Chp17: Moderate
17.1 swap a number in place.(without temporary variables)
a = a ^ b;
b = a ^ b;
a = a ^ b;
17.3 Write a function which computes the number of trailing zeros in n factorial.
To count the number of zeros, we only need to count the pairs of multiples of 5 and 2. There will always be more multiples of 2 than 5 though, so, simply counting the number of multiples of 5 is sufficient.
public int count(int num){
int count = 0;
if(num < 0) return -1;
for(int i = 5; num /i > 0; i *= 5)
count += num / i;
return count;
}
17.4 Write a method which finds the maximum of two numbers. You should not use if-else or any other comparison operator.
public int flip(int bit){
return 1 ^ bit;
}
public int sign(int a){
return flip((a >> 31) & 0x1);
}
public int getMax(int a, int b){
int c = a - b;
int sa = sign(a); //if a >= 0 : 1 other : 0
int sb = sign(b); //if b >= 1 : 1 other : 0
int sc = sign(c); //depends on whether a - b overflows, like a = INT_MAX, b < 0
int use_sign_a = sa ^ sb;
int use_sign_c = flip(sa ^ sb);
int k = use_sign_a * sa + use_sign_c * sc;
int q = flip(k);
return a * k + b * q;
}
17.9 Design a method to find the frequency of occurrences of any given word in a book.
The first question that you should ask is if you will be doing this operation once or repeatedly.
Solution: Single Query
go through the book, word by word, count the number of times that words appears. O(n)
Solution: Repetitive Queries
do pre-processing on the book! create a hash table which maps from a word to its frequency.
Hashtable<String, Integer> setupDic(String[] book){
Hashtable<String, Integer> table = new Hashtable<String, Integer>();
for(String word : book){
word = word.toLowerCase();
if(word.trim() != ""){
if(!table.containsKey(word)) table.put(word, 0);
table.put(word, table.get(word) + 1);
}
}
return table;
}
int getFrequency(Hashtable<String, Integer> table, String word){
if(table == null || word == null) return -1;
word = word.toLowerCase();
if(table.containsKey(word)) return table.get(word);
return 0;
}
17.11 Implement a method rand7() given rand5(). Given a method that generates a random number between 0 and 4, write a method that generates a random number between 0 and 6.
Nondeterministic Number of Calls
public int rand7(){
while(true){
int num = 5 * rand5() + rand5();
if(num < 21) return num % 7;
}
}
Chp17: Moderate的更多相关文章
- Moderate 加入空格使得可辨别单词数量最多 @CareerCup
递归题目,注意结合了memo的方法和trie的应用 package Moderate; import java.util.Hashtable; import CtCILibrary.AssortedM ...
- found 12 vulnerabilities (7 moderate, 5 high) run `npm audit fix` to fix them, or `npm audit` for details
npm 安装包之后,如果出现类似下面的信息 found 12 vulnerabilities (7 moderate, 5 high) run `npm audit fix` to fix them, ...
- 题解——ATCoder AtCoder Grand Contest 017 B - Moderate Differences(数学,构造)
题面 B - Moderate Differences Time limit : 2sec / Memory limit : 256MB Score : 400 points Problem Stat ...
- Atcoder B - Moderate Differences
http://agc017.contest.atcoder.jp/tasks/agc017_b B - Moderate Differences Time limit : 2sec / Memory ...
- CCI_chapter 19 Moderate
19 1 Write a function to swap a number in place without temporary variables void swap(int &a, i ...
- [图形学] Chp17 OpenGL光照和表面绘制函数
这章学了基本光照模型,物体的显示受到以下效果影响:全局环境光,点光源(环境光漫反射分量,点光源漫反射分量,点光源镜面反射分量),材质系数(漫反射系数,镜面反射系数),自身发光,雾气效果等.其中点光源有 ...
- Atcoder | AT2665 【Moderate Differences】
又是一道思路特别清奇的题qwq...(瞪了一上午才发现O(1)的结论...差点还想用O(n)解决) 问题可以转化为是否能够由\(f_{1}=a\)通过\(\pm x \in[c,d]\)得到\(f_{ ...
- Atcoder #017 agc017 B.Moderate Differences 思维
LINK 题意:给出最左和最右两个数,要求往中间填n-2个数,使得相邻数间差的绝对值$∈[L,R]$ 思路:其实也是个水题,比赛中大脑宕机似的居然想要模拟构造一个数列,其实我们只要考虑作为结果的数,其 ...
- Fedora 24中的日志管理
Introduction Log files are files that contain messages about the system, including the kernel, servi ...
随机推荐
- opencv 手写选择题阅卷 (二)字符识别
opencv 手写选择题阅卷 (二)字符识别 选择题基本上只需要识别ABCD和空五个内容,理论上应该识别率比较高的,识别代码参考了网上搜索的代码,因为参考的网址比较多,现在也弄不清是参考何处的代码了, ...
- While readingiphone真机无法显示图片,而模拟器可以正常显示
可能,很多开发IOS程序的遇到过在模拟器里,加载图片都是正常的,但是在真机里就会出现图片资源不能加载的问题. 其中一种原因是,在Simulator里面,例如:图片资源名称为:a.PNG,在代码你里,你 ...
- How to Fix Missing TortoiseSVN File Status Icons in Windows
For many Windows-based developers, Subversion and TortoiseSVN is a great source control solution. It ...
- solr6安装部署
难得写篇自己的原创文档了,哈哈哈,原谅我知识浅薄,积淀太少 一.涉及到的软件和环境jdk1.8.0_92,tomcat8,zookeeper3.4.8,solr6.1.0(solr6需要jdk8以上环 ...
- jQuery插件使用大全
1.jQuery datepicker日历插件使用说明 http://wenku.baidu.com/view/12804e1e59eef8c75fbfb3e3 2.jqueryFileUpload插 ...
- 用委托在listbox中异步显示信息,解决线程间操作无效,从不是创建控件的线程访问它
//创建一个委托,是为访问listbox控件服务的. public delegate void UpdateTxt(string msg); //定义一个委托变量 public UpdateTxt u ...
- 在Linux中三种让crontab每秒执行任务的方法
第一种方法: 1.创建脚本文件 cat phplog.sh 2.编辑脚本内容 #!/bin/bash while : ;do /home/scripts.sh 2>/dev/null & ...
- jQuery select的操作代码
jQuery對select的操作的实际应用代码. //改變時的事件 复制代码代码如下: $("#testSelect").change(function(){ //事件發生 j ...
- Linux之父Linus Torvalds:讨厌C++
"Linux内核的创始人Linus Torvalds最近在一封邮件中说明了内核开发需要使用C语言而非C++的理由.在庞大的项目中,人们对不是自己开发的模块并不了解,能快速理解其他模块中函数的 ...
- 批量kill mysql processlist进程
如果大批量的操作能够通过一系列的select语句产生,那么理论上就能对这些结果批量处理.但是mysql并没用提供eval这样的对结果集进行分析操作的功能.所以只能现将select结果保存到临时文件中, ...