Answer / majid - iran

#include<iostream>
#include<conio.h>
#include<time.h>

using namespace std;

int **m;
long q=0;
long num=0;
long number=0;

long fact( long n )
{
return ( n > 0 ? n*fact(n-1) : 1 );
}

long C( long n , long r )
{
return ( fact(n)/(fact(r)*fact(n-r)) );
}

bool isPrinted(int *a, long n)
{
bool is=true;
for(long i=0;i<num;i++)
{
is=true;
for(long j=0;j<n;j++)
{
if(a[j]!=m[i][j])
{
is=false;
}
}
if(is)return true;
}
return false;
}

void print(int *a,long n)
{
int swap=0;
for(long i=0;i<n;i++)
{
for(long j=i;j<n;j++)
{
if(a[j]<a[i])
{
swap=a[i];
a[i]=a[j];
a[j]=swap;
}
}
}
if(!isPrinted(a,n))
{
number++;
cout<<number<<".{";
for(long i=0;i<n;i++)
{
cout<<a[i];
m[q][i]=a[i];
if(i!=n-1)
{
cout<<",";
}
}
q++;
cout<<"}"<<endl;
}
}

void printAll(int *a,int *b,long n,long k)
{
if(q<num)
{
if(k==1)
{
print(a,n);
}
if(k==0)
{
print(a,n);
}
else if(k>0)
{
int *c, *d;
long p=0;
for(long j=0;j<k;j++)
{
p=0;
c=new int[k-1];
for(long t=0;t<k;t++)
{
if(t!=j)
{
c[p]=b[t];
p++;
}
}
d=new int[n];
for(long i=-1;i<n;i++)
{
for(long t=0;t<n;t++)
{
d[t]=a[t];
}
if(i!=-1)
{
d[i]=b[j];
}
printAll(d,c,n,k-1);
}
}
delete c , d;
}
}
}

int main()
{
time_t start , end;
long n = 0 , k = 0;
cout<<"Enter n:";
cin>>n;
cout<<"Enter k:";
cin>>k;
int *a=new int[n];
for(long i=0;i<n;i++)
{
cin>>a[i];
}
int *b=new int[k];
int *c=new int[n-k];
for(long i=0;i<n;i++)
{
if(i<k)
{
b[i]=a[i];
}
else
{
c[n-i-1]=a[i];
}
}
system("cls");
cout<<"S={";
for( long i = 0 ; i < n ; i++ )
{
cout<<a[i];
if( i != n-1 )
cout<<",";
}
cout<<"}"<<endl;
num=C(n,k);
m=new int*[C(n,k)];
for(long i=0;i<C(n,k);i++)
{
m[i]=new int[k];
}
start=clock();
printAll(b,c,k,n-k);
delete a , b , c;
end=clock();
cout<<"All of subsets printed in "<<float(end-
start)/float(CLK_TCK)<<" seconds."<<endl;
getch();
return 0;
}

Complexity T(n) What is the time complexity T(n) of the following portions of code? For simplicity, you may assume that n is a power of 2. That is, n = 2k for some positive integer k. a) ? for (i = 1; i <= n; i++) { j = n; cout << i << ? ? j << ? ? << endl; } b) ? for (i = 0; i <= n; i += 2) { j = n; cout << i << ? ? j << ? ? << endl; } c) ? for (i = n; i >= 1; i = i/2) { j = n; cout << i << ? ? j << ? ? << endl; } d) for (i = 1; i <= n; i++) { j = n; while (j >= 0) { cout << i << ? ? j << ? ? << endl; j = j - 2; } }

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