$//$ C$++$ code

#include

#include

#include

#include

#include

#include

using namespace std;

#define n $100$

int A$[$n$]$;

auto printall $=[]($int p$)\{$

cout endl;

for $($int i $=0$; i $$ p; i$++)\{$

for $($int j $=0$; j $$ n $/$ p; j$++)\{$

cout $$ std::setw$(6)$ A$[$i $*$ n $/$ p $+$ j$]$;

$\}$

cout $$ endl;

$\}$

cout $$ endl;

$\}$;

int main$()$

$\{$

srand$($time$($NULL$))$;

for $($int i $=0$; i $$ n; i$++)$

A$[$i$]=$ i;$//$ rand$()\%10$;

int p $=4$; $//$ Assuming that p indicates # of processors

double k $=$ n $/($double$)$p;

printall$($p$)$;

$//$ Stage $1$: each Processor i computes a local prefix sum

$//$ of a subarray of size n$/$p $=$ log$($n$)=$ k

for $($int i $=0$; i $$ p; i$++)\{$

for $($int j $=1$; j $$ k; j$++)\{$

int sourcePos $=$ i $*$ k $+$ j $-1$;

int destPos $=$ i $*$ k $+$ j;

A$[$destPos$]+=$ A$[$sourcePos$]$;

$\}$

$\}$

printall$($p$)$;

$//$ Stage $2$: Prefix summation using only the rightmost value

$//$ of each subarray $($O$($log$($n$/$k$)))$

int i $=0$;

for $($int j $=$ pow$(2,$ i$)$; j p; j$++)\{$

int destPos $=$ j $*$ k $-1$;

int sourcePos $=($j $-$ pow$(2,$ i$))*$ k $-1$;

if $($destPos $>0$ && sourcePos $>0)$

A$[$destPos$]+=$ A$[$sourcePos$]$;

$\}$

printall$($p$)$;

$//$ Stage $3$: each Proc i adds the value computed in Step $2$ by Proc i$-1$ to

$//$ each subarray element except for the last one

for $($int i $=1$; i $$ p; i$++)\{$

int sourcePos $=$ i $*$ k $-1$;

for $($int j $=0$; j $$ k $-1$; j$++)\{$

int destPos $=$ i $*$ k $+$ j;

A$[$destPos$]+=$ A$[$sourcePos$]$;

$\}$

$\}$

printall$($p$)$;

return $0$;

$\}$

PLEASE PROVIDE $2$ CODE. ONE NON$-$THREAD CODE. AND ONE THREADED CODE. FOR THREADED CODE, PLEASE USE $4$ THREAD. THANKS