C 语言实例 – 判断Armstrong数(阿姆斯壮数)

最后更新于:2022-03-27 02:17:34

C 语言实例 – 判断Armstrong数(阿姆斯壮数)

C 语言实例 – 判断Armstrong数(阿姆斯壮数) 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+3^3
1634=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: 371
371 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 1000
100 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: 371
371 不是素数
371 是 Armstrong 数。

C 语言实例 – 判断Armstrong数(阿姆斯壮数) C 语言实例