for 3 set .. set value stored in array a[3] and p is the sum
for( i=0;i<=a[0];i++)
{
for(j=0;j<=a[1];j++)
{
for(k=a[2];k>=0;k--)
{
if((i+j+k)<p) // improve running time
break;
if((i+j+k)==p)
cout<<i<<j<<k;
}
}
}
With regards,
Praveen Raj
DCE-IT 3rd yr
9999735993
praveen0...@gmail.com
On Mon, Oct 24, 2011 at 3:00 AM, Piyush Kapoor <pkjee2...@gmail.com> wrote:
> Suppose u choose ith element from the Kth set,then
> dp[K][Sum]=sum(from i=0 to number of elements in the Kth set)
> dp[K-1][Sum-(ith element of Kth set)]
>
> On Sun, Oct 23, 2011 at 3:31 PM, cegprakash <cegprak...@gmail.com> wrote:
>
>> hi i recently came across this problem..
>>
>> there are K sets
>> each sets can contain n numbers from 0 to n
>> we've to choose exactly one number from each set
>> the sum of all the elements that we chose should be equal to P.
>> we have to find how many such possibilities are there to choose so..
>>
>> for example
>>
>> assume there are 3 sets containing 1,2,3 elements in them
>> so the first set contains 0 and 1
>> second set contains 0,1 and 2
>> third set contains 0,1,2 and 3
>>
>> assume P=2
>>
>> in this case there are 5 possibilities
>>
>> (0,0,2), (0,1,1), (0,2,0), (1,0,1), (1,1,0)
>>
>> i'm struggling for a DP solution!! help me out
>>
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>
>
> --
> *Regards,*
> *Piyush Kapoor,*
> *2nd year,CSE
> IT-BHU*
>
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