我目前的编码曾经是一个goto但我被告知不再使用goto,因为它不赞成。我有麻烦改变它说一段时间循环。我对C#和一般编程都很陌生,所以这对我来说是一些全新的东西。任何帮助,将不胜感激。实际问题是输入两个数字并找到最低公倍数。
以下是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;
}
}
答案 0 :(得分:33)
这是一个更有效和简洁的最小公倍数计算实现,利用它与最大公因数(又名最大公约数)的关系。这个最大公因子函数使用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;
}
答案 1 :(得分:11)
试试这个:
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();
}
}
输出:
Enter 2 numbers to find LCM
8
12
LCM of 8 and 12 is 24
答案 2 :(得分:1)
试试这个
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
答案 3 :(得分:0)
这是一个递归解决方案。这可能是一些面试问题。我希望它有所帮助
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;
}
}
答案 4 :(得分:0)
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;
}
}
}
答案 5 :(得分:0)
这是寻找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);
}
答案 6 :(得分:-1)
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;
}
}