@another app...
slightly different code for the while loop...
while(++iter < N)
{
for (int i = 0; i < 9 ; ++i)
{
A[i] = ( A[i] * (i + iter) ) / iter ;
totalCount +=A[i];
}
}
On Dec 27, 9:36 pm, Lucifer <[email protected]> wrote:
> @above
> A slight correction in the above code, inside the while loop..
>
> while(++iter < N)
> {
> for (int i = 8; i > 0 ; --i)
> {
> A[i]+= A[i-1];
> totalCount +=A[i];
> }
> totalCount+=1;
>
> }
>
> On Dec 27, 9:34 pm, Lucifer <[email protected]> wrote:
>
>
>
>
>
>
>
> > As you are looking for a dp app...
>
> > @ Non-Decreasing Digits
>
> > Say the no. of digits is N..
> > Lets take an array A[9] to store the individual intermediate counts...
>
> > if (N > 1)
> > {
> > int iter = 0;
> > int totalCount = 10; // this is when N = 1 ...
>
> > for (int i=0; i < 9; ++i)
> > A[i] = 1;
>
> > while(++iter < N)
> > {
> > for (int i = 8; i > 0 ; --i)
> > {
> > A[i]+= A[i-1];
> > totalCount +=A[i];
> > }
> > }
> > printf("%d", totalCount);}
>
> > else
> > printf("%d", N==1 ? 10 : 0);
>
> > On Dec 27, 6:58 pm, kumar rajat <[email protected]> wrote:
>
> > > Hi
> > > I have jus started to learn DP and I have coded the standard
> > > algorithms(LCS,etc).
> > > I have been trying these problems in SPOJ:
>
> > >http://www.spoj.pl/problems/NOVICE63/http://www.spoj.pl/problems/NY10E/
>
> > > I understand these may be solved elegantly using DP,but I dont get to
> > > code the same.
> > > Can any1 help me how to solve these types of problems using DP?
> > > Any help regarding the same will be greatly appreciated.
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