C程序 查找1-1000之间阿姆斯特朗数字

一个有n个数字的正整数在以下情况下被称为阶数为n的阿姆斯特朗数（阶数是数字的数量）：

**xyz… = pow(x,n) + pow(y,n) + pow(z,n) + ….**

在这里，我们将建立一个C程序来打印1到1000之间的阿姆斯特朗数字

输入:

lowerlimit = 1
higherlimit = 1000

输出:

1 2 3 4 5 6 7 8 9 153 370 371 407 

示例:

// C Program to Display Armstrong
// numbers between 1 to 1000
#include <math.h>
#include <stdio.h>
  
int power(int x, unsigned int y)
{
    int res = 1; // Initialize result
  
    while (y > 0) {
        // If y is odd, multiply
        // x with result
        if (y & 1)
            res = res * x;
  
        // y must be even now
        y = y >> 1; // y = y/2
        x = x * x; // Change x to x^2
    }
    return res;
}
  
int main()
{
    printf("Armstrong Numbers between 1-1000 are:\n");
  
    for (int i = 1; i < 1000; ++i) {
        int x = i;
        int n = 0;
  
        // number of digits calculation
        while (x != 0) {
            x /= 10;
            ++n;
        }
        int powersum = 0;
        x = i;
  
        // finding the sum of nth
        // power of its digits
        while (x != 0) {
            int digit = x % 10;
            powersum += power(digit, n);
            x /= 10;
        }
        // checking if the obtained
        // number is armstrong or not
        if (powersum == i)
  
            // printing the result
            printf("%d ", i);
    }
    printf("\n");
    return 0;
}

输出

Armstrong Numbers between 1-1000 are:
1 2 3 4 5 6 7 8 9 153 370 371 407