小算法 ： 水仙花数

水仙花数是指一个 n 位数 ( n≥3 )，它的每个位上的数字的 n 次幂之和等于它本身。（例如：1^3 + 5^3 + 3^3 = 153）

初步代码：

	int nMax = 9999999;
	int nResult[9999];
	int nCount = 0;
	for( int i = 0; i < nMax; i++ )
	{
		if( i < 10 )
		{
			if( i == i * i * i )
			{
				nResult[nCount] = i;
				nCount++;
			}	
		}
		else if( i < 100 )
		{
			int nG = i % 10;
			int nS = i / 10 % 10;
			if( i == nG * nG * nG + nS * nS * nS )
			{
				nResult[nCount] = i;
				nCount++;
			}
		}
		else if( i < 1000 )
		{
			int nG = i % 10;
			int nS = i / 10 % 10;
			int nB = i / 100 % 10;
			if( i == nG * nG * nG + nS * nS * nS + nB * nB * nB )
			{
				nResult[nCount] = i;
				nCount++;
			}
		}
		else if( i < 10000 )
		{
						
			int nG = i % 10;
			int nS = i / 10 % 10;
			int nB = i / 100 % 10;
			int nQ = i / 1000 % 10;
			if( i == nG * nG * nG * nG + nS * nS * nS * nS + nB * nB * nB * nB + nQ * nQ * nQ * nQ)
			{
				nResult[nCount] = i;
				nCount++;
			}
		}
		else if( i < 100000 )
		{
						
			int nG = i % 10;
			int nS = i / 10 % 10;
			int nB = i / 100 % 10;
			int nQ = i / 1000 % 10;
			int nW = i / 10000 % 10;
			if( i == nG * nG * nG * nG * nG + nS * nS * nS * nS * nS + nB * nB * nB * nB * nB + nQ * nQ * nQ * nQ * nQ \
				+ nW * nW * nW * nW * nW )
			{
				nResult[nCount] = i;
				nCount++;
			}
		}
		else if( i < 1000000 )
		{
						
			int nG = i % 10;
			int nS = i / 10 % 10;
			int nB = i / 100 % 10;
			int nQ = i / 1000 % 10;
			int nW = i / 10000 % 10;
			int nSW = i /100000 % 10;
			if( i ==	nG * nG * nG * nG * nG * nG + 
						nS * nS * nS * nS * nS * nS + 
						nB * nB * nB * nB * nB * nB + 
						nQ * nQ * nQ * nQ * nQ * nQ + 
						nW * nW * nW * nW * nW * nW + 
						nSW * nSW * nSW * nSW * nSW * nSW )
			{
				nResult[nCount] = i;
				nCount++;
			}
		}
		else if( i < 10000000 )
		{
						
			int nG = i % 10;
			int nS = i / 10 % 10;
			int nB = i / 100 % 10;
			int nQ = i / 1000 % 10;
			int nW = i / 10000 % 10;
			int nSW = i /100000 % 10;
			int nBW = i /1000000 % 10;

			if( i ==	nG * nG * nG * nG * nG * nG * nG + 
						nS * nS * nS * nS * nS * nS * nS +
						nB * nB * nB * nB * nB * nB * nB + 
						nQ * nQ * nQ * nQ * nQ * nQ * nQ +
						nW  * nW  * nW  * nW  * nW  * nW  * nW + 
						nSW * nSW * nSW * nSW * nSW * nSW * nSW+
						nBW * nBW * nBW * nBW * nBW * nBW * nBW )
			{
				nResult[nCount] = i;
				nCount++;
			}
		}
	}

	CString str;
	for( int j = 0; j < nCount; j++ )
	{
		CString strTmp;
		strTmp.Format ( "%d\r\n", nResult[j] );
		str += strTmp;
	}

	AfxMessageBox( str );

}

　　显然，上面的代码略显蠢笨，就像笨勤的郭靖，但却很认真。当然，如果要快速求出5位数以内的所有水仙花数，上面上家是最简单也是最快的。

　　但是，如果要判断任意一个整数是否是水仙花数，就不是很适用了。参考如下代码，由百度得到（我进行了整理，调试和存在的问题的修改）：

    long n;
    long p;
    long c,a,j,s[30],i,q;
    p=0;
    a=10;
    scanf("%d",&n);
    q=n;

    c = n; //JQB ADD
    
    //计算出位数
    for(i=1;c>10 ;++i)
    {
        c=n/a;
        a=a*10;
    }
    
    printf("i=%d，a=%d \n",i,a);

    //得到每一位数，并依次放入数组中
    for (j=1;a>=10 ;++j) 
    {
        s[j]=n/(a/10);
        n=n-s[j]*(a/10);
        a=a/10;
        printf("j=%d,a=%d\n",j,a);
    }

   //计算n位数的每一位的n次方的和
    for (j=1;j<=i ;j++) 
    {
        p+=pow(s[j],i);printf("p=%d,i=%d\n",p,i);
    }
    
    if (p==q) 
    {
        printf("%d 为水仙花数",q);
    }
    else
    {
        printf("%d 该数不是水仙花数",q);
    }