Consider $(\var{n[0]})\times (\var{n[1]})$. Let's move the negatives out the front to get $--\var{-n[0]}\times \var{-n[1]}$, we can just do the multiplication and get $--\var{ans1}$, but this is the same as $\var{ans1}$.

In essence, we work out the numbers and the signs separately.

In general, two negatives multiplied results in a positive. 

Any even number of negative symbols will be the same as a positive symbol and any odd number of negative symbols will be the same as a negative symbol. For example,

$$-\times-\times-\times-\times-=+\times-=-$$

This is the same process as the last question. The lack of brackets doesn't affect this question.

Consider $(\var{n[2]})\times (\var{n[3]})$. Let's move the negatives out the front to get $--\var{-n[2]}\times \var{-n[3]}$, we can just do the multiplication and get $--\var{ans2}$, but this is the same as $\var{ans2}$.

In essence, we work out the numbers and the signs separately.

In general, two negatives multiplied results in a positive. 

Any even number of negative symbols will be the same as a positive symbol and any odd number of negative symbols will be the same as a negative symbol. For example,

$$-\times-\times-\times-\times-=+\times-=-$$

Recall that $(\var{m[0]})(\var{m[1]})(\var{m[2]})$ is a way of writing $(\var{m[0]})\times(\var{m[1]})\times(\var{m[2]})$.

Let's move the negatives out the front to get $---\var{-m[0]}\times \var{-m[1]}\times \var{-m[2]}$, we can just do the multiplication and get $---\var{-ans3}$, but this is the same as $\var{ans3}$.

In essence, we work out the numbers and the signs separately.

In general, two negatives multiplied results in a positive. 

Any even number of negative symbols will be the same as a positive symbol and any odd number of negative symbols will be the same as a negative symbol. For example,

$$-\times-\times-\times-\times-=+\times-=-$$