blocks|key|1464881|text|模(mod，%25)是余数运算符。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1464882|2%252+=+0+(2/2+=+1+remainder+0)
1%252+=+1+(1/2+=+0+remainder+1)
4%252+=+0+(4/2+=+2+remainder+0)
5%252+=+1+(5/2+=+2+remainder+1)|code-block|syntax|javascript|1464883|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|$]]

Modulo (mod, %) is the Remainder operator. 

<pre><code>2%2 = 0 (2/2 = 1 remainder 0)
1%2 = 1 (1/2 = 0 remainder 1)
4%2 = 0 (4/2 = 2 remainder 0)
5%2 = 1 (5/2 = 2 remainder 1)
</code></pre>

blocks|key|1464937|text|mod指的是除以时的校准器。所以2除以4等于0，剩下2。因此，2mod4等于2。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1464938|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

mod means the reaminder when divided by. So 2 divided by 4 is 0 with 2 remaining. Therefore 2 mod 4 is 2.

blocks|key|1250146|text|Modulo是余数，而不是除法。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1250147|2+/+4+=+0R2
2+%25+4+=+2|code-block|syntax|javascript|1250148|符号%25通常用来代替单词mod作为模运算符。|style|CODE|1250149|对于x+%25+4，您将获得下表(针对1-10)|1250150|+x+x%254
------
+1++1
+2++2
+3++3
+4++0
+5++1
+6++2
+7++3
+8++0
+9++1
10++2|1250151|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Modulo_operation^0|0|6|0|0|0|2|1|B|3|0|2|5|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@$A|Z|B|10|1|11]]|C|$]]|$1|D|3|E|5|F|7|12|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|13|8|@$A|14|B|15|K|L]|$A|16|B|17|K|L]]|9|@]|C|$]]|$1|M|3|N|5|6|7|18|8|@$A|19|B|1A|K|L]]|9|@]|C|$]]|$1|O|3|P|5|F|7|1B|8|@]|9|@]|C|$G|H]]|$1|Q|3|-4|5|6|7|1C|8|@]|9|@]|C|$]]]|R|$S|$5|T|U|V|C|$W|X]]]]

<a href="http://en.wikipedia.org/wiki/Modulo_operation">Modulo</a> is the remainder, not division.

<pre><code>2 / 4 = 0R2
2 % 4 = 2
</code></pre>

The sign <code>%</code> is often used for the modulo operator, in lieu of the word <code>mod</code>.

For <code>x % 4</code>, you get the following table (for 1-10)

<pre><code> x x%4
------
 1 1
 2 2
 3 3
 4 0
 5 1
 6 2
 7 3
 8 0
 9 1
10 2
</code></pre>

blocks|key|1465004|text|Mod只是意味着你在执行除法后取剩余的部分。因为4变成了2的0次，所以你最终得到的是2的余数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1465005|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Mod just means you take the remainder after performing the division. Since 4 goes into 2 zero times, you end up with a remainder of 2.

blocks|key|1250236|text|2/4=0，余数为2|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1250237|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

blocks|key|1465132|text|如果你在时钟上从4开始(就像从零开始)，顺时针旋转2个“小时”，你就会到达2，就像顺时针旋转6个“小时”也会得到2+(6+mod+4+==+2就像2+mod+4+==+2一样)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1465133|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

For a visual way to think about it, picture a clock face that, in your particular example, only goes to 4 instead of 12. If you start at 4 on the clock (which is like starting at zero) and go around it clockwise for 2 "hours", you land on 2, just like going around it clockwise for 6 "hours" would also land you on 2 (6 mod 4 == 2 just like 2 mod 4 == 2).

blocks|key|1250337|text|当你将2除以4时，你会得到0，剩下2或剩余。模是除数后的余数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1250338|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

When you divide 2 by 4, you get 0 with 2 left over or remaining. Modulo is just the remainder after dividing the number.

blocks|key|1250401|text|模运算符的计算结果是两个整数操作数的除法的余数。这里有几个例子：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1250402|23+%25+10+evaluates+to+3+(because+23/10+is+2+with+a+remainder+of+3)
50+%25+50+evaluates+to+0+(50/50+is+1+with+a+remainder+of+0)
9+%25+100+evaluates+to+9+(9/100+is+0+with+a+remainder+of+9)|code-block|syntax|javascript|1250403|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|$]]

The modulo operator evaluates to the remainder of the division of the two integer operands. Here are a few examples:

<pre><code>23 % 10 evaluates to 3 (because 23/10 is 2 with a remainder of 3)
50 % 50 evaluates to 0 (50/50 is 1 with a remainder of 0)
9 % 100 evaluates to 9 (9/100 is 0 with a remainder of 9)
</code></pre>

blocks|key|1250472|text|我认为你对如何读取模方程感到困惑。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1250473|当我们写一个像2/4这样的除法方程时，我们就是用2除以4。|offset|length|style|CODE|1250474|当一个模等式被写成像2+%25+4这样的时候，我们是在除以2+by+4+(想2除以4)并返回余数。|1250475|entityMap^0|0|7|3|0|A|5|R|6|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|M|8|@$D|N|E|O|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|P|8|@$D|Q|E|R|F|G]|$D|S|E|T|F|G]]|9|@]|A|$]]|$1|J|3|-4|5|6|7|U|8|@]|9|@]|A|$]]]|K|$]]

I think you are getting confused over how the modulo equation is read.

When we write a division equation such as <code>2/4</code> we are dividing 2 by 4.

When a modulo equation is wrote such as <code>2 % 4</code> we are dividing <code>2 by 4</code> (think 2 over 4) and returning the remainder.

blocks|key|1250531|text|MOD是余数运算符。这就是为什么2模4给出2作为余数。4*0=0，然后2-0=2。为了更清楚，尝试用6+mod+4或8+mod+3做同样的事情。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1250532|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

MOD is remainder operator. That is why 2 mod 4 gives 2 as remainder. 4*0=0 and then 2-0=2. To make it more clear try to do same with 6 mod 4 or 8 mod 3.

blocks|key|1250599|text|如果你使用香蕉和一群人，那就容易多了。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1250600|假设你有一个香蕉和一组6个人，你可以这样表达：1+mod+6+/+1+%25+6+/+1+modulo+6。|offset|length|style|CODE|1250601|你需要为小组中的每个人提供6个香蕉，才能得到良好的食物和幸福。|1250602|因此，如果你有一个香蕉并需要与6个人分享，但如果你每个组成员有1个香蕉，即6个人，那么你将有1个香蕉(剩余的，不分享给组中的任何人)，2个香蕉也是如此。然后你将有2个香蕉作为剩余(不共享任何东西)。|1250603|但是当你得到6个香蕉时，你应该感到高兴，因为在6个人的小组中，每个成员有1个香蕉，当你在6个人上分享所有6个香蕉时，剩余的香蕉是0或没有剩下的香蕉。|1250604|现在，对于7个香蕉和6个人在组中，你将有7+mod+6+=+1，这是因为你给了6个人每人1个香蕉，剩下的是1个香蕉。|1250605|对于12+mod+6或6个人共享的12个香蕉，每个人将有两个香蕉，剩下的就是0。|1250606|entityMap^0|0|N|7|X|5|15|A|0|0|0|0|K|B|0|2|8|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|U|8|@$D|V|E|W|F|G]|$D|X|E|Y|F|G]|$D|Z|E|10|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|11|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|12|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|13|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|14|8|@$D|15|E|16|F|G]]|9|@]|A|$]]|$1|P|3|Q|5|6|7|17|8|@$D|18|E|19|F|G]]|9|@]|A|$]]|$1|R|3|-4|5|6|7|1A|8|@]|9|@]|A|$]]]|S|$]]

Much easier if u use bananas and a group of people. 

Say you have 1 banana and group of 6 people, this you would express: <code>1 mod 6</code> / <code>1 % 6</code> / <code>1 modulo 6</code>. 

You need 6 bananas for each person in group to be well fed and happy. 

So if you then have 1 banana and need to share it with 6 people, but you can only share if you have 1 banana for each group member, that is 6 persons, then you will have 1 banana (remainder, not shared on anyone in group), the same goes for 2 bananas. Then you will have 2 banana as remainder (nothing is shared). 

But when you get 6 bananas, then you should be happy, because then there is 1 banana for each member in group of 6 people, and the remainder is 0 or no bananas left when you shared all 6 bananas on 6 people. 

Now, for 7 bananas and 6 people in group, you then will have <code>7 mod 6 = 1</code>, this because you gave 6 people 1 banana each, and 1 banana is the remainder. 

For <code>12 mod 6</code> or 12 bananas shared on 6 people, each one will have two bananas, and the remainder is then 0.

blocks|key|1250657|text|这是欧几里得算法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1250658|e.g|1250659|A+mod+b=k*b%2Bc+=>+a+mod+b=+c，其中k是整数，c是答案|1250660|4模2=2*2%2B0+=>+4模2=0|1250661|27模5=5*5%2B2+=>+27模5=2|1250662|所以你的答案是|1250663|2模4=0*4%2B2+=>+2模4=2|1250664|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Q|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|R|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|S|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|T|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|U|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|V|8|@]|9|@]|A|$]]|$1|N|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|O|$]]

This is Euclid Algorithm.

e.g

a mod b = k * b + c => a mod b = c, where k is an integer and c is the answer

4 mod 2 = 2 * 2 + 0 => 4 mod 2 = 0

27 mod 5 = 5 * 5 + 2 => 27 mod 5 = 2

so your answer is

2 mod 4 = 0 * 4 + 2 => 2 mod 4 = 2

blocks|key|1465541|text|模是数学除法表达式的余数，表示为整数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1465542|因此，假设你在屏幕上的位置90有一个像素，屏幕是100像素宽，再加上20，它将绕到位置10。Why...because+90+%2B+20+=+110，因此110+%25+100+=+10。|1465543|对我来说，我认为模是小数的整数表示。此外，如果您反向执行表达式，并将余数处理为小数，然后将其添加到除数，它将给出您原始的答案。|1465544|示例：|1465545|++++100
(A)+---++=++14+mod+2
+++++7

++++123
(B)+---++=++8+mod+3
+++++15

+++++3
(C)+---++=++0+mod+3
+++++4|code-block|syntax|javascript|1465546|反向工程可实现：|1465547|++++++++++++++++++++++++2++++++14(7)++++2+++++++98++++2+++++100
+(A)+14+mod+2++=++14+%2B+---++=++-----+%2B+---++=++---+%2B+---++=+---
++++++++++++++++++++++++7++++++++7++++++7+++++++7+++++7++++++7

++++++++++++++++++++++3++++++8(15)++++3++++++120++++3++++++123
+(B)+8+mod+3++=++8+%2B+---++=++-----+%2B+---++=++---+%2B+---++=++---
++++++++++++++++++++++15+++++++15+++++15++++++15++++15++++++15

++++++++++++++++++++++3+++++++3
+(B)+0+mod+3++=++0+%2B+---++=++---
++++++++++++++++++++++4+++++++4|1465548|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|U|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|V|8|@]|9|@]|A|$]]|$1|H|3|I|5|J|7|W|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$]]|$1|O|3|P|5|J|7|Y|8|@]|9|@]|A|$K|L]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

Modulo is the remainder, expressed as an integer, of a mathematical division expression. 

So, lets say you have a pixel on a screen at position 90 where the screen is 100 pixels wide and add 20, it will wrap around to position 10. Why...because 90 + 20 = 110 therefore 110 % 100 = 10.

For me to understand it I consider the modulo is the integer representation of fractional number. Furthermore if you do the expression backwards and process the remainder as a fractional number and then added to the divisor it will give you your original answer.

Examples:

<pre><code> 100
(A) --- = 14 mod 2
 7

 123
(B) --- = 8 mod 3
 15

 3
(C) --- = 0 mod 3
 4
</code></pre>

Reversed engineered to:

<pre><code> 2 14(7) 2 98 2 100
 (A) 14 mod 2 = 14 + --- = ----- + --- = --- + --- = ---
 7 7 7 7 7 7

 3 8(15) 3 120 3 123
 (B) 8 mod 3 = 8 + --- = ----- + --- = --- + --- = ---
 15 15 15 15 15 15

 3 3
 (B) 0 mod 3 = 0 + --- = ---
 4 4
</code></pre>

blocks|key|1465585|text|适用于：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1465586|2+mod+4|offset|length|style|CODE|1465587|我们可以使用这个我想了一下的小公式，也许它已经在我不知道的地方定义了，但对我来说是有效的，它真的很有用。|1465588|A+mod+B+=+C，其中C是答案|1465589|K+*+B+-+A+=+%7CC%7C式中，K是B适合A的次数|1465590|2+mod+4将为：|1465591|0+*+4+-+2+=+%7CC%7C|1465592|C+=+%7C-2%7C+=>+2|1465593|希望它能为您工作:)|1465594|entityMap^0|0|0|7|0|0|0|B|0|0|F|0|0|7|0|0|F|0|0|D|0|0^^$0|@$1|2|3|4|5|6|7|X|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Y|8|@$D|Z|E|10|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|11|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|12|8|@$D|13|E|14|F|G]]|9|@]|A|$]]|$1|L|3|M|5|6|7|15|8|@$D|16|E|17|F|G]]|9|@]|A|$]]|$1|N|3|O|5|6|7|18|8|@$D|19|E|1A|F|G]]|9|@]|A|$]]|$1|P|3|Q|5|6|7|1B|8|@$D|1C|E|1D|F|G]]|9|@]|A|$]]|$1|R|3|S|5|6|7|1E|8|@$D|1F|E|1G|F|G]]|9|@]|A|$]]|$1|T|3|U|5|6|7|1H|8|@]|9|@]|A|$]]|$1|V|3|-4|5|6|7|1I|8|@]|9|@]|A|$]]]|W|$]]

For: 

<code>2 mod 4</code>

We can use this little formula I came up with after thinking a bit, maybe it's already defined somewhere I don't know but works for me, and its really useful.

<code>A mod B = C</code> where C is the answer

<code>K * B - A = |C|</code> where K is how many times B fits in A

<code>2 mod 4</code> would be:

<code>0 * 4 - 2 = |C|</code>

<code>C = |-2| =&gt; 2</code>

Hope it works for you :)

blocks|key|1465624|text|Mod操作与提醒一起工作。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1465625|这就是所谓的模运算。|1465626|+a==b(mod+m)+
+then+m%7C(a-b)
+a-b=km+
+a=b%2Bkm
+So,+2=2%2B0*4|code-block|syntax|javascript|1465627|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

<h2>Mod operation works with reminder.</h2>

This is called modular arithmetic.

<pre><code> a==b(mod m) 
 then m|(a-b)
 a-b=km 
 a=b+km
 So, 2=2+0*4
</code></pre>

blocks|key|1465683|text|要回答模数x+%25+y，您需要问两个问题：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1465684|A-+y在x中运行了多少次，没有余数？对于2%254，这是0。|1465685|B-你需要添加多少才能从那里返回到x？要从0返回到2，你需要2-0，即2。|1465686|这些问题可以归结为这样一个问题:您需要在x除以y的整数结果上加多少，才能得到x|1465687|所谓整数，是指只对整数感兴趣，而不是对小数感兴趣。|1465688|小数除法余数(例如.283849)对模数不感兴趣，因为模数只处理整数。|1465689|entityMap^0|5|5|0|3|1|5|1|0|H|1|0|K|1|N|1|12|1|0|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@$9|S|A|T|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|Z|8|@$9|10|A|11|B|C]]|D|@]|E|$]]|$1|J|3|K|5|6|7|12|8|@$9|13|A|14|B|C]|$9|15|A|16|B|C]|$9|17|A|18|B|C]]|D|@]|E|$]]|$1|L|3|M|5|6|7|19|8|@]|D|@]|E|$]]|$1|N|3|O|5|6|7|1A|8|@]|D|@]|E|$]]|$1|P|3|-4|5|6|7|1B|8|@]|D|@]|E|$]]]|Q|$]]

To answer a modulo <code>x % y</code>, you ask two questions:

A- How many times <code>y</code> goes in <code>x</code> without remainder ? For 2%4 that's 0.

B- How much do you need to add to get from that back to <code>x</code> ? To get from 0 back to 2 you'll need 2-0, i.e. 2.

These can be summed up in one question like so:
How much will you need to add to the integer-ish result of the division of <code>x</code> by <code>y</code>, to get back at <code>x</code>?

By integer-ish it is meant only whole numbers and not fractions whatsoever are of interest.

A fractional division remainder (e.g. .283849) is not of interest in modulo because modulo only deals with integer numbers.

blocks|key|1465704|text|这可能是介绍modr()函数的好时机。它同时返回除法的整个部分和剩余部分。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1465705|print("\n+17+//+3+=",17//3,"+#+Does+the+same+thing+as+int(17/3)")
print("+17+%25++3+=",17%253,"+#+Modulo+division+gives+the+remainder.")
whole,+remain+=+divmod(17,3)
print("+divmod(17,3)+returns+->",divmod(17,3),end="")
print("+because+3+goes+into+17,",whole,"times+with+a+remainder+of",remain,end=".\n\n")|code-block|syntax|javascript|1465706|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 could be a good time to mention the modr() function. It returns both the whole and the remainder parts of a division.

<pre><code>print("\n 17 // 3 =",17//3," # Does the same thing as int(17/3)")
print(" 17 % 3 =",17%3," # Modulo division gives the remainder.")
whole, remain = divmod(17,3)
print(" divmod(17,3) returns -&gt;",divmod(17,3),end="")
print(" because 3 goes into 17,",whole,"times with a remainder of",remain,end=".\n\n")
</code></pre>

blocks|key|1465746|text|我的方法是，2%254可以解释为小于或等于2的4的最高因子是0，因此2(2%254的左操作数)减去(-)0是2|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1465747|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

The way I go about it is, 2%4 can be interpreted as what is the highest factor of 4 that is less or equal to 2, and that is 0, therefore 2 (the left operand from 2%4) minus(-) 0 is 2

I'm embarrassed to ask such a simple question. My term does not start for two more weeks so I can't ask a professor, and the suspense would kill me.

Why does 2 mod 4 = 2?

Why does 2 mod 4 = 2?

我很尴尬地问了这么简单的问题。我的学期还有两周才开始，所以我不能问教授，这样的悬念会要了我的命。为什么2mod4= 2？

问为什么2mod4= 2？
EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问为什么2mod4= 2？EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问为什么2mod4= 2？
EN