打印从 1 到 100 的数字并用字符串替换一些数字 (fizzbuzz)

Question

我正在从这里尝试 fizzbuzz 程序： 为什么程序员不能.. 编程？< /a>

"编写一个程序，打印从 1 到 100 的数字。但是对于三的倍数，打印“Fizz”而不是数字；对于五的倍数，打印“Buzz”。对于同时是三和五的倍数的数字打印“菲兹巴兹”。”

protected void btn1_Click(object sender, EventArgs e)
    {
        for (int i = 1; i < 101; i++)
        {
            if (i % 3 == 0 & i % 5 == 0)
            {
                Response.Write("fizzbuzz" + ",");
            }
            else if (i % 3 == 0)
            {
                Response.Write("fizz" + ",");
            }
            else if (i % 5 == 0)
            {
                Response.Write("buzz" + ",");
            }
            else
            {
                i = i + 0;
            }

            Response.Write(i +",");
        }

    }

我能够产生某种结果，例如：

1,2,嘶嘶声,3,4,嗡嗡声,5,嘶嘶声,6,7,8,嘶嘶声,9,嗡嗡声,10,11,嘶嘶声,12,13,14,嘶嘶声,15,16,17 ,嘶嘶声,18,19,嗡嗡声,20,嘶嘶声,21,22,23,嘶嘶声,24,嗡嗡声,25,26,嘶嘶声,27,28,29,嘶嘶声,30,31,32,嘶嘶声,33,34 、嗡嗡声、35、嘶嘶声、36、37、38、嘶嘶声、39 等等..

打印了 fizz 一词，但它没有取代 3，并且打印了 fizzbuzz，但它没有替换 15 等等...

Answer 1

无论您是否满足 if 条件，您仍然会在代码末尾打印 i 。

特别看看你的 for 循环：

for (int i = 1; i < 101; i++)         
{             
if (i % 3 == 0 & i % 5 == 0)             
{                 
Response.Write("fizzbuzz" + ",");             
}             
else if (i % 3 == 0)             
{                 
Response.Write("fizz" + ",");             
}             
else if (i % 5 == 0)             
{                 
Response.Write("buzz" + ",");             
}             
else             
{                 
i = i + 0;             
}              
Response.Write(i +",");   //look here you print i
} 

所以你需要移动最后一个 Response.Write(i + ",");在最后一个 else 条件中。查找此类错误的最简单方法是使用调试器并调试程序。然后您将很容易看到输出是什么。因此，一定要使用调试器并设置断点/监视并观察会发生什么。您的代码应更改为：

  for (int i = 1; i < 101; i++)         
    {             
    if (i % 3 == 0 & i % 5 == 0)             
    {                 
    Response.Write("fizzbuzz" + ",");             
    }             
    else if (i % 3 == 0)             
    {                 
    Response.Write("fizz" + ",");             
    }             
    else if (i % 5 == 0)             
    {                 
    Response.Write("buzz" + ",");             
    }             
    else             
    {                 
    Response.Write(i +",");   //look here you print i
    }              
    } 

请注意，您的 for 循环已通过增加 i 来处理此问题，从而删除了 i=i+1 。

编辑

Not sure if this is easier but here is another way to do this using lambda's:

            List<int> t;
            t = Enumerable.Range(1, 100).ToList();

            var fizzBuzz = t.Where(num => num % 3 == 0 && num % 5 == 0);
            var fizz = t.Where(num => num % 3 == 0);
            var buzz = t.Where(num => num % 5 == 0);
            var notFizzBuzz = t.Where(num => num % 3 != 0 && num % 5 !=0);

            //print fizzBuzz elements
            Console.WriteLine("Printing fizzBuzz elements...");
            foreach (int i in fizzBuzz)
                Console.WriteLine(i);

            //print fizz elements
            Console.WriteLine("Printing fizz elements...");
            foreach (int i in fizz)
                Console.WriteLine(i);

            //print buzz elements
            Console.WriteLine("Printing buzz elements...");
            foreach (int i in buzz)
                Console.WriteLine(i);

            //print other elements
            Console.WriteLine("Printing all others...");
            foreach (int i in notFizzBuzz)
                Console.WriteLine(i);

Answer 2

尝试这些更改

protected void btn1_Click(object sender, EventArgs e)
{
    for (int i = 1; i < 101; i++)
    {
        if (i % 3 == 0 & i % 5 == 0)
        {
            Response.Write("fizzbuzz" + ",");
        }
        else if (i % 3 == 0)
        {
            Response.Write("fizz" + ",");
        }
        else if (i % 5 == 0)
        {
            Response.Write("buzz" + ",");
        }
        else
        {
            Response.Write(i +",");
        }           
    }
}

您的 i = i + 0 显然没有执行任何操作，因为您将 0 添加到 i 的值。

而且无论 if/else 块的结果如何（放在它后面），您都会将数字打印到响应中，因此应该将其移至 else 中（意味着仅在 if 或 else if 不匹配时才打印。

Answer 3

将 Response.Write(i +","); 移动到最后的 else 中

Answer 4

protected void btn1_Click(object sender, EventArgs e)
{
    for (int i = 1; i < 101; i++)
    {
        if (i % 3 == 0 & i % 5 == 0)
        {
            Response.Write("fizzbuzz" + ",");
        }
        else if (i % 3 == 0)
        {
            Response.Write("fizz" + ",");
        }
        else if (i % 5 == 0)
        {
            Response.Write("buzz" + ",");
        }
        else
        {
            i = i + 0; //this is totally useless
            Response.Write(i + ",");

        }

        //Response.Write(i +","); //This will always write the number, even if you wrote fizz or buzz
    }

}

Answer 5

它的另一个简单实现：

for (int i = 1; i <= 100; i++)
        {
            Console.WriteLine((i % 3 == 0) ? ((i % 5 == 0) ? "FizzBuzz" : "Fizz") : ((i % 5 == 0) ? "Buzz" : i.ToString()));
        }
Console.ReadKey();

Answer 6

    public static void PrintMod3And5FromInterval(int start, int end)
    {
        if (end < start)
        {
            Console.WriteLine("End number should be higher than start.");
        }
        else
        { 
            string result = "";
            for (int x = start; x <= end; x++)
            {
                if (x % 3 == 0)
                    result += "fizz";
                if (x % 5 == 0)
                    result += "buzz";
                if (result == "")
                    Console.WriteLine(x);
                else
                    Console.WriteLine(result);
                result = "";
            }
        }
    }
    static void Main(string[] args)
    {
        PrintMod3And5FromInterval(1, 100);
        Console.Read();
    }

Answer 7

这是我最初的解决方案...

for (let number = 1; number <= 100; number ++) {
if (number % 3 === 0 && number % 5 === 0) {
    console.log(number + "fizzbuzz");
} else if (number % 5 === 0) {
console.log(number + "buzz");
} else if (number % 3 === 0)
console.log(number + "fizz");
else  {
console.log(number);
}
}     

但这一个要短得多...

for (let n = 1; n <= 100; n++) {
  let output = "";
    if (n % 3 == 0) output += "Fizz";
    if (n % 5 == 0) output += "Buzz";
  console.log(output || n);
}