Розбір

Let $a,b,c$ represent the number of passengers in the first, second and third minibuses respectively. For the number of passengers in minibuses to be the equal after the transfer, the sum $a+b+c$ must be divisible by $3$.

Let $d=(a+b+c)/3$ be the number of passengers that should be in each minibus after the transfer. Then it is necessary to transfer from each minibus a certain number of passengers so that exactly $d$ passengers remain in each. This tranfer is only possible if a minibus initially contains more than $d$ passengers. For example, if $a>d$, then $a-d$ passengers should be transferred from the first minibus. Similarly, $b-d$ passengers should be transferred from the second minibus (if $b>d$), and $c-d$ passengers from the third minibus (if $c>d$).

Algorithm realization

Read the input data.

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

If the total number of people is not divisible by $3$, then print IMPOSSIBLE.

if((a + b + c) % 3 != 0)
  puts("IMPOSSIBLE");
else
{
  res = 0;

Compute the number $d$ of people in minibuses after the transfer.

  d = (a + b + c) / 3;

In the variable $res$ count the number of transfered people.

  if (a > d) res += a - d;
  if (b > d) res += b - d;
  if (c > d) res += c - d;

Print the answer.

  printf("%d\n",res);
}

Python realization

Read the input data.

a, b, c = map(int, input().split())

If the total number of people is not divisible by $3$, then print IMPOSSIBLE.

if (a + b + c) % 3 != 0:
  print("IMPOSSIBLE")
else:
  res = 0

Compute the number $d$ of people in minibuses after the transfer.

  d = (a + b + c) // 3

In the variable $res$ count the number of transfered people.

  if a > d: res += a – d
  if b > d: res += b – d
  if c > d: res += c – d

Print the answer.

  print(res)