I need help on making the C$++$ Code

Programming Project $12\mathrm{.}17\mathrm{.}1$: $.$

The median m of a sequence of n elements is the element that would fall in the middle if the sequence was sorted. That is$,$ e $<=$ m for half the elements, and m $<=$ e for the others. Clearly, one can obtain the median by sorting the sequence, but one can do quite a bit better with the following algorithm that finds the kth element of a sequence between a $($inclusive$)$ and b $($exclusive$).($For the median, use k $=$ n$/2,$ a $=0,$ and b $=$ n$.)$

select$($k$,$ a$,$ b$)$

Pick a pivot p in the subsequence between a and b$.$

Partition the subsequence elements into three subsequences: the elements p

Let n$1,$ n$2,$ n$3$ be the sizes of each of these subsequences.

if k $<$ n$1$

return select$($k$,0,$ n$1).$

else if $($k $>$ n$1+$ n$2)$

return select$($k$,$ n$1+$ n$2,$ n$).$

else

return p$.$

Implement this algorithm and measure how much faster it is for computing the median of a random large sequence, when compared to sorting the sequence and taking the middle element.

#include

#include

#include

#include

using namespace std;

int select$($int k$,$ int a$,$ int b$)$

$\{$

int n$1=$a$,$ n$2=$b;

int n$3=$ n$1+$n$2$;

int p$=0$;

if$($kt

$\{$

return select$($k$,0,$ n$1)$;

p$++$;

$\}$

else if $($k $>$ n$1+$ n$2)$

$\{$

return select$($k$,$ n$1+$ n$2,$ n$3)$;

p$++$;

$\}$

else

$\{$

return p;

$\}$

$\}$

int main$()\{$

srand$($time$(0))$;

int n $=($rand$()\%10)$;

int k $=$ n$/2$;

int a $=0$;

int b $=$ n;

int jack $=$ select$($k$,$ a$,$ b$)$;

cout $<<$ n$<$