blocks|key|1843144|text|为什么不干脆：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1843145|++++int+decimalMaskA+=+1;
++++int+decimalMaskB+=+1;
++++while+(checkCount+>+0)
++++{+++
++++++++if+(checkCount+%25+2+==+0)
++++++++{
++++++++++++count2+=+count2+%2B+(checkCount+%25+10)*decimalMaskA;
++++++++++++decimalMaskA+*=+10;
++++++++}
++++++++else
++++++++{
++++++++++++countRest+=+countRest+%2B+(checkCount+%25+10)*decimalMaskB;
++++++++++++decimalMaskB+*=+10;
++++++++}
++++++++checkCount+/=+10;
++++}|code-block|syntax|javascript|1843146|count2和countRest将包含这些数字(135和24)，而不是计数。|offset|length|style|CODE|1843147|entityMap^0|0|0|0|6|7|9|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]|$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|N|$]]

Why not just:

<pre><code> int decimalMaskA = 1;
 int decimalMaskB = 1;
 while (checkCount &gt; 0)
 { 
 if (checkCount % 2 == 0)
 {
 count2 = count2 + (checkCount % 10)*decimalMaskA;
 decimalMaskA *= 10;
 }
 else
 {
 countRest = countRest + (checkCount % 10)*decimalMaskB;
 decimalMaskB *= 10;
 }
 checkCount /= 10;
 }
</code></pre>

<code>count2</code> and <code>countRest</code> will contain those numbers (135 and 24) instead of counts.

blocks|key|2271911|text|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2271912|当两个INT变量都包含数组中的所有数字时，我如何构建两个INT变量文件呢？|blockquote|2271913|2271914|2271915|我不能肯定，但我想你是在问如何将一个数字按照重要性的递减顺序组合起来。|2271916|要将一个数字“附加”到一个数字上，你可以将该数字乘以10，然后将该数字添加到该数字上。要创建“集合”数字，您可以对数组中的每个数字执行此操作，|2271917|int[]+digits+=+...
int+num+=+digits.Aggregate(0,+(numSoFar,+digit)+=>+10+*+numSoFar+%2B+digit);|code-block|syntax|javascript|2271918|作为一个循环，如下所示：|2271919|int+num+=+0;

foreach(int+digit+in+digits)
{
++++num+=+10+*+num+%2B+digit;
}|2271920|entityMap^0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|-4|4|5|6|U|7|@]|8|@]|9|$]]|$1|A|3|B|4|C|6|V|7|@]|8|@]|9|$]]|$1|D|3|-4|4|5|6|W|7|@]|8|@]|9|$]]|$1|E|3|-4|4|5|6|X|7|@]|8|@]|9|$]]|$1|F|3|G|4|5|6|Y|7|@]|8|@]|9|$]]|$1|H|3|I|4|5|6|Z|7|@]|8|@]|9|$]]|$1|J|3|K|4|L|6|10|7|@]|8|@]|9|$M|N]]|$1|O|3|P|4|5|6|11|7|@]|8|@]|9|$]]|$1|Q|3|R|4|L|6|12|7|@]|8|@]|9|$M|N]]|$1|S|3|-4|4|5|6|13|7|@]|8|@]|9|$]]]|T|$]]

<blockquote>
 How can I build two INT varabile
 when each one contains all of the
 numbers together from the array?
</blockquote>

I can't say for sure, but I think you're asking how to assemble a number given each of its digits in decreasing order of significance.

To 'append' a digit to a number, you can multiply the number by 10 and then add the digit to that. To create the 'assembled' number, you can perform this operation for each digit in the array, 

<pre><code>int[] digits = ...
int num = digits.Aggregate(0, (numSoFar, digit) =&gt; 10 * numSoFar + digit);
</code></pre>

As a loop, this would look like:

<pre><code>int num = 0;

foreach(int digit in digits)
{
 num = 10 * num + digit;
}
</code></pre>

blocks|key|223184|text|如果我理解你要找的是什么，这将会起到作用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|223185|++++++++//Setup+a+sample+array
++++++++int[]+a+=+new+int[2];
++++++++a[0]+=+2;
++++++++a[1]+=+4;

++++++++//Convert+each+item+to+a+string,+convert+that+to+a+string+array,+join+the+strings+and+turn+into+an+int
++++++++int+output+=+int.Parse(String.Join("",+a.Select(s+=>+s.ToString()).ToArray()));|code-block|syntax|javascript|223186|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 I understand what you're looking for this will do the trick:

<pre><code> //Setup a sample array
 int[] a = new int[2];
 a[0] = 2;
 a[1] = 4;

 //Convert each item to a string, convert that to a string array, join the strings and turn into an int
 int output = int.Parse(String.Join("", a.Select(s =&gt; s.ToString()).ToArray()));
</code></pre>

blocks|key|2277394|text|这对我来说很有效：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2277395|int+number+=+12345;

string+result1+=+"";
string+result2+=+"";

string+numberString+=+number.ToString();

for+(int+i+=+0;+i+<+numberString.Length;+i%2B%2B+)
{
++++if+(numberString[i]+%25+2+==+0)
++++{
++++++++result1+=+result1+%2B+numberString[i];
++++}
++++else
++++{
++++++++result2+=+result2+%2B+numberString[i];
++++}
}

int+evenNumbers+=+int.Parse(result1);
int+oddNumbers+=+int.Parse(result2);|code-block|syntax|javascript|2277396|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 works for me:

<pre><code>int number = 12345;

string result1 = "";
string result2 = "";

string numberString = number.ToString();

for (int i = 0; i &lt; numberString.Length; i++ )
{
 if (numberString[i] % 2 == 0)
 {
 result1 = result1 + numberString[i];
 }
 else
 {
 result2 = result2 + numberString[i];
 }
}

int evenNumbers = int.Parse(result1);
int oddNumbers = int.Parse(result2);
</code></pre>

blocks|key|1843217|text|另一个解决方案可能是..。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1843218|var+num1+=+94321;
var+oddFinal+=+0;
var+evenFinal+=+0;

var+odd+=+new+List<int>();
var+even+=+new+List<int>();

while(+num1>0+)
{
++++if(+num1+%25+2+==+0+)
++++++++odd.Add(+num1+%25+10+);
++++else
++++++++even.Add(+num1+%25+10+);

++++num1+=+num1+/+10;
}++++

for+(int+i+=+0;+i+<+odd.Count;+i%2B%2B)
{
++++oddFinal+%2B=+odd[i]+*+(int)+Math.Pow(10,i);
}

for+(int+i+=+0;+i+<+even.Count;+i%2B%2B)
{
++++evenFinal+%2B=+even[i]+*+(int)+Math.Pow(10,i);
}|code-block|syntax|javascript|1843219|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|$]]

Another solution could be...

<pre><code>var num1 = 94321;
var oddFinal = 0;
var evenFinal = 0;

var odd = new List&lt;int&gt;();
var even = new List&lt;int&gt;();

while( num1&gt;0 )
{
 if( num1 % 2 == 0 )
 odd.Add( num1 % 10 );
 else
 even.Add( num1 % 10 );

 num1 = num1 / 10;
} 

for (int i = 0; i &lt; odd.Count; i++)
{
 oddFinal += odd[i] * (int) Math.Pow(10,i);
}

for (int i = 0; i &lt; even.Count; i++)
{
 evenFinal += even[i] * (int) Math.Pow(10,i);
}
</code></pre>

blocks|key|1843234|text|尝试使用LINQ，|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1843235|int+num+=+92345;
++++++++++++string+strNum+=+Convert.ToString(num);

++++++++++++var+divisibleby2+=+from+c+in+strNum
+++++++++++++++++++++++++++++++where+int.Parse(c.ToString())+%25+2+==+0
+++++++++++++++++++++++++++++++select+c.ToString();

++++++++++++var+notDivisibleby2+=+from+c+in+strNum
+++++++++++++++++++++++++++++++where+int.Parse(c.ToString())+%25+2+!=+0
++++++++++++++++++++++++++++++++++select+c.ToString();

++++++++++++int+int_divisibleby2num+=+int.Parse(String.Join("",+divisibleby2.ToArray()));
++++++++++++int+int_Notdivisibleby2num+=+int.Parse(String.Join("",+notDivisibleby2.ToArray()));|code-block|syntax|javascript|1843236|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|$]]

Try this with LINQ,

<pre><code>int num = 92345;
 string strNum = Convert.ToString(num);

 var divisibleby2 = from c in strNum
 where int.Parse(c.ToString()) % 2 == 0
 select c.ToString();

 var notDivisibleby2 = from c in strNum
 where int.Parse(c.ToString()) % 2 != 0
 select c.ToString();

 int int_divisibleby2num = int.Parse(String.Join("", divisibleby2.ToArray()));
 int int_Notdivisibleby2num = int.Parse(String.Join("", notDivisibleby2.ToArray()));
</code></pre>

blocks|key|2277431|text|这会将数字12345拆分为数字135和24。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2277432|int+checkCount+=+12345;
int+even+=+0;
int+odd+=+0;
int+reverseEven+=+0;
int+reverseOdd+=+0;

while+(checkCount+>+0)+{
++++int+current+=+checkCount+%25+10;
++++if+(current+%25+2+==+0)+{
++++++++reverseEven+=+10+*+reverseEven+%2B+current;
++++}+else+{
++++++++reverseOdd+=+10+*+reverseOdd+%2B+current;
++++}

++++checkCount+/=+10;
}


while+(reverseEven+>+0)+{
++++even+=+10+*+even+%2B+reverseEven+%25+10;
++++reverseEven+/=+10;
}

while+(reverseOdd+>+0)+{
++++odd+=+10+*+odd+%2B+reverseOdd+%25+10;
++++reverseOdd+/=+10;
}

Console.WriteLine("even:+{0}",+even);
Console.WriteLine("odd:+{0}",+odd);|code-block|syntax|javascript|2277433|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 splits number 12345 to numbers 135 and 24.

<pre><code>int checkCount = 12345;
int even = 0;
int odd = 0;
int reverseEven = 0;
int reverseOdd = 0;

while (checkCount &gt; 0) {
 int current = checkCount % 10;
 if (current % 2 == 0) {
 reverseEven = 10 * reverseEven + current;
 } else {
 reverseOdd = 10 * reverseOdd + current;
 }

 checkCount /= 10;
}


while (reverseEven &gt; 0) {
 even = 10 * even + reverseEven % 10;
 reverseEven /= 10;
}

while (reverseOdd &gt; 0) {
 odd = 10 * odd + reverseOdd % 10;
 reverseOdd /= 10;
}

Console.WriteLine("even: {0}", even);
Console.WriteLine("odd: {0}", odd);
</code></pre>

blocks|key|3914327|text|每个程序员每天至少应该写一次古怪的代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3914328|++++++++int+checkCount+=+12345,+numEven,+numOdd;
++++++++Boolean+result;

++++++++result+=+int.TryParse(checkCount.ToString().Replace("0",+"").Replace("2",+"").Replace("4",+"").Replace("6",+"").Replace("8",+""),+out+numOdd);
++++++++result+=+int.TryParse(checkCount.ToString().Replace("1",+"").Replace("3",+"").Replace("5",+"").Replace("7",+"").Replace("9",+""),+out+numEven);|code-block|syntax|javascript|3914329|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|$]]

Every programmer should write wacky code at least once a day:

<pre><code> int checkCount = 12345, numEven, numOdd;
 Boolean result;

 result = int.TryParse(checkCount.ToString().Replace("0", "").Replace("2", "").Replace("4", "").Replace("6", "").Replace("8", ""), out numOdd);
 result = int.TryParse(checkCount.ToString().Replace("1", "").Replace("3", "").Replace("5", "").Replace("7", "").Replace("9", ""), out numEven);
</code></pre>

I should create two new numbers from a one, the first group will contain digits which are divisible by 2 and the other group will contain the others.

<pre><code>int checkCount = 94321, num1 = 94321, count2 = 0, countRest = 0;

 while (checkCount &gt; 0)
 { 
 if (checkCount % 2 == 0)
 count2++;
 else
 countRest++;
 checkCount /= 10;
 }

 int[] a = new int[count2];
 int[] b = new int[countRest];

 int k2 = 0, kRest = 0;

 for (int j = 0; j &lt; a.Length + b.Length; j++) 
 {
 if (num1 % 2 == 0)
 {
 a[k2] = num1 % 10;
 k2++;
 }
 else
 {
 b[kRest] = num1 % 10;
 kRest++;
 }
 num1 /= 10;
 }
</code></pre>

I created two arrays with the numbers I should use, now how can I build two INT varabile when each one contains all of the numbers together from the array?

Example:
If I have this number - 12345 so
var = 24, other var = 135

If you have another solution without arrays I think it will be better.
Thank you.

Creating 2 numbers from a one #C

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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我应该从1创建两个新数字，第一组将包含可被2整除的数字，另一组将包含其他数字。int checkCount = 94321, num1 = 94321, count2 = 0, countRest = 0;        while (checkCount > 0)        {               if ...

问从1 #C创建2个数字
EN

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问从1 #C创建2个数字EN