C 语言实例Armstrong 数，就是n位数的各位数的n次方之和等于该数，如：
153=1^3+5^3+3^31634=1^4+6^4+3^4+4^4
实例
#include <stdio.h>int main(){ int number, originalNumber, remainder, result = 0; printf("输入三位数: "); scanf("%d", &number); originalNumber = number; while (originalNumber != 0) { remainder = originalNumber%10; result += remainder*remainder*remainder; originalNumber /= 10; } if(result == number) printf("%d 是 Armstrong 数",number); else printf("%d 不是 Armstrong 数",number); return 0;}
运行结果：
输入三位数: 371371 是 Armstrong
实例 - 两数之间的 Armstrong 数
#include <stdio.h>#include <math.h> int main(){ int low, high, i, temp1, temp2, remainder, n = 0, result = 0; printf("输入两个整数: "); scanf("%d %d", &low, &high); printf("%d 和 %d 之间的 Armstrong 数为: ", low, high); for(i = low + 1; i < high; ++i) { temp2 = i; temp1 = i; // 计算 while (temp1 != 0) { temp1 /= 10; ++n; } while (temp2 != 0) { remainder = temp2 % 10; result += pow(remainder, n); temp2 /= 10; } if (result == i) { printf("%d ", i); } n = 0; result = 0; } return 0;}
运行结果：
输入两个整数: 100 1000100 和 1000 之间的 Armstrong 数为: 153 370 371 407
实例 - 使用函数判断Armstrong 数
#include <stdio.h>#include <math.h> int checkPrimeNumber(int n);int checkArmstrongNumber(int n); int main(){ int n, flag; printf("输入正整数: "); scanf("%d", &n); // 检测素数 flag = checkPrimeNumber(n); if (flag == 1) printf("%d 是素数。\n", n); else printf("%d 不是素数\n", n); // 检测 Armstrong 数 flag = checkArmstrongNumber(n); if (flag == 1) printf("%d 是 Armstrong 数。", n); else printf("%d 不是 Armstrong 数。",n); return 0;} int checkPrimeNumber(int n){ int i, flag = 1; for(i=2; i<=n/2; ++i) { // 非素数条件 if(n%i == 0) { flag = 0; break; } } return flag;} int checkArmstrongNumber(int number){ int originalNumber, remainder, result = 0, n = 0, flag; originalNumber = number; while (originalNumber != 0) { originalNumber /= 10; ++n; } originalNumber = number; while (originalNumber != 0) { remainder = originalNumber%10; result += pow(remainder, n); originalNumber /= 10; } // 判断条件 if(result == number) flag = 1; else flag = 0; return flag;}
输出结果为：
输入正整数: 371371 不是素数371 是 Armstrong 数。