网页拼图游戏代码（拼图游戏网页制作）

今天给各位分享网页拼图游戏代码的知识，其中也会对拼图游戏网页制作进行解释，如果能碰巧解决你现在面临的问题，别忘了关注本站，现在开始吧！

本文目录一览：

• 1、同求funcode平台下拼图游戏的C语言代码
• 2、急求拼图游戏的VB代码和程序?
• 3、急求用JAVA编写的图形化界面拼图小游戏代码！

同求funcode平台下拼图游戏的C语言代码

//第一题：#includevoidfun(char*a[],char*b[]){intl;/*获取较短数组的长度*/if(strlen(*a)usingnamespacestd;classCircle{public:voidinput(double);//输入voidprint();//输出doublegetArea();//计算半径private:doubler;//半径};voidCircle::input(doublea){r=a;}doubleCircle::getArea(){returnr*r*3.14159;}voidCircle::print(){couta;c.input(a);c.print();return0;}

急求拼图游戏的VB代码和程序?

首先这是个非常复杂的编程，其次是这个编程很复杂，最后是这是个很艰难的决定。

应该是这样，代码如下：

Private Sub DadiZs()

Dim t As Double, Itp As Double, X0 As Double, N As Double, L0 As Double

Dim V As Double, ll As Double, W As Double, M As Double

Lat = Radian(Lat)

Lon = Radian(Lon)

L0 = Radian(Lo)

If Tq = 0 Then

a = 6378245

b = 6356863.01877305

ep = 0.006693421622966

ep1 = 0.006738525414683

f = (a - b) / a

c = a ^ 2 / b

d = b ^ 2 / a

X0 = 111134.8611 * (Lat * 180# / Pi) - (32005.7799 * Sin(Lat) + 133.9238 * (Sin(Lat)) ^ 3 + 0.6973 * (Sin(Lat)) ^ 5 + 0.0039 * (Sin(Lat)) ^ 7) * Cos(Lat)

'X0 = 111134.8611 * (Lat * 180# / Pi) - (32005.7798 * Sin(Lat) + 133.9238 * (Sin(Lat)) ^ 3 + 0.6972 * (Sin(Lat)) ^ 5 + 0.0039 * (Sin(Lat)) ^ 7) * Cos(Lat)

Else

a = 6378140

b = 6356755.28815753

ep = 0.006694384999588

ep1 = 0.006739501819473

f = (a - b) / a

c = a ^ 2 / b

d = b ^ 2 / a

X0 = 111133.0047 * (Lat * 180 / Pi) - (32009.8575 * Sin(Lat) + 133.9602 * (Sin(Lat)) ^ 3 + 0.6976 * (Sin(Lat)) ^ 5 + 0.0039 * (Sin(Lat)) ^ 7) * Cos(Lat)

End If

ll = Lon - L0

t = Tan(Lat)

Itp = ep1 * Cos(Lat) ^ 2

W = Sqr(1 - ep * Sin(Lat) ^ 2)

V = Sqr(1 + ep1 * Cos(Lat) ^ 2)

M = c / V ^ 3

N = a / W

'x = X0 + N * t * (Cos(Lat)) ^ 2 * ll ^ 2 / 2 + N * t * (5 - t * t + 9 * Itp + 4 * Itp * Itp) * (Cos(Lat)) ^ 4 * ll ^ 4 / 24 + N * t * (61 - 58 * t ^ 2 + t ^ 4 + 270 * Itp - 330 * t ^ 2 * Itp) * (Cos(Lat)) ^ 6 * ll ^ 6 / 720 + N * t * (1385 - 3111 * t ^ 2 + 543 * t ^ 4 - t ^ 6) * Cos(Lat) ^ 8 * ll ^ 8 / 40320

x = X0 + N * t * (Cos(Lat)) ^ 2 * ll ^ 2 / 2 + N * t * (5 - t * t + 9 * Itp ^ 2 + 4 * Itp ^ 4) * (Cos(Lat)) ^ 4 * ll ^ 4 / 24 + N * t * (61 - 58 * t ^ 2 + t ^ 4 + 270 * Itp ^ 2 - 330 * t ^ 2 * Itp ^ 2) * (Cos(Lat)) ^ 6 * ll ^ 6 / 720 + N * t * (1385 - 3111 * t ^ 2 + 543 * t ^ 4 - t ^ 6) * Cos(Lat) ^ 8 * ll ^ 8 / 40320

y = N * Cos(Lat) * ll + N * (1 - t * t + Itp) * (Cos(Lat)) ^ 3 * ll ^ 3 / 6 + N * (5 - 18 * t * t + t ^ 4 + 14 * Itp - 58 * Itp * t * t) * (Cos(Lat)) ^ 5 * ll ^ 5 / 120 + N * (61 - 479 * t ^ 2 + 179 * t ^ 4 - t ^ 6) * Cos(Lat) ^ 7 * ll ^ 7 / 5040

r = Sin(Lat) * ll + Sin(Lat) * (Cos(Lat)) ^ 2 * ll ^ 3 * (1 + 3 * Itp + 2 * Itp ^ 2) / 3 + Sin(Lat) * (Cos(Lat)) ^ 4 * ll ^ 5 * (2 - t * t) / 15

r = Degree(r)

y = y + 500000#

End Sub

Private Sub DadiFs()

Dim t As Double, Itp As Double, X0 As Double, Bf As Double, N As Double

Dim v As Double, ll As Double, W As Double, M As Double, L0 As Double

L0 = Radian(Lo)

X0 = x * 0.000001

y = y - 500000#

If Tq = 0 Then

a = 6378245

b = 6356863.01877305

ep = 0.006693421622966

ep1 = 0.006738525414683

f = (a - b) / a

c = a ^ 2 / b

d = b ^ 2 / a

If X0 3 Then

Bf = 9.04353301294 * X0 - 0.00000049604 * X0 ^ 2 - 0.00075310733 * X0 ^ 3 - 0.00000084307 * X0 ^ 4 - 0.00000426055 * X0 ^ 5 - 0.00000010148 * X0 ^ 6

ElseIf X0 6 Then

Bf = 27.11115372595 + 9.02468257083 * (X0 - 3) - 0.00579740442 * (X0 - 3) ^ 2 - 0.00043532572 * (X0 - 3) ^ 3 + 0.00004857285 * (X0 - 3) ^ 4 + 0.00000215727 * (X0 - 3) ^ 5 - 0.00000019399 * (X0 - 3) ^ 6

End If

Else

a = 6378140

b = 6356755.28815753

ep = 0.006694384999588

ep1 = 0.006739501819473

f = (a - b) / a

c = a ^ 2 / b

d = b ^ 2 / a

If X0 3 Then

Bf = 9.04369066313 * X0 - 0.00000049618 * X0 ^ 2 - 0.00075325505 * X0 ^ 3 - 0.0000008433 * X0 ^ 4 - 0.00000426157 * X0 ^ 5 - 0.0000001015 * X0 ^ 6

ElseIf X0 6 Then

Bf = 27.11162289465 + 9.02483657729 * (X0 - 3) - 0.00579850656 * (X0 - 3) ^ 2 - 0.00043540029 * (X0 - 3) ^ 3 + 0.00004858357 * (X0 - 3) ^ 4 + 0.00000215769 * (X0 - 3) ^ 5 - 0.00000019404 * (X0 - 3) ^ 6

End If

End If

Bf = Bf * Pi / 180#

t = Tan(Bf)

Itp = ep1 * Cos(Bf) ^ 2

W = Sqr(1 - ep * Sin(Bf) ^ 2)

v = Sqr(1 + ep1 * Cos(Bf) ^ 2)

M = c / v ^ 3

N = a / W

Lat = Bf - 0.5 * v ^ 2 * t * ((y / N) ^ 2 - (5 + 3 * t * t + Itp - 9 * Itp * t * t) * (y / N) ^ 4 / 12 + (61 + 90 * t * t + 45 * t ^ 4) * (y / N) ^ 6 / 360)

ll = ((y / N) - (1 + 2 * t * t + Itp) * (y / N) ^ 3 / 6 + (5 + 28 * t * t + 24 * t ^ 4 + 6 * Itp + 8 * Itp * t * t) * (y / N) ^ 5 / 120) / Cos(Bf)

r = y * t / N - y ^ 3 * t * (1 + t * t - Itp) / (3 * N ^ 3) + y ^ 5 * t * (2 + 5 * t * t + 3 * t ^ 4) / (15 * N ^ 5)

Lat = Degree(Lat)

Lon = Degree(L0 + ll)

r = Degree(r)

End Sub

Public Const Pi = 3.14159265358979, p = 206264.806

Public Cktq As String

Public Function Radian(a As Double) As Double

Dim Ro As Double

Dim c As Double

Dim Fs As Double

Dim Ib As Integer

Dim Ic As Integer

If a 0 Then a = -a: t = 1

Ro = Pi / 180#

Ib = Int(a)

c = (a - Ib) * 100#

Ic = Int(c + 0.000000000001)

Fs = (c - Ic) * 100#

If t = 1 Then Radian = -(Ib + Ic / 60# + Fs / 3600#) * Ro Else Radian = (Ib + Ic / 60# + Fs / 3600#) * Ro

End Function

Public Function Degree(a As Double) As Double

Dim Bo As Double

Dim Fs As Double

Dim Im As Integer

Dim Id As Integer

If a 0 Then a = -a: t = 1

Bo = a

Call DMS(Bo, Id, Im, Fs)

If t = 1 Then Degree = -(Id + Im / 100# + Fs / 10000#) Else Degree = Id + Im / 100# + Fs / 10000#

End Function

Public Sub DMS(a As Double, Id As Integer, Im As Integer, Fs As Double)

Dim Bo As Double

Dim c As Double

c = a

c = 180# / Pi * c

Id = Int(c)

Bo = (c - Id) * 60

Im = Int(Bo)

Fs = (Bo - Im) * 60

End Sub

Public Function Qw(a As Double, Ws As Integer) As Double

Qw = Int(a * 10 ^ Ws + 0.5) / 10 ^ Ws

End Function

急求用JAVA编写的图形化界面拼图小游戏代码！

个人见解，总体需要两个二维数组（一个存储正确图片排列 Array1 String[][]，一个随机生成图片排列Array2 String[][]），一个一维数组来存储图片的名称Array3 String[]，。

（1）如何实现图片移动

使用带图片的按钮（button =new button（getImage（Array[2][4]））），然后通过单击事件来更改按钮的图片来源。 把被点击的按钮的图片路径更新到空白按钮，并且把被点击的按钮图片更新的成空白。其实就是变换两个的二维数组成员的值。更新Array2中的值，然后重绘按钮

如 Array[2][3]=“3.image”

Array[2][4]=“”

图片3.image右移

Array[2][3]=“”

Array[2][4]=“3.image”

（2）如何判断被单击的网格与空白的网格是否相邻

后台使用一个二维数组Array2来做映射。通过二维数组的下标来判断，如Array[2][3]可以知道Array[2][4]是它右边的那个。

（3）如何实现图片的随机摆放

比如有9个图片，你可以命名1-9，然后初始化一个长度为9的一维String 数组Array3来存储图片的名称，

使用随机函数给二维数组Array2赋值，如Array2[2][3]=Array3[random()],这里要判断这个图片是否已被使用过，可以通过遍历Array2来确定当前Array3这个值是否已经在Array2中了

最后通过Array1 和Array2来比较，用户的拼图是否正确。

语言组织能力有限。讲不太清楚。

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