给定数字x，请确定给定数字是否为阿姆斯特朗数字。
n位数字的正整数称为if的n阶阿姆斯特朗数(order是位数)。
abcd... = pow(a,n) + pow(b,n) + pow(c,n) + pow(d,n) + ....
例子：
Input : 153
Output : Yes
153 is an Armstrong number.
1*1*1 + 5*5*5 + 3*3*3 = 153
Input : 120
Output : No
120 is not a Armstrong number.
1*1*1 + 2*2*2 + 0*0*0 = 9
Input : 1253
Output : No
1253 is not a Armstrong Number
1*1*1*1 + 2*2*2*2 + 5*5*5*5 + 3*3*3*3 = 723
Input : 1634
Output : Yes
1*1*1*1 + 6*6*6*6 + 3*3*3*3 + 4*4*4*4 = 1634
C++
// C++ program to determine whether the number is
// Armstrong number or not
#include
using namespace std;
/* Function to calculate x raised to the power y */
int power(int x, unsigned int y)
{
if( y == 0)
return 1;
if (y%2 == 0)
return power(x, y/2)*power(x, y/2);
return x*power(x, y/2)*power(x, y/2);
}
/* Function to calculate order of the number */
int order(int x)
{
int n = 0;
while (x)
{
n++;
x = x/10;
}
return n;
}
// Function to check whether the given number is
// Armstrong number or not
bool isArmstrong(int x)
{
// Calling order function
int n = order(x);
int temp = x, sum = 0;
while (temp)
{
int r = temp%10;
sum += power(r, n);
temp = temp/10;
}
// If satisfies Armstrong condition
return (sum == x);
}
// Driver Program
int main()
{
int x = 153;
cout << isArmstrong(x) << endl;
x = 1253;
cout << isArmstrong(x) << endl;
return 0;
}
Please refer complete article on Program for Armstrong Numbers for more details!Want to learn from the best curated videos and practice problems, check out the C Foundation Course for Basic to Advanced C.