky official blog
T
The Exploit Database - CXSecurity.com
S
Securelist
cs.CL updates on arXiv.org
cs.CL updates on arXiv.org
B
Blog RSS Feed
L
Lohrmann on Cybersecurity
Spread Privacy
Spread Privacy
博客园 - 司徒正美
Stack Overflow Blog
Stack Overflow Blog
博客园 - Franky
The GitHub Blog
The GitHub Blog
B
Blog
F
Fortinet All Blogs
I
InfoQ
C
Check Point Blog
Webroot Blog
Webroot Blog
博客园 - 叶小钗
P
Privacy International News Feed
Latest news
Latest news
Forbes - Security
Forbes - Security
博客园 - 三生石上(FineUI控件)
N
News | PayPal Newsroom

博客园 - Zero Lee

调用栈(call stack) 关于STL allocator Calculate maximum sum of any subarray set Convert one binary search tree to double-linked list 设计包含min函数的栈 类模板的模板友元函数定义 一道百度的面试题解答 非printf形式的十六进制和二进制打印(雅虎面试题) 一道腾讯面试题 (转)C++中extern “C”含义深层探索 selection algorithm to select nth small elements based on partition 删除与某个字符相邻且相同的字符 产生全排列的方法解析 一组数的全排列和组合程序实现 求一个正整数的平方根程序实现 [转]多线程队列的算法优化 [转载] STL allocator的介绍和一个基于malloc/free的allocator的简单实现 如何将一片内存链接成链表 One simple counted object pointer
Calcuate power n of x recursively
Zero Lee · 2012-06-17 · via 博客园 - Zero Lee

Given x and n, calculate its power n:

 1 int power(int x, int n)
 2 {
 3    if (n==0) 
 4      return 1;
 5    else (n%2==0) 
 6      return power(x*x, n/2);
 7    else 
 8      return x*power(x*x, n/2);
 9 }
10   
11