optitape(daa)

#include<iostream.h>
#include<iomanip.h>

int fact(int);

int main()
{
    cout<<"enter total number of tapes\n";
    int n,j,a[10],c,b[10],sum;
    cin>>n;
    cout<<"enter the time taken for indivisual tapes to play\n";
    for(j=0;j<=n-1;j++)
        cin>>a[j];
    cout<<"the tapes are\n";
    for(int k=1;k<=n;k++)
        cout<<setw(3)<<"T"<<k;
    cout<<"\nthe combinations are:\n";
    k=0;
    for(int i=0;i<=(fact(n)-1)/2;i++)
    {
        sum=0;
        c=n;
        for(j=i+1;j<=n;j++)
        {
            cout<<setw(3)<<"T"<<j;
            sum=sum+(c--)*a[j-1];
        }
        for(j=1;j<=i%n;j++)
        {
            cout<<setw(3)<<"T"<<j;
            sum=sum+(c--)*a[j-1];
        }
        cout<<setw(14)<<" Time taken is="<<sum<<endl;
        b[k]=sum;
        k++;
        sum=0;
        c=n;
        for(j=i%n;j>=1;j--)
        {
            cout<<setw(3)<<"T"<<j;
            sum=sum+(c--)*a[j-1];
        }
        for(j=n;j>=i+1;j--)
        {
            cout<<setw(3)<<"T"<<j;
            sum=sum+(c--)*a[j-1];
        }
        cout<<setw(14)<<" Time taken is="<<sum<<endl;
        b[k]=sum;
        k++;
    }
    int t;
    k--;
    t=b[0];
    for(j=0;j<=5;j++)
    {
        if(t>b[j])
            t=b[j];
    }
    cout<<"the minimal optimum solution is:\n"<<t<<endl;
    return 0;
}

int fact(int x)
{
    int facto=1;
    if(x==1||x==0)
        return 1;
    else
    {

        return x*fact(x-1);
    }
}
OUTPUT:
enter total number of tapes
3
enter the time taken for indivisual tapes to play
5
10
3
the tapes are
 T1  T2  T3
the combinations are:
 T1  T2  T3 Time taken is=38
 T3  T2  T1 Time taken is=34
 T2  T3  T1 Time taken is=41
 T1  T3  T2 Time taken is=31
 T3  T1  T2 Time taken is=29
 T2  T1  T3 Time taken is=43
the minimal optimum solution is:
29