#include <iostream.h>
#include<conio.h>
main()
{
int s=-1,m=1,n,max=0,e=-1,sum=0,i,a[20];
cout<<"enter n:";
cin>>n;
cout<<"enter elements:";
for(i=0;i<n;i++)
cin>>a[i];
for(i=0;i<n;i++)
{
if(m==1)
{
if(a[i]>0)
s=i;
m=0;
}
if(a[i]+sum>0)
sum+=a[i];
else
{

m=1;
sum=0;
}
if(sum>max)
{
max=sum;
e=i;
}
}
cout<<"max sum is:"<<max;
cout<<"\n\nmax sum producing sub array is:";
if(s!=-1&&e!=-1)
for(i=s;i<=e;i++)
cout<<a[i]<<"\t";
getch();
return 0;
}

#include <iostream.h>
#include<conio.h>
/*prints subarray producing maximum sum..Efficiency is O(n)*/
main()
{
int flag=0,n,i,c=0,sum=0,max=0,a[25],s=-1,e=-1;
cout<<"enter n:";
cin>>n;
cout<<"enter elements:";
for(i=1;i<=n;i++)
cin>>a[i];
for(i=1;i<=n;i++)
{
if(a[i]+sum>0)
sum+=a[i];
else
{
sum=0;
if(c!=0)
c=1;
}
if(sum>max)
{
max=sum;
e=i;
s=e-c;
c++;
}
}
cout<<"max sum is:"<<max;
cout<<"\n\nmax sum producing sub array is:";
if(s!=-1&&e!=-1)
for(i=s;i<=e;i++)
cout<<"\t"<<a[i];
getch();
return 0;
}

/*prints subarray producing maximum sum.Efficiency is O(n)*/
#include <iostream.h>
#include<conio.h>
main()
{
int n,i,c=0,sum=0,max=0,a[25],s=-1,e=-1;
cout<<"enter n:";
cin>>n;
cout<<"enter elements:";
for(i=1;i<=n;i++)
cin>>a[i];             
for(i=1;i<=n;i++)
{
if(a[i]+sum>0)                       
{
sum+=a[i];                          
if(sum>=max)
s=i-c;
c++;
}
else
{
sum=0;
c=0;
}
if(sum>max)
{
max=sum;
e=i;
}
}
cout<<"max sum is:"<<max;
cout<<"\n\nmax sum producing sub array is:";
if(s!=-1&&e!=-1)
for(i=s;i<=e;i++)
cout<<"\t"<<a[i];                            
getch();
return 0;
}

Actually I have posted 3 answers..Bcoz after posting twice I
found there are some logical errors.Last one is fully
correct..If not please let me know..

i think ths program doesnt not work when the sum is 
negative.. cant we not work around it by setting the max as 
somelarge negative value and comparing with that??

I came to this website looking for the solution and finally
ended up writing it myself. Hope this is useful for someone
else.

Here is a working solution. Set the array "a" and N_ELEMENTS
accordingly. Some cases are not covered but should be an
easy fix.

#include <stdio.h>
#define N_ELEMENTS 7

int main() {
    int a[N_ELEMENTS] = {-1, 2, -3, 2, 0, 5, -11 }; // if
you change the array, make sure you change N_ELEMENTS
    int i = 0;

    while(a[i] < 0 && i<N_ELEMENTS) {
        i++;
    }

    if (a[i] < 0) {

        printf ("DEBUG: array with only negative numbers.
Print the smallest negative number as the sum and we are
done.\n");

    }

    int sum_p=0, sum_n = 0;
    int largest_sum = 0;

    while (i<N_ELEMENTS) {
        if (a[i] > 0) {
            sum_p += a[i];

        }
        else {
            sum_n += a[i];

        }

        if (sum_p+sum_n > largest_sum) {
            largest_sum = sum_p + sum_n;

        }

        if (sum_p+sum_n <= 0) {
            // find the next positive number
            while(a[i] < 0 && i<N_ELEMENTS) {
                i++;
            }
            if (a[i] < 0 || i == N_ELEMENTS) {
                break;

            }

            sum_p = 0;
            sum_n = 0;

        } else {
            i++;
        }
    }

    printf ("DEBUG: The largest consecutive sum = %d\n",
largest_sum);

}

Guys..Answer 3 is correct..When there are -ve
numbers..maximum sum is 0.U should not consider any elements
at all..Coz null set which has maximum sum.i.e 0.So the 4th
ans is right.