我是C#的初学者,试图以彩票形式申请。
有类型,首先你有5个提示(otos bool)和5个提示(hatos bool)。
并且有很多类型的抽奖数量(tiz,harminc,kilencven,negyvenot)。
我尝试使用此代码Array.Equals
扫描抽奖后的数字:
for (int i = 0; i <= 4; i++)
{
for (int y = 0; y <= 4; y++)
{
if (Array.Equals(lottoszamok[i], lottoszamok[y]))
lottoszamok[i] = r.Next (1, ?);
}
}
但是在这个时候,这个数字也将被自己扫描,所以它总是相等的。
顺便说一句,这是我的代码:
if (otos == true)
{
for (int i = 0; i <= 5; i++)
{
if (tiz == true)
{
lottoszamok[i] = r.Next(1, 10);
}
else if (harminc == true)
{
lottoszamok[i] = r.Next(1, 30);
}
else if (kilencven == true)
{
lottoszamok[i] = r.Next(1, 90);
}
else if (negyvenot == true)
{
lottoszamok[i] = r.Next(1, 45);
}
else if (egyeni == true)
{
lottoszamok[i] = r.Next(1, (egyeniertek + 1));
}
}
}
if (hatos == true)
{
for (int i = 0; i <= 6; i++)
{
if (tiz == true)
{
lottoszamok[i] = r.Next(1, 10);
}
else if (harminc == true)
{
lottoszamok[i] = r.Next(1, 30);
}
else if (kilencven == true)
{
lottoszamok[i] = r.Next(1, 90);
}
else if (negyvenot == true)
{
lottoszamok[i] = r.Next(1, 45);
}
else if (egyeni == true)
{
lottoszamok[i] = r.Next(1, (egyeniertek + 1));
}
}
}
答案 0 :(得分:2)
如果您尝试从1..n
范围中选择数字而不重复,则需要将这些数字“洗牌”出来:
int[] allPossibleNumbers = Enumerable.Range(1, maxNumber).ToArray();
int[] picked = new int[numberToPick];
for (int i = 0; i < numberToPick; i++)
{
int index = r.Next(i, maxNumber);
picked[i] = allPossibleNumbers[index];
allPossibleNumbers[index] = allPossibleNumbers[i];
}
其中numberToPick
5
otos
或6
hatos
,maxNumber
取决于tiz
,{{ 1}},harminc
,kilencven
,negyvenot
和egyeni
。
如果您的egyeniertek
很大而且您只想选择一些数字,则以下内容并不要求整个范围同时存在于内存中:
maxNumber
答案 1 :(得分:1)
试试这个!
if (i!=y && Array.Equals(lottoszamok[i], lottoszamok[y]))
答案 2 :(得分:0)
private void TenNumbersRandomly()
{
int[] a = new int[10];
Random r = new Random();
int x;
for (int i = 0; i < 10; i++)
{
x= r.Next(1, 11);
for (int j = 0; j <= i ; j++)
{
while (a[j] == x)
{
x = r.Next(1, 11);
j = 0;
}
}
a[i] = x;
tb1.Text += a[i]+"\n";
}
}
Private Sub Button1_Click(sender As Object, e As EventArgs) Handles Button1.Click
Dim x As Integer, i As Integer, j As Integer
x = Int(Rnd() * 10) + 1
Label1.Text = ""
Dim a(9) As Integer
For i = 0 To 9
x = Int(Rnd() * 10) + 1
For j = 0 To i
While (a(j) = x)
x = Int(Rnd() * 10) + 1
j = 0
End While
Next j
a(i) = x
Label1.Text += a(i).ToString() + " "
Next i
答案 3 :(得分:0)
我是这样做的,如果你想要你可以像方法一样交换。
static void SwapInts(int[] array, int position1, int position2)
{
// Swaps elements in an array.
int temp = array[position1]; // Copy the first position's element
array[position1] = array[position2]; // Assign to the second element
array[position2] = temp; // Assign to the first element
}
static void Main()
{
Random rng = new Random();
int n = int.Parse(Console.ReadLine());
int[] intarray = new int[n];
for (int i = 0; i < n; i++)
{
// Initialize array
intarray[i] = i + 1;
}
// Exchange resultArray[i] with random element in resultArray[i..n-1]
for (int i = 0; i < n; i++)
{
int positionSwapElement1 = i + rng.Next(0, n - i);
SwapInts(intarray, i, positionSwapElement1);
}
for (int i = 0; i < n; i++)
{
Console.Write(intarray[i] + " ");
}
}
}