[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 = *** *
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*
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* **
*
6
10

Comments

Great

Answer 2

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* = a + *


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*
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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;
}