ACM/ICPC Asia Regional Shenyang Online 1007/hdu 5898 Digit DP

Odd-even number

Time limit:2000/1000 MS (java/others) Memory limit:65536/65536 K (java/others)
Total submission (s): 388 Accepted Submission (s): 212

Problem Descriptionfor A number,if the length of continuous odd digits is even and the length of continuous even digits are Odd,we call it odd-even number. Now we want to know the amount of odd-even number between l,r (1<=l<=r<= 9*10^18).

Inputfirst Line a t,then t cases.every line contains, integers L and R.

Outputprint the output for each case is on one line in the format as shown below.

Sample INPUT2 1 100 110 220

Sample outputcase #1:29Case #2:36

Source ACM/ICPC Asia Regional Shenyang Online test instructions: A number that has an even-length segment on every odd number of digits, and even-numbered bits that form odd-length segments. He's good. Ask [L, R] for a number of good numbers. DP[I][J] indicates that the state of the first digit is J. There are 4 states, 1-odd length is odd, 2-odd length is even, 3-even length is odd, 4-even length is even plus a state flag=0 means that the current bit is preceded by 0 (note that transitions between states have been memorized)

1 /******************************2 code by drizzle3 blog:www.cnblogs.com/hsd-/4 ^ ^    ^ ^5 o o6 ******************************/7 //#include <bits/stdc++.h>8#include <iostream>9#include <cstring>Ten#include <cmath> One#include <cstdio> A #definell Long Long - #defineMoD 1000000007 - #definePI ACOs (-1.0) the using namespacestd; - intT; - ll L,r; - intnum[ -]; +ll dp[ -][5]; - /* + 1 Kiki A 2 Odd-even at 3 even odd - 4 Even - */ -ll Dfs (intPosintStatusintflag) - { -     if(pos<1) in     { -         if(status==2|| status==3) to              return 1; +         Else -              return 0; the     } *     if(!flag&&Dp[pos][status]) $         returnDp[pos][status];Panax Notoginseng     intEnd=flag? Num[pos]:9The current one can take 0~9 if it's preceded by a leading zero. -ll ans=0; the      for(intI=0; i<=end;i++) +     { A         if(!status) the         { +             if(!i) -             { $Ans=dfs (pos-1,0,0); $             } -             Else if(i&1) -             { theAns+=dfs (pos-1,1, flag&&i==end); -             }Wuyi             Else the             { -Ans+=dfs (pos-1,3, flag&&i==end); Wu             } -  About         } $         Else -         { -             if(status==1){ -                 if(i&1){ AAns+=dfs (pos-1,2, flag&&i==end); +                 } the             } -             Else if(status==2){ $                 if(i&1){ theAns+=dfs (pos-1,1, flag&&i==end); the                 } the                 Else{ theAns+=dfs (pos-1,3, flag&&i==end); -                 } in             } the             Else if(status==3){ the                 if(i&1){ AboutAns+=dfs (pos-1,1, flag&&i==end); the                 } the                 Else theAns+=dfs (pos-1,4, flag&&i==end); +             } -             Else{ the                 if(! (i&1))Bayi                 { theAns+=dfs (pos-1,3, flag&&i==end); the                 } -             } -         } the     } thedp[pos][status]=ans; the     returnans; the } - ll Slove (ll x) the { theMemset (DP,0,sizeof(DP)); the     intlen=0;94      while(x) the     { thenum[++len]=x%Ten; theX/=Ten;98     } About     returnDFS (Len,0,1); - }101 intMain ()102 {103scanf"%d",&t);104          for(intI=1; i<=t;i++) the         {106scanf"%i64d%i64d",&l,&R);107printf"Case #%d:%i64d\n", I,slove (R)-slove (l1));108         }109     return 0; the}

ACM/ICPC Asia Regional Shenyang Online 1007/hdu 5898 Digit DP