标签:i++ new 个数 交换 use ogr 排列 next ali
int number1=10,number2=20,temp=0;
temp=number1;
number1=number2;
number2=temp;
通过 *和/
int number1=10,number2=20;
number1=number1*number2; //200
number2=number1/number2; //10
number1= number1/number2; //20
通过 +和-
int number1=10,number=20;
number1=number1+number2; //30
number2=number1-number2; //10
number1=number1-number2; //20
string name1="Dotnet", name2="C#";
name1=name1+name2; //DotnetC#
name2=name1.Substring(0,name1.Length-name2.Length); //Dotnet
name1=name1.Substring(name2.Length); //C#
斐波那契数列是按以下顺序排列的数字序列:
在C#中实现斐波那契数列的方法
迭代打印斐波那契数列的方法:
int firstNumber=0;SecondNumber=1;nextNumber;
Console.Write("输入打印的元素数量:");
int numberOfElement=int.Parse(Console.ReadLine());
if(numberOfElements>2)
{
Console.Write($"{firstNumber} {SecondNumber}");
for(int i=2;i<numberOfElements;i++)
{
nextNumber=firstNumber+SecondNumber;
Console.Write($"{nextNumber} ");
firstNumber=SecondNumber;
SecondNumber=nextNumber;
}
}
递归打印斐波那契数列的方法:
Console.Write("输入打印的元素数量:");
int numberOfElement=int.Parse(Console.ReadLine());
FibonacciSeries(0,1,1,numberOfElement);
//递归打印方法
FibonacciSeries(int firstNumber,int secondNumber,int Counter,int number)
{
Console.Write($"{firstNumber} ");
if(counter>number)
FibonacciSeries(secondNumber,firstNumber+secondNumber,counter+1);
}
static void Main(string[] args)
{
Console.Write("请输入需要获取的斐波那契数值的编号: ");
int number = int.Parse(Console.ReadLine());
//因为序列是从0开始, 所以我们计算某一位时,需要-1;
number = number - 1;
Console.Write(NthFibonacciNumber(number));
Console.ReadKey();
}
private static int NthFibonacciNumber(int number)
{
int firstNumber = 0, secondNumber = 1, nextNumber = 0;
//如果输入0,则返回0
if (number == 0)
return firstNumber;
for (int i = 2; i <= number; i++)
{
nextNumber = firstNumber + secondNumber;
firstNumber = secondNumber;
secondNumber = nextNumber;
}
return secondNumber;
}
质数又称素数, 一个大于1的自然数, 除了1和它自身外, 不能被其他自然数整除的数叫做质数。,例如: 2,3,5,7,11,13,17,19,23... 是质数。
Console.Write("输入数字: ");
int number = int.Parse(Console.ReadLine());
bool IsPrime = true;
for (int i = 2; i < number / 2; i++)
{
if (number % i == 0)
{
IsPrime = false;
break;
}
}
if (IsPrime)
Console.Write($"{number}是质数");
else
Console.Write($"{number}不是质数");
Console.ReadKey();
Console.Write("输入起始数字: ");
int startNumber = int.Parse(Console.ReadLine());
Console.Write("输入结束数字: ");
int endNumber = Convert.ToInt32(Console.ReadLine());
for (int i = startNumber; i <= endNumber; i++)
{
int counter = 0;
for (int j = 2; j <= i / 2; j++)
{
if (i % j == 0)
{
counter++;
break;
}
}
if (counter == 0 && i != 1)
{
Console.Write("{0} ", i);
}
}
Console.ReadKey();
}
回文数
是一组数字, 反转该数字之后与原来相同。例如: 121, 343, 7887 等。
回文字符串
一组文字, 返回该文字之后与原来相同。例如: ABA、我爱我等。
static void Main(string[] args)
{
Console.Write("请输入要检查的数字 : ");
int number = int.Parse(Console.ReadLine());
int remineder, sum = 0;
int temp = number;
while (number > 0)
{
//得到余数
remineder = number % 10;
//总和*10, 然后与余数相加
sum = (sum * 10) + remineder;
//除以10,得到商
number = number / 10;
}
if (temp == sum)
Console.WriteLine($"{temp}是回文");
else
Console.WriteLine($"{temp}不是回文");
Console.ReadKey();
}
static void Main(string[] args)
{
Console.Write("输入要检查的回文: ");
string name = Console.ReadLine();
string reverse = string.Empty;
for (int i = name.Length - 1; i >= 0; i--)
reverse += name[i];
if (name == reverse)
Console.WriteLine($"{name}是回文");
else
Console.WriteLine($"{name}不是回文");
Console.ReadKey();
}
static void Main()
{
Console.Write("输入要检查的回文: ");
string name = Console.ReadLine();
string reverse = string.Empty;
foreach (char c in name)
reverse = c + reverse;
if (name.Equals(reverse, StringComparison.OrdinalIgnoreCase))
Console.WriteLine($"{name}是回文");
else
Console.WriteLine($"{name}不是回文");
Console.ReadKey();
}
static void Main()
{
Console.Write("输入要检查的回文: ");
string name = Console.ReadLine();
char[] nameArray = name.ToCharArray();
Array.Reverse(nameArray);
string reverse = new string(nameArray);
if (name.Equals(reverse, StringComparison.OrdinalIgnoreCase))
Console.WriteLine($"{name}是回文");
else
Console.WriteLine($"{name}不是回文");
Console.ReadKey();
}
标签:i++ new 个数 交换 use ogr 排列 next ali
原文地址:https://www.cnblogs.com/zh7791/p/14713954.html