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

Posted 2020-08-09 半根毛线code

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）

  1 /******************************
  2 code by drizzle
  3 blog: www.cnblogs.com/hsd-/
  4 ^ ^    ^ ^
  5  O      O
  6 ******************************/
  7 //#include<bits/stdc++.h>
  8 #include<iostream>
  9 #include<cstring>
 10 #include<cmath>
 11 #include<cstdio>
 12 #define ll long long
 13 #define mod 1000000007
 14 #define PI acos(-1.0)
 15 using namespace std;
 16 int t;
 17 ll l,r;
 18 int num[25];
 19 ll dp[25][5];
 20 /*
 21 1 奇奇
 22 2 奇偶
 23 3 偶奇
 24 4 偶偶
 25 */
 26 ll dfs(int pos,int status,int flag)
 27 {
 28     if(pos<1)
 29     {
 30         if(status==2||status==3)
 31              return 1;
 32         else
 33              return 0;
 34     }
 35     if(!flag&&dp[pos][status])
 36         return dp[pos][status];
 37     int end=flag ? num[pos] : 9;//如果之前都是前导零 则当前这位可以取0~9
 38     ll ans=0;
 39     for(int i=0;i<=end;i++)
 40     {
 41         if(!status)
 42         {
 43             if(!i)
 44             {
 45                 ans=dfs(pos-1,0,0);
 46             }
 47             else if(i&1)
 48             {
 49                 ans+=dfs(pos-1,1,flag&&i==end);
 50             }
 51             else
 52             {
 53                 ans+=dfs(pos-1,3,flag&&i==end);
 54             }
 55 
 56         }
 57         else
 58         {
 59             if(status==1){
 60                 if(i&1){
 61                     ans+=dfs(pos-1,2,flag&&i==end);
 62                 }
 63             }
 64             else if(status==2){
 65                 if(i&1){
 66                     ans+=dfs(pos-1,1,flag&&i==end);
 67                 }
 68                 else{
 69                     ans+=dfs(pos-1,3,flag&&i==end);
 70                 }
 71             }
 72             else if(status==3){
 73                 if(i&1){
 74                     ans+=dfs(pos-1,1,flag&&i==end);
 75                 }
 76                 else
 77                     ans+=dfs(pos-1,4,flag&&i==end);
 78             }
 79             else{
 80                 if(!(i&1))
 81                 {
 82                     ans+=dfs(pos-1,3,flag&&i==end);
 83                 }
 84             }
 85         }
 86     }
 87     dp[pos][status]=ans;
 88     return ans;
 89 }
 90 ll slove(ll x)
 91 {
 92     memset(dp,0,sizeof(dp));
 93     int len=0;
 94     while(x)
 95     {
 96         num[++len]=x%10;
 97         x/=10;
 98     }
 99     return dfs(len,0,1);
100 }
101 int main()
102 {
103         scanf("%d",&t);
104         for(int i=1;i<=t;i++)
105         {
106             scanf("%I64d %I64d",&l,&r);
107             printf("Case #%d: %I64d\\n",i,slove(r)-slove(l-1));
108         }
109     return 0;
110 }