#include <iostream>
using namespace std;
int main()
{
int input[10];
int n;
cout<<"enter n"<<endl;
cin>>n;
int output[10];
cout<<"enter input array"<<endl;
for(int i=0;i<n;i++)
cin>>input[i];
int a[n],b[n];
a[0]=1;
for(int i=1;i<n;i++)
{
a[i]=a[i-1]*input[i-1];
}
b[n-1]=1;
for(int i=n-2;i>=0;i--)
{
b[i]=b[i+1]*input[i+1];
}
for(int i=0;i<n;i++)
{
output[i]=a[i]*b[i];
cout<<output[i]<<endl;
}
return 0;
}
On Sun, Jun 26, 2011 at 9:38 PM, ross <jagadish1...@gmail.com> wrote:
> Given an array A , of N integers ( In no particular order), fill up an
> auxilary array B such that B[i] contains the product of
> all elements in A other than A[i].
> Constraints:
> O(n) Time,
> Can this be done with O(1) space?
> Division is *not* allowed .
>
> eg: A 1 2 3 4 5
> B 120 60 40 30 24
>
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