blocks|key|1891041|text|你可能想得太多了。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1891042|if+(x+%25+10)
{
+++..+code+for+1..9+..
}+else
{
+++..+code+for+0,+10,+20+etc.
}|code-block|syntax|javascript|1891043|第一行if+(x+%25+10)之所以有效，是因为(a)+10的倍数的值被计算为'0'，其他数字的余数为‘0’，(b)+if中的值为0被认为是false，其他任何值都是true。|offset|length|style|CODE|1891044|编辑：|1891045|要在20岁时来回切换，请使用相同的技巧。这一次，关键数字是10|1891046|if+(((x-1)/10)+&+1)
{
++..+code+for+10,+30,+..
}+else
{
+++..+code+for+20,+40,+etc.
}|1891047|x/10返回0到9之间的任意数字作为0，返回10到19作为1，依此类推。测试偶数或奇数--+&+1+--会告诉您它是偶数还是奇数。因为您的范围实际上是"11到20"，所以在测试之前减去1。|1891048|entityMap^0|0|0|3|B|1M|2|1X|5|2A|4|0|0|T|2|0|0|0|4|I|1|T|1|1A|3|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@$I|Z|J|10|K|L]|$I|11|J|12|K|L]|$I|13|J|14|K|L]|$I|15|J|16|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|17|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|18|8|@$I|19|J|1A|K|L]]|9|@]|A|$]]|$1|Q|3|R|5|D|7|1B|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|1C|8|@$I|1D|J|1E|K|L]|$I|1F|J|1G|K|L]|$I|1H|J|1I|K|L]|$I|1J|J|1K|K|L]]|9|@]|A|$]]|$1|U|3|-4|5|6|7|1L|8|@]|9|@]|A|$]]]|V|$]]

You might be overthinking this.

<pre><code>if (x % 10)
{
 .. code for 1..9 ..
} else
{
 .. code for 0, 10, 20 etc.
}
</code></pre>

The first line <code>if (x % 10)</code> works because (a) a value that is a multiple of 10 calculates as '0', other numbers result in their remainer, (b) a value of 0 in an <code>if</code> is considered <code>false</code>, any other value is <code>true</code>.

Edit:

To toggle back-and-forth in twenties, use the same trick. This time, the pivotal number is <code>10</code>:

<pre><code>if (((x-1)/10) &amp; 1)
{
 .. code for 10, 30, ..
} else
{
 .. code for 20, 40, etc.
}
</code></pre>

<code>x/10</code> returns any number from 0 to 9 as <code>0</code>, 10 to 19 as <code>1</code> and so on. Testing on even or odd -- the <code>&amp; 1</code> -- tells you if it's even or odd. Since your ranges are actually "11 to 20", subtract 1 before testing.

blocks|key|1891085|text|你基本上自己解释了答案，但这是以防万一的代码。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1891086|if((x+%25+10)+==+0)+{
++//+Do+this
}
if((x+>+10+&&+x+<+21)+%7C%7C+(x+>+30+&&+x+<+41)+%7C%7C+(x+>+50+&&+x+<+61)+%7C%7C+(x+>+70+&&+x+<+81)+%7C%7C+(x+>+90+&&+x+<+101))+{
++//+Do+this
}|code-block|syntax|javascript|1891087|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|$]]

You basically explained the answer yourself, but here's the code just in case.

<pre><code>if((x % 10) == 0) {
 // Do this
}
if((x &gt; 10 &amp;&amp; x &lt; 21) || (x &gt; 30 &amp;&amp; x &lt; 41) || (x &gt; 50 &amp;&amp; x &lt; 61) || (x &gt; 70 &amp;&amp; x &lt; 81) || (x &gt; 90 &amp;&amp; x &lt; 101)) {
 // Do this
}
</code></pre>

blocks|key|1677389|text|对于第一个，要检查一个数字是否为多次使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1677390|if+(num+%25+10+==+0)+//+It's+divisible+by+10|code-block|syntax|javascript|1677391|对于第二个：|1677392|if(((num+-+1)+/+10)+%25+2+==+1+&&+num+<=+100)|1677393|但这是相当密集的，而且您最好只是显式地列出这些选项。|1677394|​|1677395|1677396|现在你已经对你正在做的事情有了一个更好的想法，我把第二个写成：|1677397|1677398|+++int+getRow(int+num)+{
++++++return+(num+-+1)+/+10;
+++}

+++if+(getRow(num)+%25+2+==+0)+{
+++}|1677399|这是相同的逻辑，但通过使用函数，我们对它的含义有了更清晰的理解。|1677400|entityMap^0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|12|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|13|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|14|8|@]|9|@]|A|$]]|$1|P|3|Q|5|6|7|15|8|@]|9|@]|A|$]]|$1|R|3|N|5|6|7|16|8|@]|9|@]|A|$]]|$1|S|3|T|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|U|3|V|5|6|7|18|8|@]|9|@]|A|$]]|$1|W|3|-4|5|6|7|19|8|@]|9|@]|A|$]]]|X|$]]

For the first one, to check if a number is a multiple of use:

<pre><code>if (num % 10 == 0) // It's divisible by 10
</code></pre>

For the second one:

<pre><code>if(((num - 1) / 10) % 2 == 1 &amp;&amp; num &lt;= 100)
</code></pre>

But that's rather dense, and you might be better off just listing the options explicitly.

<hr/>
Now that you've given a better idea of what you are doing, I'd write the second one as:

<pre><code> int getRow(int num) {
 return (num - 1) / 10;
 }

 if (getRow(num) % 2 == 0) {
 }
</code></pre>

It's the same logic, but by using the function we get a clearer idea of what it means.

blocks|key|1677500|text|这是为未来的访问者准备的，而不是初学者。对于更一般的类似算法的解决方案，您可以获取一个起始值和结束值的列表，并检查传递的值是否在其中一个值内：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1677501|template<typename+It,+typename+Elem>
bool+in_any_interval(It+first,+It+last,+const+Elem+&val)+{
++++return+std::any_of(first,+last,+[&val](const+auto+&p)+{
++++++++return+p.first+<=+val+&&+val+<=+p.second;
++++});
}|code-block|syntax|javascript|1677502|为简单起见，我使用了一个多态的lambda+(C%2B%2B14)而不是一个显式的pair参数。这也应该坚持使用<和==来与标准算法保持一致，但只要Elem为其定义了<=，它就会像这样工作。无论如何，它可以像这样使用：|offset|length|style|CODE|1677503|std::pair<int,+int>+intervals[]{
++++{11,+20},+{31,+40},+{51,+60},+{71,+80},+{91,+100}
};

const+int+num+=+15;
std::cout+<<+in_any_interval(std::begin(intervals),+std::end(intervals),+num);|1677504|这里有一个here的实例。|1677505|entityMap|0|LINK|mutability|MUTABLE|url|http://coliru.stacked-crooked.com/a/17eb0445853f1f19^0|0|0|11|4|1G|1|1I|2|1Y|4|27|2|0|0|5|4|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@$I|11|J|12|K|L]|$I|13|J|14|K|L]|$I|15|J|16|K|L]|$I|17|J|18|K|L]|$I|19|J|1A|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|1B|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|1C|8|@]|9|@$I|1D|J|1E|1|1F]]|A|$]]|$1|Q|3|-4|5|6|7|1G|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X]]]]

This is for future visitors more so than a beginner. For a more general, algorithm-like solution, you can take a list of starting and ending values and check if a passed value is within one of them:

<pre><code>template&lt;typename It, typename Elem&gt;
bool in_any_interval(It first, It last, const Elem &amp;val) {
 return std::any_of(first, last, [&amp;val](const auto &amp;p) {
 return p.first &lt;= val &amp;&amp; val &lt;= p.second;
 });
}
</code></pre>

For simplicity, I used a polymorphic lambda (C++14) instead of an explicit <code>pair</code> argument. This should also probably stick to using <code>&lt;</code> and <code>==</code> to be consistent with the standard algorithms, but it works like this as long as <code>Elem</code> has <code>&lt;=</code> defined for it. Anyway, it can be used like this:

<pre><code>std::pair&lt;int, int&gt; intervals[]{
 {11, 20}, {31, 40}, {51, 60}, {71, 80}, {91, 100}
};

const int num = 15;
std::cout &lt;&lt; in_any_interval(std::begin(intervals), std::end(intervals), num);
</code></pre>

There's a live example <a href="http://coliru.stacked-crooked.com/a/17eb0445853f1f19">here</a>.

blocks|key|1677554|text|第一个很简单。您只需对num值应用模运算符：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1677555|if+(+(+num+%25+10+)+==+0)|code-block|syntax|javascript|1677556|由于C%2B%2B会将每个不为0的数字都计算为真，因此您还可以这样写：|1677557|if+(+!+(+num+%25+10+)+)++//+Does+not+have+a+residue+when+divided+by+10|1677558|对于第二个，我认为这更容易理解：|1677559|这个模式每20次重复一次，所以你可以计算模20。除了那些可以被20整除的元素外，所有你想要的元素都将在一行中。|1677560|要获得这些，只需使用num-1或更好的num%2B19来避免处理负数。|1677561|if+(+(+(+num+%2B+19+)+%25+20+)+>+9+)|1677562|这是假设模式永远重复，因此对于111-120，它将再次应用，以此类推。否则，您需要将数量限制为100：|1677563|if+(+(+(+(+num+%2B+19+)+%25+20+)+>+9+)+&&+(+num+<=+100+)+)|1677564|entityMap^0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|12|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|13|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|14|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|15|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|16|8|@]|9|@]|A|$]]|$1|U|3|V|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|W|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|X|$]]

The first one is easy. You just need to apply the modulo operator to your num value:

<pre><code>if ( ( num % 10 ) == 0)
</code></pre>

Since C++ is evaluating every number that is not 0 as true, you could also write:

<pre><code>if ( ! ( num % 10 ) ) // Does not have a residue when divided by 10
</code></pre>

For the second one, I think this is cleaner to understand:

The pattern repeats every 20, so you can calculate modulo 20.
All elements you want will be in a row except the ones that are dividable by 20.

To get those too, just use num-1 or better num+19 to avoid dealing with negative numbers.

<pre><code>if ( ( ( num + 19 ) % 20 ) &gt; 9 )
</code></pre>

This is assuming the pattern repeats forever, so for 111-120 it would apply again, and so on. Otherwise you need to limit the numbers to 100:

<pre><code>if ( ( ( ( num + 19 ) % 20 ) &gt; 9 ) &amp;&amp; ( num &lt;= 100 ) )
</code></pre>

blocks|key|1677589|text|在代码中加上几个很好的注释，它就可以写得非常简洁和可读性很好。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1677590|//+Check+if+it's+a+multiple+of+10
if+(num+%25+10+==+0)+{+...+}

//+Check+for+whether+tens+digit+is+zero+or+even+(1-10,+21-30,+...)
if+((num+/+10)+%25+2+==+0)+{+...+}
else+{+...+}|code-block|syntax|javascript|1677591|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|$]]

With a couple of good comments in the code, it can be written quite concisely and readably.

<pre><code>// Check if it's a multiple of 10
if (num % 10 == 0) { ... }

// Check for whether tens digit is zero or even (1-10, 21-30, ...)
if ((num / 10) % 2 == 0) { ... }
else { ... }
</code></pre>

blocks|key|1677646|text|正如其他人指出的那样，使条件更简洁不会加快编译或执行速度，也不一定有助于可读性。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1677647|它有助于使您的程序更加灵活，以防您稍后决定希望在6x6的棋盘上使用幼儿版本的游戏，或者在40x50的棋盘上使用高级版本(可以整夜玩)。|1677648|因此，我会将其编码如下：|1677649|//+What+is+the+size+of+the+game+board?
#define+ROWS++++++++++++10
#define+COLUMNS+++++++++10

//+The+numbers+of+the+squares+go+from+1+(bottom-left)+to+(ROWS+*+COLUMNS)
//+(top-left+if+ROWS+is+even,+or+top-right+if+ROWS+is+odd)
#define+firstSquare+++++1
#define+lastSquare++++++(ROWS+*+COLUMNS)
//+We+haven't+started+until+we+roll+the+die+and+move+onto+the+first+square,
//+so+there+is+an+imaginary+'square+zero'
#define+notStarted(num)+(num+==+0)
//+and+we+only+win+when+we+land+exactly+on+the+last+square
#define+finished(num)+++(num+==+lastSquare)
#define+overShot(num)+++(num+>+lastSquare)

//+We+will+number+our+rows+from+1+to+ROWS,+and+our+columns+from+1+to+COLUMNS
//+(apologies+to+C+fanatics+who+believe+the+world+should+be+zero-based,+which+would
//++have+simplified+these+expressions)
#define+getRow(num)+++(((num+-+1)+/+COLUMNS)+%2B+1)
#define+getCol(num)+++(((num+-+1)+%25+COLUMNS)+%2B+1)

//+What+direction+are+we+moving+in?
//+On+rows+1,+3,+5,+etc.+we+go+from+left+to+right
#define+isLeftToRightRow(num)++++((getRow(num)+%25+2)+==+1)
//+On+rows+2,+4,+6,+etc.+we+go+from+right+to+left
#define+isRightToLeftRow(num)++++((getRow(num)+%25+2)+==+0)

//+Are+we+on+the+last+square+in+the+row?
#define+isLastInRow(num)++++(getCol(num)+==+COLUMNS)

//+And+finally+we+can+get+onto+the+code

if+(notStarted(mySquare))
{
++//+Some+code+for+when+we+haven't+got+our+piece+on+the+board+yet
}
else
{
++if+(isLastInRow(mySquare))
++{
++++//+Some+code+for+when+we're+on+the+last+square+in+a+row
++}


++if+(isRightToLeftRow(mySquare))
++{
++++//+Some+code+for+when+we're+travelling+from+right+to+left
++}
++else
++{
++++//+Some+code+for+when+we're+travelling+from+left+to+right
++}
}|code-block|syntax|javascript|1677650|是的，它是冗长的，但它清楚地说明了游戏板上发生了什么。|1677651|如果我正在开发这款游戏以在手机或平板电脑上显示，我将使行和列变量而不是常量，这样它们就可以动态设置(在游戏开始时)以匹配屏幕大小和方向。|1677652|我还允许在游戏中的任何时候改变屏幕方向--你所需要做的就是切换行和列的值，同时保持其他所有东西(每个玩家当前所在的方块数，以及所有蛇和梯子的开始/结束方块)不变。然后你“只要”很好地画出棋盘，并为你的动画编写代码(我假设这就是你的if语句的目的)……|offset|length|style|CODE|1677653|entityMap^0|0|0|0|0|0|0|37|2|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|X|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|Y|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|Z|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|10|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|11|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|12|8|@$Q|13|R|14|S|T]]|9|@]|A|$]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

As others have pointed out, making the conditions more concise won't speed up the compilation or the execution, and it doesn't necessarily help with readability either.

It can help in making your program more flexible, in case you decide later that you want a toddler's version of the game on a 6 x 6 board, or an advanced version (that you can play all night long) on a 40 x 50 board.

So I would code it as follows:

<pre><code>// What is the size of the game board?
#define ROWS 10
#define COLUMNS 10

// The numbers of the squares go from 1 (bottom-left) to (ROWS * COLUMNS)
// (top-left if ROWS is even, or top-right if ROWS is odd)
#define firstSquare 1
#define lastSquare (ROWS * COLUMNS)
// We haven't started until we roll the die and move onto the first square,
// so there is an imaginary 'square zero'
#define notStarted(num) (num == 0)
// and we only win when we land exactly on the last square
#define finished(num) (num == lastSquare)
#define overShot(num) (num &gt; lastSquare)

// We will number our rows from 1 to ROWS, and our columns from 1 to COLUMNS
// (apologies to C fanatics who believe the world should be zero-based, which would
// have simplified these expressions)
#define getRow(num) (((num - 1) / COLUMNS) + 1)
#define getCol(num) (((num - 1) % COLUMNS) + 1)

// What direction are we moving in?
// On rows 1, 3, 5, etc. we go from left to right
#define isLeftToRightRow(num) ((getRow(num) % 2) == 1)
// On rows 2, 4, 6, etc. we go from right to left
#define isRightToLeftRow(num) ((getRow(num) % 2) == 0)

// Are we on the last square in the row?
#define isLastInRow(num) (getCol(num) == COLUMNS)

// And finally we can get onto the code

if (notStarted(mySquare))
{
 // Some code for when we haven't got our piece on the board yet
}
else
{
 if (isLastInRow(mySquare))
 {
 // Some code for when we're on the last square in a row
 }


 if (isRightToLeftRow(mySquare))
 {
 // Some code for when we're travelling from right to left
 }
 else
 {
 // Some code for when we're travelling from left to right
 }
}
</code></pre>

Yes, it's verbose, but it makes it clear exactly what's happening on the game board.

If I was developing this game to display on a phone or tablet, I'd make ROWS and COLUMNS variables instead of constants, so they can be set dynamically (at the start of a game) to match the screen size and orientation.

I'd also allow the screen orientation to be changed at any time, mid-game - all you need to do is switch the values of ROWS and COLUMNS, while leaving everything else (the current square number that each player is on, and the start/end squares of all the snakes and ladders) unchanged.
Then you 'just' have to draw the board nicely, and write code for your animations (I assume that was the purpose of your <code>if</code> statements) ...

blocks|key|1677789|text|我知道这个问题有很多答案，但不管怎样，我还是会在这里提出我的答案。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1677790|摘自Steve+McConnell的，第二版：“Stair+Step+Access+Table：|1677791|另一种表访问是阶梯式方法。这种访问方法不像索引结构那样直接，但它不会浪费那么多的数据空间。如图18-5所示，阶梯式结构的一般思想是，表中的条目对数据范围有效，而不是对不同的数据点有效。|1677792|​|1677793|📷|atomic|offset|length|1677794|1677795|图18-5阶梯法通过确定每个条目到达“楼梯”的级别来对每个条目进行分类。它点击的“步骤”决定了它的类别。|1677796|例如，如果您正在编写评分程序，则“B”条目的范围可能在75%25到90%25之间。下面是一系列你可能有一天需要编程的等级：|1677797|1677798|1677799|1677800|要使用阶梯法，您可以将每个范围的上端放入一个表中，然后编写一个循环来对照每个范围的上端检查分数。当你找到分数第一次超过范围顶端的点时，你就知道分数是多少了。使用阶梯式技术时，您必须小心处理范围的端点。下面是Visual+Basic中的代码，它根据此示例将成绩分配给一组学生：|1677801|1677802|1677803|1677804|虽然这是一个简单的示例，但您可以很容易地将其概括为处理多个学生、多个评分方案(例如，不同作业的不同分数级别的不同评分)以及评分方案中的更改。|1677805|代码完整，第二版，第426+-+428页(第18章)。|1677806|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/65bf24ff6bebf035cade2c959d068a23.png|imageAlt|1|https://ask.qcloudimg.com/http-save/yehe-900000/b9cc65fc4a3ff00b90b99adbd2b60e15.png|2|https://ask.qcloudimg.com/http-save/yehe-900000/cc2a7e106a34a8881b63c0d438d11056.png^0|0|0|0|0|0|1|0|0|0|0|0|0|0|1|1|0|0|0|0|0|1|2|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1G|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1H|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|1I|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|1J|8|@]|9|@]|A|$]]|$1|H|3|I|5|J|7|1K|8|@]|9|@$K|1L|L|1M|1|1N]]|A|$]]|$1|M|3|G|5|6|7|1O|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|1P|8|@]|9|@]|A|$]]|$1|P|3|Q|5|6|7|1Q|8|@]|9|@]|A|$]]|$1|R|3|G|5|6|7|1R|8|@]|9|@]|A|$]]|$1|S|3|I|5|J|7|1S|8|@]|9|@$K|1T|L|1U|1|1V]]|A|$]]|$1|T|3|G|5|6|7|1W|8|@]|9|@]|A|$]]|$1|U|3|V|5|6|7|1X|8|@]|9|@]|A|$]]|$1|W|3|G|5|6|7|1Y|8|@]|9|@]|A|$]]|$1|X|3|I|5|J|7|1Z|8|@]|9|@$K|20|L|21|1|22]]|A|$]]|$1|Y|3|G|5|6|7|23|8|@]|9|@]|A|$]]|$1|Z|3|10|5|6|7|24|8|@]|9|@]|A|$]]|$1|11|3|12|5|6|7|25|8|@]|9|@]|A|$]]|$1|13|3|-4|5|6|7|26|8|@]|9|@]|A|$]]]|14|$15|$5|16|17|18|A|$19|1A|1B|-4]]|1C|$5|16|17|18|A|$19|1D|1B|-4]]|1E|$5|16|17|18|A|$19|1F|1B|-4]]]]

I know that this question has so many answers, but I will thrown mine here anyway...

Taken from Steve McConnell's <a href="https://en.wikipedia.org/wiki/Code_Complete" rel="nofollow noreferrer">Code Complete</a>, 2nd Edition:
"Stair-Step Access Tables: 

Yet another kind of table access is the stair-step method. This access method isn’t as direct as an index structure, but it doesn’t waste as much data space. The general idea of stair-step structures, illustrated in Figure 18-5, is that entries in a table are valid for ranges of data rather than for distinct data points.

<img src="https://i.stack.imgur.com/fta0S.png" alt="Enter image description here">

Figure 18-5 The stair-step approach categorizes each entry by determining the level at which it hits a “staircase.” The “step” it hits determines its category.

For example, if you’re writing a grading program, the “B” entry range might be from 75 percent to 90 percent. Here’s a range of grades you might have to program someday:

<img src="https://i.stack.imgur.com/kOyaa.png" alt="Enter image description here">

To use the stair-step method, you put the upper end of each range into a table and then write a loop to check a score against the upper end of each range. When you find the point at which the score first exceeds the top of a range, you know what the grade is. With the stair-step technique, you have to be careful to handle the endpoints of the ranges properly. Here’s the code in Visual Basic that assigns grades to a group of students based on this example:

<img src="https://i.stack.imgur.com/Ay77k.png" alt="Enter image description here">

Although this is a simple example, you can easily generalize it to handle multiple students, multiple grading schemes (for example, different grades for different point levels on different assignments), and changes in the grading scheme."

Code Complete, 2nd Edition, pages 426 - 428 (Chapter 18).

blocks|key|1677814|text|您可以尝试以下操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1677815|//+Multiple+of+10
if+((num+%25+10)+==+0)
{
+++//+Do+something
}
else+if+(((num+/+10)+%25+2)+!=+0)
{
++++//+11-20,+31-40,+51-60,+71-80,+91-100
}
+else
{
++++//+Other+case
}|code-block|syntax|javascript|1677816|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|$]]

You can try the following:
<pre><code>// Multiple of 10
if ((num % 10) == 0)
{
 // Do something
}
else if (((num / 10) % 2) != 0)
{
 // 11-20, 31-40, 51-60, 71-80, 91-100
}
 else
{
 // Other case
}
</code></pre>

I have a few loops that I need in my program. I can write out the pseudo code, but I'm not entirely sure how to write them logically.
I need -
<pre><code>if (num is a multiple of 10) { do this }

if (num is within 11-20, 31-40, 51-60, 71-80, 91-100) { do this }
else { do this } //this part is for 1-10, 21-30, 41-50, 61-70, 81-90
</code></pre>
This is for a snakes and ladders board game, if it makes any more sense for my question.
I imagine the first if statement I'll need to use modulus. Would <code>if (num == 100%10)</code> be correct?
The second one I have no idea. I can write it out like <code>if (num &gt; 10 &amp;&amp; num is &lt; 21 || etc.)</code>, but there has to be something smarter than that.

Determining if a number is either a multiple of ten or within a particular set of ranges

我的程序中有一些我需要的循环。我可以写出伪代码，但我不完全确定如何从逻辑上编写它们。我需要-if (num is a multiple of 10) { do this }if (num is within 11-20, 31-40, 51-60, 71-80, 91-100) { do this }else { do...

问确定一个数字是10的倍数还是在一组特定范围内
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

问确定一个数字是10的倍数还是在一组特定范围内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

问确定一个数字是10的倍数还是在一组特定范围内
EN