C# 最小公倍数
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原文地址: http://stackoverflow.com/questions/13569810/
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Least Common Multiple
提问by Tristan M.
I have the current coding which used to be a goto but I was told to not use goto anymore as it is frowned upon. I am having troubles changing it into for say a while loop. I am fairly new to C# and programming in general so some of this is completely new stuff to me. Any help would be appreciated. The actual question is input two numbers and find the lowest common multiple.
我有当前的编码，它曾经是一个 goto，但我被告知不要再使用 goto，因为它不受欢迎。我在将其更改为例如 while 循环时遇到了麻烦。我对 C# 和一般编程相当陌生，所以其中一些对我来说是全新的东西。任何帮助，将不胜感激。实际问题是输入两个数字并找到最小的公倍数。
Here is the original with goto:
这是带有 goto 的原文：
BOB:
if (b < d)
{
a++;
myInt = myInt * a;
b = myInt;
myInt = myInt / a;
if (b % myInt2 == 0)
{
Console.Write("{0} ", h);
Console.ReadLine();
}
}
if (d < b)
{
c++;
myInt2 = myInt2 * c;
d = myInt2;
myInt2 = myInt2 / c;
if (d % myInt == 0)
{
Console.Write("{0} ", t);
Console.ReadLine();
}
else
{
goto BOB;
}
}
else
{
goto BOB;
}
}
采纳答案by user1211929
Try This:
尝试这个：
using System;
public class FindLCM
{
public static int determineLCM(int a, int b)
{
int num1, num2;
if (a > b)
{
num1 = a; num2 = b;
}
else
{
num1 = b; num2 = a;
}
for (int i = 1; i < num2; i++)
{
if ((num1 * i) % num2 == 0)
{
return i * num1;
}
}
return num1 * num2;
}
public static void Main(String[] args)
{
int n1, n2;
Console.WriteLine("Enter 2 numbers to find LCM");
n1 = int.Parse(Console.ReadLine());
n2 = int.Parse(Console.ReadLine());
int result = determineLCM(n1, n2);
Console.WriteLine("LCM of {0} and {1} is {2}",n1,n2,result);
Console.Read();
}
}
Output:
输出：
Enter 2 numbers to find LCM
8
12
LCM of 8 and 12 is 24
回答by Tilak
Try this
尝试这个
int number1 = 20;
int number2 = 30;
for (tempNum = 1; ; tempNum++)
{
if (tempNum % number1 == 0 && tempNum % number2 == 0)
{
Console.WriteLine("L.C.M is - ");
Console.WriteLine(tempNum.ToString());
Console.Read();
break;
}
}
// output -> L.C.M is - 60
回答by AffluentOwl
Here's a more efficient and concise implementation of the Least Common Multiple calculation which takes advantage of its relationship with the Greatest Common Factor (aka Greatest Common Divisor). This Greatest Common Factor function uses Euclid's Algorithm which is more efficient than the solutions offered by user1211929 or Tilak.
这是最小公倍数计算的更有效和简洁的实现，它利用了它与最大公因数（又名最大公约数）的关系。这个最大公因数函数使用 Euclid 算法，它比 user1211929 或 Tilak 提供的解决方案更有效。
static int gcf(int a, int b)
{
while (b != 0)
{
int temp = b;
b = a % b;
a = temp;
}
return a;
}
static int lcm(int a, int b)
{
return (a / gcf(a, b)) * b;
}
For more information see the Wikipedia articles on computing LCMand GCF.
回答by Vlada
Here is one recursive solution. It might be on some interview question. I hope it helps
这是一种递归解决方案。它可能是在一些面试问题上。我希望它有帮助
public static int GetLowestDenominator(int a, int b, int c = 2)
{
if (a == 1 | b == 1) {
return 1;
}
else if (a % c == 0 & b % c == 0)
{
return c;
}
else if (c < a & c < b)
{
c += 1;
return GetLowestDenominator(a, b, c);
}
else
{
return 0;
}
}
回答by Arthur
int num1, num2, mull = 1;
num1 = int.Parse(Console.ReadLine());
num2 = int.Parse(Console.ReadLine());
for (int i = 1; i <= num1; i++)
{
for (int j = 1; j <= num2; j++)
{
if (num1 * j == num2 * i)
{
mull = num2 * i;
Console.Write(mull);
return;
}
}
}
回答by Amit Kumar Verma
Here is much optimized solution for finding LCM.
这是用于查找 LCM 的优化解决方案。
private static int lcmOfNumbers(int num1, int num2)
{
int temp = num1 > num2 ? num1 : num2;
int counter = 1;
while (!((temp* counter++) % num1 == 0 && (temp* counter++) % num2 == 0)) {
}
return temp* (counter-2);
}
回答by Ahsan
int n1 = 13;
int n2 = 26;
for (int i = 2; i <= n1; i++)
{
if (n1 % i == 0 && n2 % i == 0)
{
Console.WriteLine("{0} is the LCM of {1} and
{2}",i,n1,n2);
break;
}
}
