blocks|key|3344694|text|如果C#有模(可能是%25运算符)，就使用它。(我假设x和y是整数。)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3344695|entityMap^0|A|1|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

If C# has modulo (presumably the <code>%</code> operator), use that. (I'm assuming x and y are integers.)

blocks|key|1989084|text|你在找这个吗？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1989085|++++UInt16+x+=+123;
++++UInt16+y+=+456;

++++UInt32+Point;

++++//+encode+y+high,+x+low
++++Point+=+(UInt32)(x+%7C+(y+<<+0x10));

++++//+decode
++++x+=+(UInt16)(Point+&+0x0000FFFF);
++++y+=+(UInt16)((Point+&+0xFFFF0000)+>>+0x10);|code-block|syntax|javascript|1989086|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|$]]

Are you looking for this?

<pre><code> UInt16 x = 123;
 UInt16 y = 456;

 UInt32 Point;

 // encode y high, x low
 Point = (UInt32)(x | (y &lt;&lt; 0x10));

 // decode
 x = (UInt16)(Point &amp; 0x0000FFFF);
 y = (UInt16)((Point &amp; 0xFFFF0000) &gt;&gt; 0x10);
</code></pre>

blocks|key|1277765|text|如果你知道30,000+>y>+-30,000，你可以这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1277766|x+=+(id+%2B+30000)+/+65536
y+=+id+-+(x+*+65536)|code-block|syntax|javascript|1277767|如果65536+>y>+-65536，则：|1277768|id+=+x+*+131072+%2B+y

x+=+(id+%2B+65536)+/+131072
y+=+id+-+(x+*+131702)|1277769|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

If you know that 30,000 > y > -30,000, you could do this:

<pre><code>x = (id + 30000) / 65536
y = id - (x * 65536)
</code></pre>

If 65536 > y > -65536 then:

<pre><code>id = x * 131072 + y

x = (id + 65536) / 131072
y = id - (x * 131702)
</code></pre>

blocks|key|1277796|text|你在评论中说y可以取一个从-65336到655536的值。在这种情况下，您应该使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1277797|id+=+x+*+131072+%2B+y;+//+131072+=+2%5E17|code-block|syntax|javascript|1277798|否则，假设id为100000，则可以将其转换为两个不同的坐标：|1277799|100000+==+1+*+65536+%2B+34464++//+x+=+1,+y+=+34464
100000+==+2+*+65536+-+31072++//+x+=+2,+y+=+-31072|1277800|因此，如果y的范围是-65536到65535，那么您应该使用|1277801|id+=+x+*+131072+%2B+y;|1277802|要转换回来，您应该使用grc提出的解决方案：|offset|length|1277803|x+=+(id+%2B+65536)+/+131072;
y+=+id+-+(x+*+131072);|1277804|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/users/645956/grc^0|0|0|0|0|0|0|B|3|0|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|14|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|15|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|16|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|18|8|@]|9|@$Q|19|R|1A|1|1B]]|A|$]]|$1|S|3|T|5|D|7|1C|8|@]|9|@]|A|$E|F]]|$1|U|3|-4|5|6|7|1D|8|@]|9|@]|A|$]]]|V|$W|$5|X|Y|Z|A|$10|11]]]]

You said in the comments that y can take a value from -65336 to 655536. In that case you should use:

<pre><code>id = x * 131072 + y; // 131072 = 2^17
</code></pre>

For otherwise let, id be 100000, then this could be translated to two different coordinates:

<pre><code>100000 == 1 * 65536 + 34464 // x = 1, y = 34464
100000 == 2 * 65536 - 31072 // x = 2, y = -31072
</code></pre>

So if the range of y is -65536 up to and including 65535, then you should use

<pre><code>id = x * 131072 + y;
</code></pre>

and to convert back, you should use the solution that <a href="https://stackoverflow.com/users/645956/grc">grc</a> proposed:

<pre><code>x = (id + 65536) / 131072;
y = id - (x * 131072);
</code></pre>

I'm calculating a ID for a grid cell like this:

<pre><code>id = x * 65536 + y;
</code></pre>

Trying to calculate x and y for a ID I do this:

<pre><code>x = id / 65536;
y = id - (x * 65536);
</code></pre>

This works, as long as Y is bigger than 0. If Y is negative, I have to calculate it like this:

<pre><code>x = (int)Math.Ceiling((double)id / 65536.0);
y = id - (x * 65536);
</code></pre>

How do I know if I have to round up or down before knowing what y is? Is there a better way to generate one ID for a cell based on x and y? I feel like I'm missing something obvious, but after hours of tracking down strange errors to rounding problems, my brain is not working.

Calculate a ID for a cell based on x and y

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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我正在计算一个网格单元的ID，如下所示：id = x * 65536 + y;为了计算ID的x和y，我这样做：x = id / 65536;y = id - (x * 65536);只要Y大于0，它就可以工作。如果Y是负的，我必须这样计算它：x = (int)Math.Ceiling((double)id / 6553...

问根据x和y计算单元格的ID
EN

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问根据x和y计算单元格的IDEN