blocks|key|4056084|text|您可以使用中心(start_X，start_Y)在2r×2r的正方形区域上进行搜索：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4056085|std::vector<+std::pair<int>+>+circlePoints;

for(int+i+=+start_X+-+r;+i+<=+start_X+%2B+r;+i%2B%2B)
{
+++for(int+j+=+start_Y+-+r;+j+<=+start_Y+%2B+r;+j%2B%2B)
+++{
+++++++if((i-r)*(i-r)+%2B+(j-r)*(j-r)+<=+r*r)
+++++++{
+++++++++circlePoints.push_back(std::pair<int>(i,j));
+++++++}
+++}
}|code-block|syntax|javascript|4056086|entityMap^0|8|7|G|7|Q|1|T|1|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|L|$]]

You can search over a square region 2<code>r</code> by 2<code>r</code> with center (<code>start_X</code>,<code>start_Y</code>):

<pre><code>std::vector&lt; std::pair&lt;int&gt; &gt; circlePoints;

for(int i = start_X - r; i &lt;= start_X + r; i++)
{
 for(int j = start_Y - r; j &lt;= start_Y + r; j++)
 {
 if((i-r)*(i-r) + (j-r)*(j-r) &lt;= r*r)
 {
 circlePoints.push_back(std::pair&lt;int&gt;(i,j));
 }
 }
}
</code></pre>

blocks|key|647970|text|你会得到一个圆的方程式：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|647971|​|647972|📷|atomic|offset|length|647973|647974|其中a和b是中心点坐标。所有满足这个方程的x&y点都是圆的一部分。要查看某个点(x1，y1)是否为，请检查是否|647975|((x1+-+start_X)+*+(x1+-+start_X)+%2B+(y1+-+start_Y)+*+(y1+-+start_Y))+<=+r+*+r|code-block|syntax|javascript|647976|<=符号也包括位于圆内的点。您可以安全地将点范围限制在区间start_X+-+r；startX+%2B+r和start_Y+-+r；startY+%2Br中。|647977|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/9a6d176988cf08dec52ac88485273040.png|imageAlt^0|0|0|0|1|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|11|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|12|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|13|8|@]|9|@$G|14|H|15|1|16]]|A|$]]|$1|I|3|C|5|6|7|17|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|18|8|@]|9|@]|A|$]]|$1|L|3|M|5|N|7|19|8|@]|9|@]|A|$O|P]]|$1|Q|3|R|5|6|7|1A|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|1B|8|@]|9|@]|A|$]]]|T|$U|$5|V|W|X|A|$Y|Z|10|-4]]]]

You get the equation of a circle:

<img src="https://i.stack.imgur.com/CB9vG.png" alt="">

where a &amp; b are the center point coordinates. All x &amp; y points that satisfy this equation are part of the circle. To see if a certain point (x1, y1) is, check if 

<pre><code>((x1 - start_X) * (x1 - start_X) + (y1 - start_Y) * (y1 - start_Y)) &lt;= r * r
</code></pre>

The &lt;= sign is to include the points that lie inside the circle, too. You can safely limit the point ranges in the intervals [start_X - r; startX + r] and [start_Y - r; startY + r].

blocks|key|4056144|text|如果你想在不检查的情况下直接到达圆圈中的所有点，这是最好的方法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4056145|SatY+=+CenterY;//StartY+%2B+R
++++++++for+(int+i+=+StartX;+i+<+EndX;+i%2B%2B)
++++++++{
++++++++++++int+StartY+=+(int)(SatY+-+Math.Sqrt(Math.Abs((R+%2B+i+-+StartX)+*+(R+-+i+%2B+StartX))));
++++++++++++int+EndY+=+(int)(SatY+%2B+Math.Sqrt(Math.Abs((R+%2B+i+-+StartX)+*+(R+-+i+%2B+StartX))));
++++++++++++for+(int+j+=+StartY;+j+<+EndY;+j%2B%2B)
++++++++++++{
++++++++++++++++//+Do+Job
++++++++++++}
++++++++}|code-block|syntax|javascript|4056146|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

if you wanna go stright to all point in the circle without checking, this is the way.

<pre><code>SatY = CenterY;//StartY + R
 for (int i = StartX; i &lt; EndX; i++)
 {
 int StartY = (int)(SatY - Math.Sqrt(Math.Abs((R + i - StartX) * (R - i + StartX))));
 int EndY = (int)(SatY + Math.Sqrt(Math.Abs((R + i - StartX) * (R - i + StartX))));
 for (int j = StartY; j &lt; EndY; j++)
 {
 // Do Job
 }
 }
</code></pre>

blocks|key|1931667|text|这是一个完整的Java解决方案。您可以很容易地将其转换为C%2B%2B。从一个全为0的空矩阵开始，如果点(x，y)位于一个圆内，则用1填充它，否则用9填充外部的圆。(为什么是9-这样你就能清楚地看到在矩阵中绘制的圆)。下面的代码适用于奇数大小的矩阵。如果有人有更好的解决方案，请告诉我。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1931668|++private+static+void+drawCircle(int[][]+emptyMatrix,+int+diameter)+{
++++int+startX+=+diameter/2;
++++int+startY+=+diameter/2;
++++int+radius+=+diameter/2;
++++drawCircleRecursive(emptyMatrix,+diameter,+startX,+startY,+radius,+radius);

++++System.out.println("Filled+matrix:+");
++++for+(int+i+=+0;+i+<+emptyMatrix.length;+i%2B%2B)+{
++++++for+(int+j+=+0;+j+<+emptyMatrix[0].length;+j%2B%2B)+{
++++++++System.out.print(emptyMatrix[i][j]+%2B+"+");
++++++}
++++++System.out.println();
++++}
++}

++private+static+void+drawCircleRecursive(int[][]+emptyMatrix,+int+d,+int+startX,+int+startY,+int+x,+int+y)+{

++++if(x+>=+emptyMatrix.length+%7C%7C+y+>=+emptyMatrix[0].length+%7C%7C+x+<+0+%7C%7C+y+<+0+%7C%7C+emptyMatrix[x][y]+==+1)
++++++return;
++++else+if(emptyMatrix[x][y]+==+9)
++++++return;

++++int+r+=+d/2;
++++if+(((x+-+startX)+*+(x+-+startX)+%2B+(y+-+startY)+*+(y+-+startY))+<=+(r+*+r))
++++++emptyMatrix[x][y]+=+1;+
++++else+
++++++emptyMatrix[x][y]+=+9;

++++drawCircleRecursive(emptyMatrix,+d,+startX,+startY,+x%2B1,+y);++//+down
++++drawCircleRecursive(emptyMatrix,+d,+startX,+startY,+x,+y%2B1);++//+right
++++drawCircleRecursive(emptyMatrix,+d,+startX,+startY,+x-1,+y);++//up
++++drawCircleRecursive(emptyMatrix,+d,+startX,+startY,+x,+y-1);++//left
++++drawCircleRecursive(emptyMatrix,+d,+startX,+startY,+x-1,+y-1);+//+diagonal+up-left
++++drawCircleRecursive(emptyMatrix,+d,+startX,+startY,+x%2B1,+y%2B1);+//+diagonal+right-down
++++drawCircleRecursive(emptyMatrix,+d,+startX,+startY,+x%2B1,+y-1);++//+diagonal+left-down
++++drawCircleRecursive(emptyMatrix,+d,+startX,+startY,+x-1,+y%2B1);++//+diagonal+right-up


++}|code-block|syntax|javascript|1931669|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This is a complete solution in Java. You can easily convert it to C++. Start with an empty matrix predefined with all 0's and fill it with 1's if that the point(x,y) lies inside a circle else fill the outer circle with 9's. (Why 9 - just so that you see circle clearly drawn in the matrix). Below code works fine for the matrix of the odd size. Let me know if anyone has a better solution.

<pre><code> private static void drawCircle(int[][] emptyMatrix, int diameter) {
 int startX = diameter/2;
 int startY = diameter/2;
 int radius = diameter/2;
 drawCircleRecursive(emptyMatrix, diameter, startX, startY, radius, radius);

 System.out.println("Filled matrix: ");
 for (int i = 0; i &lt; emptyMatrix.length; i++) {
 for (int j = 0; j &lt; emptyMatrix[0].length; j++) {
 System.out.print(emptyMatrix[i][j] + " ");
 }
 System.out.println();
 }
 }

 private static void drawCircleRecursive(int[][] emptyMatrix, int d, int startX, int startY, int x, int y) {

 if(x &gt;= emptyMatrix.length || y &gt;= emptyMatrix[0].length || x &lt; 0 || y &lt; 0 || emptyMatrix[x][y] == 1)
 return;
 else if(emptyMatrix[x][y] == 9)
 return;

 int r = d/2;
 if (((x - startX) * (x - startX) + (y - startY) * (y - startY)) &lt;= (r * r))
 emptyMatrix[x][y] = 1; 
 else 
 emptyMatrix[x][y] = 9;

 drawCircleRecursive(emptyMatrix, d, startX, startY, x+1, y); // down
 drawCircleRecursive(emptyMatrix, d, startX, startY, x, y+1); // right
 drawCircleRecursive(emptyMatrix, d, startX, startY, x-1, y); //up
 drawCircleRecursive(emptyMatrix, d, startX, startY, x, y-1); //left
 drawCircleRecursive(emptyMatrix, d, startX, startY, x-1, y-1); // diagonal up-left
 drawCircleRecursive(emptyMatrix, d, startX, startY, x+1, y+1); // diagonal right-down
 drawCircleRecursive(emptyMatrix, d, startX, startY, x+1, y-1); // diagonal left-down
 drawCircleRecursive(emptyMatrix, d, startX, startY, x-1, y+1); // diagonal right-up


 }
</code></pre>

Which algorithm to use to get points of filled circle?

<pre><code>int start_X = 30; // center point
int start_Y = 30;

int r = 5;

// current point
int x; 
int y;

if(?==true)
{
map2D[x][y] = 1; // for filled circle points
}
</code></pre>

Filled circle in matrix(2D array)

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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使用哪种算法来获取实心圆的点？int start_X = 30; // center pointint start_Y = 30;int r = 5;// current pointint x; int y;if(?==true){map2D[x][y] = 1; // for filled circle points...

问矩阵中的实心圆(二维数组)
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