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

  1. 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 ...

  2. 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, ...

  3. 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 ...

  4. Atcoder B - Moderate Differences

    http://agc017.contest.atcoder.jp/tasks/agc017_b B - Moderate Differences Time limit : 2sec / Memory ...

  5. CCI_chapter 19 Moderate

    19 1  Write a function to swap a number in place without temporary variables void swap(int &a, i ...

  6. [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 ...

  7. 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_{ ...

  8. 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 ...

  9. 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

  1. 将mysql的查询结果输出到文件

    在sql命令中我们可以查询到前数行的表,同时也可以将查询结果输出到txt文档 语句:select * from tablename into outfile 'filename.txt'; 例如:se ...

  2. FIR滤波器(1)- 基础知识

    FIR滤波器广泛应用于数字信号处理中,主要功能就是将不感兴趣的信号滤除,留下有用信号.FIR滤波器是全零点结构,系统永远稳定:并且具有线性相位的特征,在有效频率范围内所有信号相位上不失真.在无线通信收 ...

  3. 数字图象处理MATLAB学习

    diagram = imread('C:\Users\Administrator\Desktop\Compressed\fiter\lena256.jpg') %diagram = rgb2gray( ...

  4. WebStorm 快捷键收藏

    快捷键 Ctrl+/ 或 Ctrl+Shift+/ 注释(// 或者/-/ ) Ctrl+X 或 Ctrl+Y 删除一行 Shift+F6 重构-重命名 Alt+~vcs操作 ​ Alt+~ 7关闭重 ...

  5. CenterOS中安装Redis及开机启动设置

    Redis安装 从官方下载最新Redis进行安装,官网地址:http://redis.io/download $ wget http://download.redis.io/releases/redi ...

  6. linux查看硬件信息的命令(图文)

    发布:脚本学堂/Linux命令  编辑:JB02   2013-12-23 21:48:18  [大 中 小] 转自:http://www.jbxue.com/LINUXjishu/14996.htm ...

  7. 有关Mysql连接问题

    问题一——Mysql number Error 2003 MySQL连接错误.错误代码10061.10061一般是Mysql服务没启动.或Mysql服务器无法连接 . 在程序栏找到Mysql\Mysq ...

  8. PHP获取时间日期的多种方法

    分享下PHP获取时间日期的多种方法. <?php echo "今天:".date("Y-m-d")."<br>";     ...

  9. posix 消息队列

    注意 在涉及到posix消息的函数时, gcc 编译时要加-lrt参数, 如 gcc -lrt unpipc.c mqpack.c send.c -o send gcc -lrt unpipc.c m ...

  10. Ubuntu下编程环境GNU安装

    ubuntu下C编程   环境搭建 其实,linux下写C也是很容易的.IDE的话用 eclipse 集成 CDT 模块就行了.当然这属于重量级的了,就如同VC++之于windows一样.那有没有像T ...