算法15---数论3---水仙花数

Posted 2020-08-13 taoliu_alex

算法15---数论3---水仙花数

简单地说。三维正整数在树枝上等于其各位数字的立方之和，称为水仙花数；

同样我们也可以定义一些更高等级的水仙花数，比如4位的，例如1634，1634位1,6,3,4的四次方和。

 

 1 //
 2 /*
 3     题目：水仙花数
 4     author taoliu——alex  2016.10
 5 
 6     主要实现：
 7     1 找出给定位数内的水仙花数；
 8     2 判断一个数是不是水仙花数；
 9 
10 */
11 
12 
13 
14 #include <stdio.h>
15 #include <math.h>
16 
17 
18 //输出n为的水仙花数
19 
20 void shuixianhua(int n)
21 {
22     long start,end;
23     long sum;
24 
25     start=(long)pow(10,n-1);
26     end=(long)pow(10,n)-1;
27     for (long i = start; i <=end; i++)
28     {
29         long num=i;
30         sum=0;
31         for (int j = 0; j < n; j++)
32         {
33             long temp=num%10;
34             sum=sum+(long)pow(temp,n);
35             num=(num-temp)/10;
36         }
37         if (sum==i)
38         {
39             printf("%ld\n",i);
40         }
41 
42     }
43 }
44 
45 
46 void  shuixianhua_judge(long n)
47 {
48     int digit=1;
49     long sum=0;
50     long temp ;
51     long num=n;
52     long num2=n;
53     while (num/10!=0)
54     {
55         digit=digit+1;
56         num=num/10;
57     }
58     printf("the digit is %d\n", digit);
59 
60 
61     for (int i = 0; i < digit; i++)
62     {
63         temp=num2%10;
64         //printf("%ld\n",temp);
65         sum=sum+(long)pow(temp,digit);
66         printf("%ld\n",sum );
67         num2=(num2-temp)/10;
68     }
69     //printf("%ld\n",sum);
70     if (sum==n)
71     {
72         printf("%ld is shuixianhua number\n",n);
73     }
74     else
75         printf("%ld is NOT shuixianhua number\n",n);
76 
77 }
78 
79 
80 int main()
81 {
82      long num;
83      printf("please input the number you want to judge!\n");
84      scanf("%ld",&num);
85      shuixianhua_judge(num);
86 
87 
88      int n;
89      printf("give the digit you want to find!\n");
90      scanf("%d",&n);
91      shuixianhua(n);
92 
93      return 0;
94 }