How do I solve this mystery number problem?

Question

My problem is, I'm doing some practice while I'm sick and I don't go to school and I ran into a problem which I was solving whole last night, but looks so simple yet it is complicated, so here's the problem: /*Write a program where the user will load the three-digit integer and sums all digits of the number and calculates a mystery number with the following equation:

mn = round((a^3+ b^2+c^ )/(a+2*b+3*c)) , where a is the first digit, b is the second digit and c is the third digit

In the first line of the console must be the sum of the digits and in the second has to be the mystery number.*/

And here's where I got the problem, no matter how many times I try it just prints that mystery number is 0, but the sum is alright, although I could write the code a bit better here's the code:

    #include <math.h>
    #include <stdio.h>

    int main(){
       int n;
       int a, b, c, sum;
       double mn;

       printf("n = ");
       scanf("%d", &n);

       a = (n / 100);
       b = (n / 10) - a * 10;
       c = n - a * 100 - b * 10;

       sum = a + b + c;

       mn = round((a^3 + b^2 + c) / (a + 2 * b + 3 * c));

       printf("\nmn = %f", mn);
       printf("\nsum is = %d", sum);

       return 0;

For this code I used stdio.h and math.h. Help me. Please.

Answer 1

In mn = round((a^3 + b^2 + c) / (a + 2 * b + 3 * c)); :

• ^ is not exponentiation; a^3 does not calculate the cube of a .
• (a^3 + b^2 + c) and (a + 2 * b + 3 * c) are integer expressions, so / performs integer division, which truncates.

To calculate simple powers of integer values, simply multiply them as needed, such as a*a*a and b*b .

To perform floating-point division instead of integer division, convert either of the operands to double , as with (double) (a*a*a + b*b + c) .

Answer 2

您将方程的结果分配给tb并打印mn ，它从未初始化并且可能将 0 作为其默认值。

Answer 3

You don't seem to store anything in variable mn, instead I see you're using a variable named tb for you mystery number calculation. I also second the fact that the ^operator doesn't do exponentiation, it is a bitwise xor.