1

我有一个包含 40 个旋转图像的图像。

实际上，图像索引从 0. 0-39 开始。

这是将 0-39 转换为度数的代码

int image_direction = 0; //Can be 0-39
int facing_degrees = (int)(360.0 * (-(image_direction- 10.0))/40.0);
while(facing_degrees < 0)
    facing_degrees += 360;
while (facing_degrees > 360)
    facing_degrees -= 360;

所以是的，它也可以给出负度数以及超过 360 度数。这就是为什么有 2 个 while 循环。

现在我想扭转这个过程说我指定90度我想回到0..

我正在考虑做类似的事情

 if(degrees == 90 || degrees >= 80 && degrees <= 99)
    image_direction = 0;
 elseif(degrees == 100 || degrees >= 91 && degrees <= 109)
    image_direction = 39;
//etc.......

好吧，我不擅长数学，好吧，我忘记了这种东西。

我想知道如何反转从 image_direction 给出度数的函数以在一个简单的方程中向后运行，以避免 if 语句的巨大情况。

这是一些结果 轮换结果

Image Index = 0 Image Degree = 90
Image Index = 1 Image Degree = 81
Image Index = 2 Image Degree = 72
Image Index = 3 Image Degree = 63
Image Index = 4 Image Degree = 54
Image Index = 5 Image Degree = 45
Image Index = 6 Image Degree = 36
Image Index = 7 Image Degree = 27
Image Index = 8 Image Degree = 18
Image Index = 9 Image Degree = 9
Image Index = 10 Image Degree = 0
Image Index = 11 Image Degree = 351
Image Index = 12 Image Degree = 342
Image Index = 13 Image Degree = 333
Image Index = 14 Image Degree = 324
Image Index = 15 Image Degree = 315
Image Index = 16 Image Degree = 306
Image Index = 17 Image Degree = 297
Image Index = 18 Image Degree = 288
Image Index = 19 Image Degree = 279
Image Index = 20 Image Degree = 270
Image Index = 21 Image Degree = 261
Image Index = 22 Image Degree = 252
Image Index = 23 Image Degree = 243
Image Index = 24 Image Degree = 234
Image Index = 25 Image Degree = 225
Image Index = 26 Image Degree = 216
Image Index = 27 Image Degree = 207
Image Index = 28 Image Degree = 198
Image Index = 29 Image Degree = 189
Image Index = 30 Image Degree = 180
Image Index = 31 Image Degree = 171
Image Index = 32 Image Degree = 162
Image Index = 33 Image Degree = 153
Image Index = 34 Image Degree = 144
Image Index = 35 Image Degree = 135
Image Index = 36 Image Degree = 126
Image Index = 37 Image Degree = 117
Image Index = 38 Image Degree = 108
Image Index = 39 Image Degree = 99
Image Index = 0 Image Degree = 90
Image Index = 1 Image Degree = 81
Image Index = 2 Image Degree = 72
Image Index = 3 Image Degree = 63
Image Index = 4 Image Degree = 54
Image Index = 5 Image Degree = 45
Image Index = 6 Image Degree = 36
Image Index = 7 Image Degree = 27
Image Index = 8 Image Degree = 18
4

2 回答 2

1

嗯，为什么不使用类似于“百分比公式”的东西并稍作调整：

image_direction = (int)(((float)(degrees-90) / 360) * 40) % 40;
if (image_direction < 0)
    image_direction += 40; // for negative integers

分解公式；我们将度数除以 360 并乘以 40 以获得给定度数的正确图像，然后我们使用模运算符将度数 > +360 或 < -360 的图像方向保持在 0-39 范围内，最后，如果度数是负数，image_direction 将变成负数，因此向其添加 40 将纠正 image_direction。

希望这可以帮助。

编辑：哦，抱歉，我似乎错过了大约 90 度 == 图像索引为 0 的部分，在这里很容易添加，只需从实际度数中减去 90（上面编辑的代码）

于 2012-01-26T10:33:32.640 回答
1

从输入到输出的映射，我将它们粘贴到 Excel 中，并计算出计算度数的公式如下：

RawDegrees = 90 - (index * 9)

然后将其剪裁为 0...360（使用 while 循环环绕），您将获得最终的度数输出。通过重新排列上述内容以使索引成为主题，可以找到反转此公式的公式

    RawDegrees - 90 = - (指数 * 9) --- 1
    RawDegrees/9.0 - 10 = -index --- 2

    所以
    指数 = 10 - 度/9.0

这将为您提供以下映射

    学位发现指数
    90 0
    81 1
    72 2
    63 3
    54 4
    45 5
    36 6
    27 7
    18 8
    9 9
    0 10
    351 -29
    342 -28
    333 -27
    324 -26
    315 -25
    306 -24
    297 -23
    288 -22
    279 -21
    270 -20
    261 -19
    252 -18
    243 -17
    234 -16
    225 -15
    216 -14
    207 -13
    198 -12
    189 -11
    180 -10
    171 -9
    162 -8
    153 -7
    144 -6
    135 -5
    126 -4
    117 -3
    108 -2
    99 -1
    90 0
    81 1
    72 2
    63 3
    54 4
    45 5
    36 6
    27 7
    18 8

您需要做的就是执行环绕以将答案剪裁为 0..39，例如

index = index < 0 ? index + 40 : index;

以获得正确的输出。

顺便说一句，您的度数索引代码可以按如下方式分解以获得相同的输出（对输入 0..39 有效）

int facing_degrees = (int)(90.0 - (image_direction * 9.0));    
facing_degrees = facing_degrees < 0 ? facing_degrees + 360 : facing_degrees;
于 2012-01-26T10:47:50.667 回答