blocks|key|45131|text|您需要始终在for-i循环中设置sum+=+0。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|45132|entityMap^0|G|7|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|$]]

You need to always set <code>sum = 0</code> inside the for-i-loop.

blocks|key|3525621|text|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3525622|+Armstrong数:n位数，等于其数字的n次方和。|blockquote|3525623|3525624|3525625|从你的代码中|3525626|sum+=+sum+%2B+r*r*r;|code-block|syntax|javascript|3525627|“r*r*r”不是数字的n次方。|3525628|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|-4|4|5|6|Q|7|@]|8|@]|9|$]]|$1|A|3|B|4|C|6|R|7|@]|8|@]|9|$]]|$1|D|3|-4|4|5|6|S|7|@]|8|@]|9|$]]|$1|E|3|-4|4|5|6|T|7|@]|8|@]|9|$]]|$1|F|3|G|4|5|6|U|7|@]|8|@]|9|$]]|$1|H|3|I|4|J|6|V|7|@]|8|@]|9|$K|L]]|$1|M|3|N|4|5|6|W|7|@]|8|@]|9|$]]|$1|O|3|-4|4|5|6|X|7|@]|8|@]|9|$]]]|P|$]]

<blockquote>
 Armstrong numbers: n-digit numbers equal to sum of n-th powers of their digits.
</blockquote>

From your code 

<pre><code>sum = sum + r*r*r;
</code></pre>

'r*r*r' isn't n'th power of the number.

blocks|key|3241201|text|您可以使用log来计算n：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3241202|n+=+log(i)%2B1|code-block|syntax|javascript|3241203|然后正确计算r%5En并在求和中使用它：sum+%2B=+r%5En;。r*r*r不是计算它的正确方法。|offset|length|style|CODE|3241204|entityMap^0|0|0|6|3|I|B|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|$]]

you can calculate n using log:

<pre><code>n = log(i)+1
</code></pre>

then calculate <code>r^n</code> correctly and use it in your summation: <code>sum += r^n;</code>. r*r*r is not the correct way of calculating it.

blocks|key|45560|text|第一件事是您假设n(如在n次方中)总是3(在您的r*r*r中)。只有当您的初始值有三位数时才是正确的(如153示例)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|45561|您需要计算初始数字中的数字来计算n，然后将r*r*r替换为将r提升到n的n次方。|45562|然而，这并不能解释为什么找不到153。其原因是，除非找到匹配项，否则不会将sum重置为零。无论是否找到匹配项，都需要将其重置为零。|45563|entityMap^0|O|5|0|L|5|0|11|3|0^^$0|@$1|2|3|4|5|6|7|L|8|@$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|R|8|@$9|S|A|T|B|C]]|D|@]|E|$]]|$1|J|3|-4|5|6|7|U|8|@]|D|@]|E|$]]]|K|$]]

The first thing is that you're assuming that n (as in the nth power) is always three (in your <code>r*r*r</code>). That's only true if your initial value has three digits (as with the 153 example).

You need to count the digits in your initial number to calculate n, and then replace your <code>r*r*r</code> with raising r to the nth power.

This doesn't explain why 153 isn't found, though. The reason for that is because you aren't reseting <code>sum</code> to zero unless you find a match. You need to reset it to zero whether you found a match or not.

blocks|key|3525866|text|您的代码仅适用于n=3：sum+=+sum+%2B+r*r*r;|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3525867|您必须使用pow()函数(http://www.codecogs.com/reference/c/math.h/pow.php)来计算幂。(或者创建一个自定义的。)|3525868|entityMap|0|LINK|mutability|MUTABLE|url|http://www.codecogs.com/reference/c/math.h/pow.php^0|8|3|C|I|0|5|5|D|1E|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]|$9|S|A|T|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@$9|X|A|Y|1|Z]]|E|$]]|$1|H|3|-4|5|6|7|10|8|@]|D|@]|E|$]]]|I|$J|$5|K|L|M|E|$N|O]]]]

Your code only works with <code>n=3</code> : <code>sum = sum + r*r*r;</code>

You must use the <code>pow()</code> function (<a href="http://www.codecogs.com/reference/c/math.h/pow.php" rel="nofollow">http://www.codecogs.com/reference/c/math.h/pow.php</a>) to compute powers. (Or create a custom one.)

blocks|key|3525881|text|总结一下正确但部分的答案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3525882|//+#include+<math.h>

for+(long+int+i+=+1;+i+<=+num;+i%2B%2B)+
{+
++++long+int+n+=+0,+sum+=+0;++//++<---+here+
++++long+ing+temp+=+i;+
++++while+(+temp+!=+0+)+
++++{+
++++++++%2B%2Bn;
++++++++temp+/=+10;+
++++}+

++++temp+=+i;+
++++while+(+temp+!=+0+)+
++++{+
++++++++int+r+=+temp%2510;+
++++++++sum+%2B=+int(pow(double(r),+n));+//+<--+here
++++++++temp+/=+10;+
++++}+
++++if+(+i+==+sum+)+
++++{
++++++++cout+<<+i;
++++++++sum+=+0;+
++++}
}+|code-block|syntax|javascript|3525883|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|$]]

To summarize the right, but partial answers: 

<pre><code>// #include &lt;math.h&gt;

for (long int i = 1; i &lt;= num; i++) 
{ 
 long int n = 0, sum = 0; // &lt;--- here 
 long ing temp = i; 
 while ( temp != 0 ) 
 { 
 ++n;
 temp /= 10; 
 } 

 temp = i; 
 while ( temp != 0 ) 
 { 
 int r = temp%10; 
 sum += int(pow(double(r), n)); // &lt;-- here
 temp /= 10; 
 } 
 if ( i == sum ) 
 {
 cout &lt;&lt; i;
 sum = 0; 
 }
} 
</code></pre>

blocks|key|45594|text|@Power-inside，我看到了你的代码，很难更改和编辑你的代码，但我已经写了一个类似的代码来在给定的限制内生成Armstrong数，它工作得很好。这就是……|type|unstyled|depth|inlineStyleRanges|entityRanges|data|45595|++#include<iostream.h>
++#include<conio.h>
++++class+arm
+++++++{
++++int+a;
++++public:
+++++++void+display();
++++++++};
+++++void+arm::display()
++++++++++{
+++++++++++++cout<<"Enter+any+number+to+find+armstrong+numbers+less+than+it";
+++++++++++++cin>>a;
++++++++++++++for(int+i=a;i>=1;i--)
++++++++++++{
++++++++++++++int+d=i;
++++++++++++++int+b=i;
++++++++++++++int+c=i;
++++++++++++++int+count=0;
++++++++++++++++while(b!=0)
++++++++++++++++++++{
++++++++++++++++++++++b=b/10;
++++++++++++++++++++++count%2B%2B;
++++++++++++++++++++}
+++++++++++++++++++int+l,m;
+++++++++++++++++++m=0;
++++++++++++++++for(int+k=1;k<=count;k%2B%2B)
+++++++++++++++{
++++++++++++++++++l=c%2510;
+++++++++++++++++++c=c/10;
+++++++++++++++++++m=m%2Bl*l*l;
++++++++++++++++}



++++++++++++++if(d==m)

++++++++++++++cout<<d<<"\t";

+++++++++++++}

++++++++++++++}


++++++++++++void+main()
++++++++++++++{
+++++++++++++++++arm+k;
+++++++++++++++++k.display();
+++++++++++++++++getch();
++++++++++++++++}|code-block|syntax|javascript|45596|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|$]]

@Power-inside, I saw your code, its difficult to change your code and edit it, but I have written a similar code to generate Armstrong numbers in a given limit, and it works fine.
Here it is....

<pre><code> #include&lt;iostream.h&gt;
 #include&lt;conio.h&gt;
 class arm
 {
 int a;
 public:
 void display();
 };
 void arm::display()
 {
 cout&lt;&lt;"Enter any number to find armstrong numbers less than it";
 cin&gt;&gt;a;
 for(int i=a;i&gt;=1;i--)
 {
 int d=i;
 int b=i;
 int c=i;
 int count=0;
 while(b!=0)
 {
 b=b/10;
 count++;
 }
 int l,m;
 m=0;
 for(int k=1;k&lt;=count;k++)
 {
 l=c%10;
 c=c/10;
 m=m+l*l*l;
 }



 if(d==m)

 cout&lt;&lt;d&lt;&lt;"\t";

 }

 }


 void main()
 {
 arm k;
 k.display();
 getch();
 }
</code></pre>

The following is my simple attempt at generating <a href="http://oeis.org/A005188" rel="nofollow">Armstrong numbers</a>. But it only outputs "1". What might be wrong?

<pre><code>#include&lt;stdio.h&gt; 
#include&lt;conio.h&gt; 
#include&lt;iostream.h&gt;

int main() 
{ 
 clrscr();
 int r; 
 long int num = 0, i, sum = 0, temp; 

 cout &lt;&lt; "Enter the maximum limit to generate Armstrong number "; 
 cin &gt;&gt; num;
 cout &lt;&lt; "Following armstrong numbers are found from 1 to " &lt;&lt; num &lt;&lt; "\t \n"; 

 for(i=1;i&lt;=num;i++) 
 { 
 temp = i; 
 while( temp != 0 ) 
 { 
 r = temp%10; 
 sum = sum + r*r*r; 
 temp = temp / 10; 
 } 

 if ( i == sum ) {
 cout &lt;&lt; i;
 sum = 0; 
 }
 } 

 getch(); 

 return 0; 
}
</code></pre>

Simple C++ code to generate Armstrong numbers

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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下面是我生成的简单尝试。但它只输出"1“。可能出了什么问题？#include<stdio.h> #include<conio.h> #include<iostream.h>int main() { clrscr(); int r; long int num = 0, i, sum = 0, temp; cout << "Enter the maximum limit

问生成阿姆斯特朗数的简单C++代码
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