C++ 将笛卡尔平面上的X,Y坐标旋转θ度,单位为C++;
在我的代码中,我正在创建点的二维散点图:C++ 将笛卡尔平面上的X,Y坐标旋转θ度,单位为C++;,c++,C++,在我的代码中,我正在创建点的二维散点图: //Loop through all 13 objects for (int playerEntity = 0; playerEntity < 13; playerEntity++) { //Obtain X and Y Value Offset from Scatter Plot's origin for each object int px = positionX[playerEntity] / 20; int py
//Loop through all 13 objects
for (int playerEntity = 0; playerEntity < 13; playerEntity++) {
//Obtain X and Y Value Offset from Scatter Plot's origin for each object
int px = positionX[playerEntity] / 20;
int py = positionY[playerEntity] / 20;
//Establish X and Y origin in terms of the screen's pixels;
int originX = 500;
int originY = 1000;
//Plot the scatter plot point in respect to the origin
FillRGB(originX + px , originY + py);
}
//遍历所有13个对象
对于(int-playerEntity=0;playerEntity<13;playerEntity++){
//获取每个对象从散点图原点的X和Y值偏移
int px=位置X[播放性]/20;
int py=位置Y[游戏性]/20;
//根据屏幕像素建立X和Y原点;
int-originX=500;
int-originY=1000;
//绘制相对于原点的散点图点
FillRGB(originX+px,originY+py);
}
如何将点px和py围绕图形原点旋转一定程度?例如,如果我想将图形的点px和py围绕原点旋转48度,我该怎么做?有什么想法吗?你可以用这个公式:x=r*cosθ;y=r*sinθ。这里r是半径,这很方便,你可以用这个公式:x=r*cosθ;y=r*sinθ。这是半径,这很方便