<Codesignal> Arithmetic Expression

Easy

Consider an arithmetic expression of the form a#b=c. Check whether it is possible to replace # with one of the four signs: +, -, * or / to obtain a correct expression.

Example

• For a = 2, b = 3, and c = 5, the output should be
  arithmeticExpression(a, b, c) = true.

  We can replace # with a + to obtain 2 + 3 = 5, so the answer is true.

• For a = 8, b = 2, and c = 4, the output should be
  arithmeticExpression(a, b, c) = true.

  We can replace # with a / to obtain 8 / 2 = 4, so the answer is true.

• For a = 8, b = 3, and c = 2, the output should be
  arithmeticExpression(a, b, c) = false.

  • 8 + 3 = 11 ≠ 2;
  • 8 - 3 = 5 ≠ 2;
  • 8 * 3 = 24 ≠ 2;
  • 8 / 3 = 2.(6) ≠ 2.

  So the answer is false.

Input/Output

• [execution time limit] 0.5 seconds (c)

• [input] integer a

  Guaranteed constraints:
  1 ≤ a ≤ 20.

• [input] integer b

  Guaranteed constraints:
  1 ≤ b ≤ 20.

• [input] integer c

  Guaranteed constraints:
  0 ≤ c ≤ 20.

• [output] boolean

  • true if the desired replacement is possible, false otherwise.

[C] Syntax Tips

// Prints help message to the console
// Returns a string
char * helloWorld(char * name) {
    char * answer = malloc(strlen(name) + 8);
    printf("This prints to the console when you Run Tests");
    strcpy(answer, "Hello, ");
    strcat(answer, name);
    return answer;
}

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Solution

bool arithmeticExpression(int a,int b,int c)
{
	return a+b==c||a-b==c||a*b==c||a/b==c&&a%b==0;
}