首页分享小算法：水仙花数

小算法：水仙花数

来源：花匠小妙招时间：2024-11-06 13:27

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 ( "%drn", 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=%dn",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=%dn",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);
}