欧几里得算法求最大公约数

  • If A = 0 then GCD(A,B)=B, since the GCD(0,B)=B, and we can stop.
  • If B = 0 then GCD(A,B)=A, since the GCD(A,0)=A, and we can stop.
  • Write A in quotient remainder form (A = B⋅Q + R)
  • Find GCD(B,R) using the Euclidean Algorithm since GCD(A,B) = GCD(B,R)

这里Q是正整数.

Example:

Find the GCD of 270 and 192

  • A=270, B=192
  • A ≠0
  • B ≠0
  • Use long division to find that 270/192 = 1 with a remainder of 78. We can write this as: 270 = 192 * 1 +78
  • Find GCD(192,78), since GCD(270,192)=GCD(192,78)

A=192, B=78

  • A ≠0
  • B ≠0
  • Use long division to find that 192/78 = 2 with a remainder of 36. We can write this as:
  • 192 = 78 * 2 + 36
  • Find GCD(78,36), since GCD(192,78)=GCD(78,36)

A=78, B=36

  • A ≠0
  • B ≠0
  • Use long division to find that 78/36 = 2 with a remainder of 6. We can write this as:
  • 78 = 36 * 2 + 6
  • Find GCD(36,6), since GCD(78,36)=GCD(36,6)

A=36, B=6

  • A ≠0
  • B ≠0
  • Use long division to find that 36/6 = 6 with a remainder of 0. We can write this as:
  • 36 = 6 * 6 + 0
  • Find GCD(6,0), since GCD(36,6)=GCD(6,0)

A=6, B=0

  • A ≠0
  • B =0, GCD(6,0)=6

So we have shown:

GCD(270,192) = GCD(192,78) = GCD(78,36) = GCD(36,6) = GCD(6,0) = 6

GCD(270,192) = 6

应用:

int gcd(int a, int b) {
while(b){
int r = a % b;
a = b;
b = r;
}
return a;
}

Greatest common divisor(gcd)的更多相关文章

  1. 最大公约数Greatest Common Divisor(GCD)

    一 暴力枚举法 原理:试图寻找一个合适的整数i,看看这个整数能否被两个整形参数numberA和numberB同时整除.这个整数i从2开始循环累加,一直累加到numberA和numberB中较小参数的一 ...

  2. upc组队赛17 Greatest Common Divisor【gcd+最小质因数】

    Greatest Common Divisor 题目链接 题目描述 There is an array of length n, containing only positive numbers. N ...

  3. [UCSD白板题] Greatest Common Divisor

    Problem Introduction The greatest common divisor \(GCD(a, b)\) of two non-negative integers \(a\) an ...

  4. 845. Greatest Common Divisor

    描述 Given two numbers, number a and number b. Find the greatest common divisor of the given two numbe ...

  5. 2018CCPC桂林站G Greatest Common Divisor

    题目描述 There is an array of length n, containing only positive numbers.Now you can add all numbers by ...

  6. CCPC2018 桂林 G "Greatest Common Divisor"(数学)

    UPC备战省赛组队训练赛第十七场 with zyd,mxl G: Greatest Common Divisor 题目描述 There is an array of length n, contain ...

  7. greatest common divisor

    One efficient way to compute the GCD of two numbers is to use Euclid's algorithm, which states the f ...

  8. 最大公约数和最小公倍数(Greatest Common Divisor and Least Common Multiple)

    定义: 最大公约数(英语:greatest common divisor,gcd).是数学词汇,指能够整除多个整数的最大正整数.而多个整数不能都为零.例如8和12的最大公因数为4. 最小公倍数是数论中 ...

  9. hdu 5207 Greatest Greatest Common Divisor 数学

    Greatest Greatest Common Divisor Time Limit: 1 Sec  Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/ ...

随机推荐

  1. 专门讲讲这个MYSQL授权当中的with grant option的作用

    对象的owner将权限赋予某个用户(如:testuser1) grant select ,update on bd_corp to testuser1 [with grant option ]1.如果 ...

  2. c++(smart pointer)

    (一)首先对智能指针有一些概念性的了解 **********本部分内容摘自开源中国社区http://my.oschina.net/u/158589/blog/28994******** 1.什么是智能 ...

  3. 强大疯狂的qttools

    就是有点疑惑,为什么不整合到QT主项目中呢? 有空好好看看: https://github.com/qtproject/qttools/tree/dev/src ------------------- ...

  4. Android attrs.xml文件中属性类型format值的格式

    "reference" //引用 "color" //颜色 "boolean" //布尔值 "dimension" // ...

  5. 【转】C语言文件操作解析(三)

    原文网址:http://www.cnblogs.com/dolphin0520/archive/2011/10/07/2200454.html C语言文件操作解析(三) 在前面已经讨论了文件打开操作, ...

  6. 【KMP+DP】Count the string

    KMP算法的综合练习 DP很久没写搞了半天才明白.本题结合Next[]的意义以及动态规划考察对KMP算法的掌握. Problem Description It is well known that A ...

  7. Lenovo k860i 移植Android 4.4 cm11进度记录【下篇--实时更新中】

    2014.8.24 k860i的cm11的移植在中断了近两三个月之后又開始继续了,进度记录的日志上一篇已经没什么写的了,就完结掉它吧,又一次开一篇日志做下篇好了.近期的战况是,在scue同学的努力之下 ...

  8. 关于MyEclipse查看底层源码出现source not found的问题(MyEclipse、Eclipse配置JAD)

    一.MyEclipse 第一步:      下载jad.exe文件:jad下载地址 eclipse插件:net.sf.jadclipse_版本号.jar下载地址一 net.sf.jadclipse_版 ...

  9. 在VMware中为Linux系统安装VM-Tools的详解教程

    在安装Linux的虚拟机中,单击“虚拟机”菜单下的“安装Vmware-Tools”. 先介绍一下下面安装该工具时要用到的几个目录: /mnt 挂载目录,用来临时挂载别的文件系统,硬件设备 /tmp临时 ...

  10. [serverlet][转载: 深入理解HTTP Session]

    [serverlet][转载: 深入理解HTTP Session] 标签(空格分隔): 未分类 原创作品,允许转载,转载时请务必以超链接形式标明文章 原始出处 .作者信息和本声明.否则将追究法律责任. ...