[英]C# Program that generates 20 random numbers and search the array for a number (Continued)
[英]Program that generates 20 random numbers and search the array for a number
我想制作一个生成 20 个随机数并在数组中搜索一个数字的程序。 如果在输入中输入了 20 个随机数之一,则 output 应该说“它在这里”。 如果数字不在 ReadLine 中,它应该说“是的,它在那里”。我想知道如何使所有 20 个随机数都能够搜索。 现在的代码只能搜索右边的数字。 即使输入是 20 个随机数之一,除了右边的那个,它也会说“不,它不在这里”。
图片是我现在的output。 它显示 20 个数字。 输出1输出2
public static void Main(string[] args)
{
Random random = new Random();
int[] myIntArray = new int[100];
for (int i = 1; i <= 20; i++)
{
int x = random.Next(100);
myIntArray[i] = x;
Console.Write(myIntArray[i] + " ");
if (i == 20)
{
Console.Write("\nType a number to search for:");
int z = Convert.ToInt32(Console.ReadLine());
if (z == x)
{
Console.WriteLine("Yes it is there.");
}
else
{
Console.WriteLine("No it is not here.");
}
}
}
Console.ReadKey();
}
您当前的代码检查 20tyh 项 ( x
) 是否等于用户输入 ( z
):
if (i == 20) // 20th item only
{
int z = Convert.ToInt32(Console.ReadLine());
if (z == x) // if 20th item (x) equals to user input (z)
{
Console.WriteLine("Yes it is there.");
}
else
{
Console.WriteLine("No it is not here.");
}
}
尝试逐步解决问题:
代码:
public static void Main(string[] args)
{
// Array...
int[] myIntArray = new int[20]; // We want 20 items, not 100, right?
// Of random items
Random random = new Random();
for (int i = 0; i < myIntArray.Length; ++i)
myIntArray[i] = random.Next(100);
// Debug: let's have a look at the array:
Console.WriteLine(string.Join(" ", myIntArray));
// myIntArray is built. Time to ask user for a number
Console.Write("Type a number to search for:");
if (int.TryParse(Console.ReadLine(), out int number)) {
// number is a valid integer number, let's scan myIntArray
bool found = false;
foreach (int item in myIntArray)
if (item == number) {
found = true;
break;
}
if (found)
Console.WriteLine("Yes it is there.");
else
Console.WriteLine("No it is not here.");
}
else
Console.WriteLine("Not a valid integer");
Console.ReadKey();
}
编辑:在现实生活中,我们经常使用Linq来查询 collections(数组):
using System.Linq;
...
public static void Main(string[] args)
{
Random random = new Random();
int[] myIntArray = Enumerable
.Range(0, 20)
.Select(i => random.Next(100))
.ToArray();
Console.WriteLine(string.Join(" ", myIntArray));
Console.Write("Type a number to search for:");
if (int.TryParse(Console.ReadLine(), out int number))
if (myIntArray.Any(item => item == number))
Console.WriteLine("Yes it is there.");
else
Console.WriteLine("No it is not here.");
else
Console.WriteLine("Not a valid integer");
Console.ReadKey();
}
所以基本上你应该做一个循环,用一个数组用一个随机数初始化你的数组
for (int i = 1; i <= 20; i++)
{
int x = random.Next(100);
myIntArray[i] = x;
}
之后,您可以在另一个循环中在控制台中搜索键入的值
Console.Write("\nType a number to search for:");
bool isValueFound = false;
int z = Convert.ToInt32(Console.ReadLine());
for(int i=0; i<=20; i++){
if (z == x)
{
Console.WriteLine("Yes it is there.");
}
else
{
Console.WriteLine("No it is not here.");
}
}
所有代码:
public static void Main(string[] args)
{
Random random = new Random();
int[] myIntArray = new int[100];
for (int i = 1; i <= 20; i++)
{
int x = random.Next(100);
myIntArray[i] = x;
}
Console.Write("\nType a number to search for:");
bool isValueFound = false;
int z = Convert.ToInt32(Console.ReadLine());
for(int i=0; i<=20; i++){
if (z == x)
{
Console.WriteLine("Yes it is there.");
}
else
{
Console.WriteLine("No it is not here.");
}
}
Console.ReadKey();
}
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.