2016 ACM/ICPC Asia Regional Shenyang Online 1007/HDU 5898 数位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 Description

For a number,if the length of continuous odd digits is even and the length of continuous even digits is 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).

 

Input

First line a t,then t cases.every line contains two integers L and R.

 

Output

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

 

Sample Input

2
1 100
110 220

 

Sample Output

Case #1: 29

Case #2: 36

 

Source

2016 ACM/ICPC Asia Regional Shenyang Online

 题意：一个数字，它每个数位上的奇数都形成偶数长度的段，偶数位都形成奇数长度的段他就是好的。问[L , R]的好数个数。

 题解：用dp[i][j]表示第i位数前一位数的状态是j。状态有4种，1-奇数长度为奇，2-奇数长度为偶，3-偶数长度为奇，4-偶数长度为偶

外加一个状态flag=0表示当前这一位之前都是前导零  （注意各个状态之间的转换已经记忆化）

 /******************************
 code by drizzle
 blog: www.cnblogs.com/hsd-/
 ^ ^    ^ ^
  O      O
 ******************************/
 //#include<bits/stdc++.h>
 #include<iostream>
 #include<cstring>
 #include<cmath>
 #include<cstdio>
 #define ll long long
 #define mod 1000000007
 #define PI acos(-1.0)
 using namespace std;
 int t;
 ll l,r;
 int num[];
 ll dp[][];
 /*
 1 奇奇
 2 奇偶
 3 偶奇
 4 偶偶
 */
 ll dfs(int pos,int status,int flag)
 {
     if(pos<)
     {
         if(status==||status==)
              return ;
         else
              return ;
     }
     if(!flag&&dp[pos][status])
         return dp[pos][status];
     int end=flag ? num[pos] : ;//如果之前都是前导零 则当前这位可以取0~9
     ll ans=;
     for(int i=;i<=end;i++)
     {
         if(!status)
         {
             if(!i)
             {
                 ans=dfs(pos-,,);
             }
             else if(i&)
             {
                 ans+=dfs(pos-,,flag&&i==end);
             }
             else
             {
                 ans+=dfs(pos-,,flag&&i==end);
             }

         }
         else
         {
             if(status==){
                 if(i&){
                     ans+=dfs(pos-,,flag&&i==end);
                 }
             }
             else if(status==){
                 if(i&){
                     ans+=dfs(pos-,,flag&&i==end);
                 }
                 else{
                     ans+=dfs(pos-,,flag&&i==end);
                 }
             }
             else if(status==){
                 if(i&){
                     ans+=dfs(pos-,,flag&&i==end);
                 }
                 else
                     ans+=dfs(pos-,,flag&&i==end);
             }
             else{
                 if(!(i&))
                 {
                     ans+=dfs(pos-,,flag&&i==end);
                 }
             }
         }
     }
     dp[pos][status]=ans;
     return ans;
 }
 ll slove(ll x)
 {
     memset(dp,,sizeof(dp));
     int len=;
     while(x)
     {
         num[++len]=x%;
         x/=;
     }
     return dfs(len,,);
 }
 int main()
 {
         scanf("%d",&t);
         for(int i=;i<=t;i++)
         {
             scanf("%I64d %I64d",&l,&r);
             printf("Case #%d: %I64d\n",i,slove(r)-slove(l-));
         }
     return ;
 }