水仙花数是指一个N位正整数(N≥3),它的每个位上的数字的N次幂之和等于它本身。例如:153=1 11+555+333。本题要求编写两个函数,一个判断给定整数是否水仙花数,另一个按从小到大的顺序打印出给定区间(m,n)内所有的水仙花数。
函数narcissistic判断number是否为水仙花数,是则返回1,否则返回0。
函数PrintN则打印开区间(m, n)内所有的水仙花数,每个数字占一行。题目保证100≤m≤n≤10000。
int narcissistic( int number )
{
int a1,a2,a3,a4,flag;
if(number!=10000){
a1=number%10;
a2=number/10%10;
a3=number/100;
if(a1*a1*a1+a2*a2*a2+a3*a3*a3==number)
flag=1;
else
flag=0;}
else{
a1=number%10;
a2=number/10%10;
a3=number/10/10%10;
a4=number/1000;
if(a1*a1*a1*a1+a2*a2*a2*a2+a3*a3*a3*a3+a4*a4*a4*a4==number)
flag=1;
else
flag=0;
}
return flag;
}
void PrintN( int m, int n )
{
int i,a1,a2,a3;
for(i=m+1;i<n;i++){
if(narcissistic(i))
printf("%d\n",i);
}
}
#include <stdio.h>
#include <math.h>
int narcissistic(int number)
{
int x = number;
int n = (int)log10(x) + 1;
int sum = 0;
while (x)
{
sum += (int)pow(x % 10, n);
x /= 10;
}
return sum == number;
}
void PrintN(int m, int n)
{
for (int i = m; i <= n; i++)
{
if (narcissistic(i))
printf("%d\n", i);
}
}
int main()
{
int m, n;
scanf("%d%d", &m, &n);
PrintN(m, n);
return 0;
}
修改如下,供参考:
#include <stdio.h>
#include <math.h>
int narcissistic(int n)
{
int N, t, sum, m, k;
N = (int)log10(n) + 1;
for (sum = 0, k = 1, m = N, t = n; t > 0; k = 1, m = N, t /= 10) {
while (m--)k *= t % 10;
sum += k;
}
return sum == n;
}
void PrintN(int m, int n)
{
int i;
for (i = m; i <= n; i++)
{
if (narcissistic(i))
printf("%d\n", i);
}
}
int main()
{
int m, n;
scanf("%d%d", &m, &n);
PrintN(m, n);
return 0;
}