华为机试题：水仙花数

冷不防 2022-08-20 04:29 194阅读 0赞

<table style="margin:0px; padding:0px; width:970px; border:0px; border-collapse:collapse; border-spacing:0px; empty-cells:show; color:rgb(0,0,0); font-family:'Times New Roman',Arial,微软雅黑,宋体,Tahoma,Helvetica,SimSun,sans-serif,宋体"> 
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   <td style="margin:0px; padding:6px 3px 3px; word-break:break-all; word-wrap:break-word; border-right-style:solid; border-right-width:1px; border-right-color:rgb(238,238,238); font-size:14px; border-bottom-style:solid; border-bottom-width:1px; border-bottom-color:rgb(238,238,238)"> 描述:&nbsp;</td> 
   <td style="margin:0px; padding:6px 3px 3px; word-break:break-all; word-wrap:break-word; font-size:14px; border-bottom-style:solid; border-bottom-width:1px; border-bottom-color:rgb(238,238,238)"> <p style="margin-top:0px; margin-bottom:0px; padding-top:0px; padding-bottom:0px"> <span style="margin:0px; padding:0px; font-size:13px">水仙花数又称阿姆斯特朗数。</span></p> <p style="margin-top:0px; margin-bottom:0px; padding-top:0px; padding-bottom:0px"> <span style="margin:0px; padding:0px; font-size:13px">水仙花数是指一个n&nbsp;位数(&nbsp;n</span><span style="margin:0px; padding:0px; font-size:13px">≥</span><span style="margin:0px; padding:0px; font-size:13px">3&nbsp;)</span><span style="margin:0px; padding:0px; font-size:13px">，它的每个位上的数字的n&nbsp;次幂之和等于它本身。（例如：1^3&nbsp;+&nbsp;5^3&nbsp;+&nbsp;3^3&nbsp;=&nbsp;153）</span></p> <p style="margin-top:0px; margin-bottom:0px; padding-top:0px; padding-bottom:0px"> <span style="margin:0px; padding:0px; font-size:13px">求输入的数字是否为水仙花数</span></p> </td> 
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</table>

注意：是每位数的n次幂，并且少于三位的数，也不是水仙花数

#include <stdlib.h>
    #include <math.h>
    #include "oj.h"
    
    // 功能：判断输入 nValue 是否为水仙花数
    // 输入： nValue为正整数
    // 输出：无
    // 返回：如果输入为水仙花数，返回1，否则返回0
    unsigned int IsDaffodilNum(unsigned int  nValue)
    {
    	int data[10] = { 0 };
    	int i = 0,sum = 0;
    	unsigned int temp = nValue;
    	int bits = 0;
    
    	/*先判断数的位数，注意数小于三位不是水仙花数*/
    	while (temp)
    	{
    		bits++;
    		temp = temp / 10;
    	}
    
    	if (bits < 3)
    	{
    		return 0;
    	}
    
    	/*求出每一位的数*/
    	temp = nValue;
    	while (temp)
    	{
    		data[i++] = temp % 10;
    		temp = temp / 10;
    	}
    
    	/*判断是否是水仙花数*/
    	sum = 0;
    	for (i= 0; i < bits; i++)
    	{
    		sum += pow(static_cast<double>(data[i]), bits);
    	}
    
    	if (sum == nValue)
    	{
    		return 1;
    	}
    	return 0;
    }