Cohsider the problem of adding two n$-$bit binary integers, stored in two n$-$element arrays A and B$.$ The sum of the two integers should be stored in binary form in an $($n$+1)-$element array C$.$ The following code should perform the addition. $($I don't error check that the A and B arrays have equal length, nor that they contain only $0,1$ values to start.$)$ int Array add$1($A$,$ B$)\{$ int C $($an Array$)$; C$=$ new int Array, length $($A$,$ length $+1)$; int carry $=0$; for $($int i$=0$ ; i$$ A$.$ length; i$++)\{$ C$\_-$ i$=($A$-$i$+$B$-$i$+$carry$)$

Consider the problem of adding two $n-$bit binary integers, stored in two $n-$element arrays A and $B.$ The sum of the two integers should be stored in binary form in an $(n+1)-$element array $C.$

The following code should perform the addition. $($I don't error check that the A and $B$ arrays have equal length, nor that they contain only $0,1$ values to start.$)$To formally state what the above code accomplishes, without changing the A nor B array values it creates and returns array C such that:sum$\_($k$=0)^("$A$.$length $")2^($k$)$C$\_($k$)=$sum$\_($k$=0)^("$A$.$length$-1")2^($k$)$A$\_($k$)+$sum$\_($k$=0)^("$A$.$length $"-1)2^($k$)$B$\_($k$).$ This equation relies on the fact that the numbers are represented in least significant bit first format.

We can also write a program for most significant bit first format, such as:Because the arrays represent something different, the formal statement of what it accomplishes is different: without changing the A nor B array values it creates and returns array C such that:

${\displaystyle \underset{k=0}{\overset{A.\text{l}\text{e}\text{n}\text{g}\text{t}\text{h}}{}}}{2}^k{C}_c={\displaystyle \underset{k=0}{\overset{A.\text{l}\text{e}\text{n}\text{g}\text{t}\text{h}-1}{}}}{2}^k{A}_a+{\displaystyle \underset{k=0}{\overset{A.\text{l}\text{e}\text{n}\text{g}\text{t}\text{h}-1}{}}}{2}^k{B}_b$

You need to fill in values:

$c=$

$a=$

$b=$