ALLZR - Editorial

editorial
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PROBLEM LINK:

Practice
Contest: Division 1
Contest: Division 2
Contest: Division 3
Contest: Division 4

Author: iceknight1093
Tester: sushil2006
Editorialist: iceknight1093

DIFFICULTY:

Cakewalk

PREREQUISITES:

None

PROBLEM:

You have integers A, B, C.
You can do the following:

  • Subtract 1 from A and 2 from B.
  • Subtract 1 from B and 3 from C.

Is it possible to make all three values equal to 0?

EXPLANATION:

The only possible way we can affect A is by performing a type 1 operation.
Each such operation reduces A by 1, so we need A operations to make the first variable 0.
Note that this will reduce B by 2A.

Similarly, the only way we can change C is by performing a type 2 operation.
This will reduce C by 3 repeatedly.
So,

  • If C is not a multiple of 3, it’s impossible to make it reach 0.
  • Otherwise, we need \frac{C}{3} type 3 operations to make C equal to 0.
    In this case, B will reduce by \frac{C}{3}.

After performing the forced operations that make A and C zero, no more operations can be performed - so B must already equal 0 as well.
B has reduced by both 2A and \frac{C}{3}, so it will equal 0 only if equals the sum of these two values.

Hence, we obtain the following:

  • If C is not a multiple of 3, no solution exists.
  • Otherwise, the answer is Yes if and only if
B = 2A + \frac{C}{3}

TIME COMPLEXITY:

\mathcal{O}(1) per testcase.

CODE:

Editorialist's code (PyPy3) 
for _ in range(int(input())):
    a, b, c = map(int, input().split())
    print('Yes' if c%3 == 0 and b == 2*a + (c//3) else 'No')