[Exercise 02] Simple Discriminant Calculation

Question

Please write a simple program that performs calculation and, shows its result equals to '10'

** There is no restriction as long as the result indicates 10 from its calculation **

** However, your codes should NOT be the same as the given example **

 

Example

#include <stdio.h>
int main(){

int a, b, c, d;

scanf ("%d %d %d", &a, &b, &c);

d = a + b * c;

printf ("%d", d);

return 0;
}

Input

2 2 4

Output

2 2 4
10

Suggestion

• Try to apply Addition, Subtraction, Multiplication, Division in your codes
• Be advised that 'Order of Operation' is crucial. The relative precedence levels of operators found in many C-style languages are as follows

Comments

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char *argv[]) {
    int a, b, c, d;
    /* prompt the user
    and read input numbers */
    printf("Input three numbers separated by space: ");
    scanf("%d%d%d", &a, &b, &c);
    //computation itself
    d = a / b * c;
    //print the result
    printf("%d / %d * %d = %d", a, b, c, d);
    return 0;
}
input 2 1 5 output 2 1 5 10

Answer 1

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a, b, c;
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c = a + b;
*** ** ***
*** ********* + %d = ****** * *
* ** *

** ****
* 6
*
* **
6
10

Comments

Great

Answer 2

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5
*

Comments

Good job

Answer 3

Hidden content!
* ***** *** * * ***
* *** * * * %d * * * * ** * ** ***** * = a * b + * * * ** ***
*
**** *
* * * 1 * *** * 1 + 8 = 10

Comments

Good job

Answer 4

Hidden content!
* ** *** ** * * ** ** * ** ** * * * ** the * ***** or add * own * *** * * * * * * or *** * *** * ** * *** ** a, b, * ** * the * ** * * * * ** * ** two ** ** by ** * ***** ** *** * * **** ** * * * = a + * *** *** * ********** + %d = a, b, * * * * ** * ****

Comments

Good job

Answer 5

Hidden content!
* * * ***** *

Comments

We have already received your codes. Nicely done.

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char *argv[]) {
    int a, b, c, d;
    /* prompt the user
    and read input numbers */
    printf("Input three numbers separated by space: ");
    scanf("%d%d%d", &a, &b, &c);
    //computation itself
    d = a / b * c;
    //print the result
    printf("%d / %d * %d = %d", a, b, c, d);
    return 0;
}