1026: [Scoi2009]windy number

Time limit:1 Sec Memory limit:162 MB
submit:9670 solved:4445
[Submit] [Status] [Discuss] Description

Windy defines a windy number. A positive integer that does not contain a leading 0 and the difference of at least 2 of the adjacent two digits is called the windy number. Windy want to know,
What is the total number of windy between A and B, including A and b?

Input

Contains two integers, A B.

Output

An integer

Sample Input"Input Sample One"
1 10
"Input Sample Two"
theSample Output"Output Example One"
9
"Output Example II"
-HINT

"Data size and conventions"

100% data, meet 1 <= A <= B <= 2000000000.

The approximate meaning of windy number is: like 135,6913,13579, so the difference of every two adjacent numbers ≥2

Digital DP:

Direct speculation, it is impossible to preserve the 2000000000 state, and a more ingenious approach must be adopted.

you can set the DP array to: F[i][j],i represents the number of digits , J represents the highest number

Thus the DP equation can be obtained:

\[f[i][j]=\sum_{k=0}^{k\leq 9} f[i-1][k]\left (\left | k-j \right |\geq 2\right) \]

But this only calculates the result of the highest level J, unable to find the result in the range, then can convert the thought, find out the result of 1~b minus 1~a-1, is not the answer?

Set Len to indicate number of digits, T[i] represents the number of the I-bit

First, the ANS plus the 1~len-1 bit situation

Then ans plus the highest digit number for the 1~t[len]-1 case

Then and so on, coupled with the situation of the second highest 1~t[len-1]-1, coupled with the situation of the next highest bit 1~t[len-2]-1 ...

1#include <iostream>2#include <cstdio>3#include <cstdlib>4#include <cmath>5 using namespacestd;6 7 #defineLL Long Long8 9 intb;TenLL f[ -][Ten]; One  ALL Solve (intt) - { -LL ans=0; the     intlen=0, t[ -]; -      while(t) -     { -t[++len]=t%Ten; +T/=Ten; -     } +      for(intI=1; i<len;i++) A          for(intj=1; j<=9; j + +) atans+=F[i][j]; -      for(intI=1; i<t[len];i++) -ans+=F[len][i]; -      for(inti=len-1; i>=1; i--) -     { -          for(intj=0; j<t[i];j++) in             if(ABS (j-t[i+1]) >=2) -ans+=F[i][j]; to         if(ABS (t[i]-t[i+1]) <2) Break; +     } -     returnans; the } *  $ intMain ()Panax Notoginseng { -scanf"%d%d",&a,&b); the      for(intI=0; i<=9; i++) f[1][i]=1; +      for(intI=2; i<= the; i++) A          for(intj=0; j<=9; j + +) the              for(intk=0; k<=9; k++) +                 if(ABS (K-J) >=2) -f[i][j]+=f[i-1][k]; $Cout<<solve (b +1)-solve (a) <<Endl; $     return 0; -}

1026: [Scoi2009]windy number