Editorial

Let's find the sum of $n$ input integers — this is the number of coins that are heads-up. Let this sum be denoted as $s$. To make all the coins heads-up, you need to flip the remaining $n-s$ coins. To make all the coins tails-up, you need to flip s coins that are currently heads-up. Thus, the answer is $min(s,n-s)$.

Example

There are $s=3$ coins heads-up.

To make all the coins heads-up, you need to flip $n-s=5-3=2$ coins. To make all the coins tails-up, you need to flip $s=3$ coins. The answer to the problem is $min(s,n-s)=min(3,2)=2$ coins.

Algorithm realization

Read the input data. Compute the sum of $n$ integers in the variable $s$.

scanf("%d",&n);
for(i = 0; i < n; i++)
{
  scanf("%d",&v);
  s += v;
}

Compute and print the result — the value of $min(s,n-s)$.

if (s < n - s) printf("%d\n",s);
else printf("%d\n",n - s);