C程序显示两个区间的阿姆斯特朗数字

为了显示两个区间的阿姆斯特朗数字，我们可以使用2种不同的方法，即4种方法。

1.不使用pow()函数
2.使用pow()函数

我们将在上述所有方法中保持相同的输入，并得到相应的输出。

输入:

start = 1, end = 500

输出:

解释： 阿姆斯特朗数字是其数字的立方体之和。

i.e. 1=13 
153= 13 + 53 + 33  
370= 33 + 73 + 03 etc.

方法1：不使用pow()函数

办法A

// C program to demonstrate an armstrong number
// between the given intervals
  
#include <stdio.h>
  
int main()
{
    int s = 1, e = 500, num, n, arm = 0, i, sum;
    
    // iterating the for loop 
    // using the given intervals
    for (i = s; i <= e; i++) {
        num = i;
        sum = i;
        
        // finding the armstrong number
        while (num != 0) {
            n = num % 10;
            arm = arm + (n * n * n);
            num = num / 10;
        }
        
        // if number is equal to 
        // the arm then it is a
        // armstrong number
        if (sum == arm) {
            printf("%d\n", i);
        }
        arm = 0;
    }
    return 0;
}

输出

办法B

// C program to demonstrate an armstrong number
// between the given intervals
#include <stdio.h>
  
int main()
{
  
    int s = 1, e = 500, num1, n, arm = 0, i, num2, c;
    
    // iterating the for loop using the given intervals
    for (i = s; i <= e; i++) {
        
        num1 = i;
        num2 = i;
        
        // finding the number of digits
        while (num1 != 0) {
            num1 = num1 / 10;
            ++c;
        }
        
        // finding the armstrong number
        while (num2 != 0) {
            n = num2 % 10;
            arm = arm + (n * n * n);
            num2 = num2 / 10;
        }
        
        // if number is equal to the arm then it is a
        // armstrong number
        if (arm == i) {
            printf("%d\n", i);
        }
        arm = 0;
        c = 0;
    }
    return 0;
}

输出

方法2：使用pow()函数

办法A

// C program to demonstrate an armstrong number
// between the given intervals using pow()
  
#include <math.h>
#include <stdio.h>
int main()
{
  
    int s = 1, e = 500, num, n, arm = 0, i, sum;
    // iterating the for loop using the given intervals
    for (i = s; i <= e; i++) {
        num = i;
        sum = i;
        // finding the armstrong number
        while (num != 0) {
            n = num % 10;
            arm = arm + pow(n, 3);
            num = num / 10;
        }
        // if number is equal to the arm then it is a
        // armstrong number
        if (sum == arm) {
            printf("%d\n", i);
        }
        arm = 0;
    }
    return 0;
}

输出

办法B

// C program to demonstrate an armstrong number
// between the given intervals using pow()
  
#include <math.h>
#include <stdio.h>
int main()
{
  
    int s = 1, e = 500, num1, n, arm = 0, i, num2, c;
    // iterating the for loop using the given intervals
    for (i = s; i <= e; i++) {
        num1 = i;
        num2 = i;
        // finding the number of digits
        while (num1 != 0) {
            num1 = num1 / 10;
            ++c;
        }
        // finding the armstrong number
        while (num2 != 0) {
            n = num2 % 10;
            arm = arm + pow(n, 3);
            num2 = num2 / 10;
        }
        // if number is equal to the arm then it is a
        // armstrong number
        if (arm == i) {
            printf("%d\n", i);
        }
        arm = 0;
        c = 0;
    }
    return 0;
}

输出