Yuo might wanna check out The latest codeforces beta round problem
C .

On Jan 28, 8:34 pm, nishaanth <[email protected]> wrote:
> @All... According to the constraints(SPOJ problem) wont this dp solution
> time out ?
>
> On Tue, Jan 25, 2011 at 12:23 AM, sankalp srivastava <
>
>
>
>
>
> [email protected]> wrote:
> > Correct me if I'm wrong
>
> > dp[i][j]=how many numbers of length i with the last digit j(int base
> > 10)
> > dp[0][j]=0
> > dp[i][0]=dp[i-1][0] , last digit can't be zero otherwise the number
> > has i-1 digits , not j;
>
> > now the recursion to pass from one state to the next
>
> > dp[i][j]=dp[i-1][k]+1(for each value of k such that k is less than j ,
> > from 0 to k)
> > That is to say , the number of numbers with length i and last digit
> > j , will be equal to the number of numbers with the last digit k , for
> > each k less than j .One is added because , we must not have included
> > the 0 in the last case , but we will include the zero case here . this
> > one corresponds to zero case .
>
> > Finally , the answer will be
>
> > dp[n][j]  , 1<=j<=9 , sum up all these and u have the answer
> > For example for 2 digits
> > dp[1][1-9]=1 , as nothing preceeds them and as said in the problem ,
> > one digit numbers are always increasing/decreasing .
> > next dp[2][1]=dp[1][1](As only 1 is less than  , equal to 1 , the last
> > digit in this state)
> > dp[2][2]=dp[1][1]+dp[1][2] =2( 1,2 and 2,2)
> > Continuing this way , we will get the answer , may be 50  , though i
> > din code it .
>
> > similarly , for 3 digits
>
> > Correct me if not!
>
> > On Jan 25, 12:06 am, juver++ <[email protected]> wrote:
> > > dp[i][j] - how many numbers of length i and with the last digit j.
> > > Apply the scheme to increasing and decreasing number, then find ratio.
>
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> --
> S.Nishaanth,
> Computer Science and engineering,
> IIT Madras.

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