#include<iostream>
using namespace std;
int maxsum(int a[], int n)
{
int prevmaxsum=0;
int currmaxsum=a[0];
int last=0,temp=0;
for(int i=1;i<n;i++)
{
if(last!=(i-1))
{
prevmaxsum=currmaxsum;
currmaxsum=a[i]+prevmaxsum;
last=i;
cout<<prevmaxsum<<"\t\t\t"<<currmaxsum<<"\t\t\t"<<last<<endl;
continue;
}
if((a[i]+prevmaxsum)>currmaxsum)
{
temp=a[i]+prevmaxsum;
prevmaxsum=currmaxsum;
currmaxsum=temp;
last=i;
cout<<prevmaxsum<<"\t\t\t"<<currmaxsum<<"\t\t\t"<<last<<endl;
}
}
return currmaxsum;
}
int main()
{
int arr[6]={3, 2 ,5,7 ,10};
int num=5;
int sum=maxsum(arr,6);
cout<<sum;
getchar();
return 0;
}
On Fri, Oct 9, 2009 at 9:26 AM, ankur aggarwal <[email protected]>wrote:
>
> 2. Given an array all of whose elements are positive numbers, find the
> maximum sum of a subsequence with the constraint that no 2 numbers in the
> sequence should be adjacent in the array.
>
> i) 3 2 7 10 should return 13 (sum of 3 and 10)
>
> ii) 3 2 5 10 7 should return 15 (sum of 3, 5 and 7)
>
> >
>
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