Lightoj1205——Palindromic Numbers（数位dp+回文数）

A palindromic number or numeral palindrome is a ‘symmetrical’ number like 16461 that remains the same when its digits are reversed. In this problem you will be given two integers i j, you have to find the number of palindromic numbers between i and j (inclusive).

Input 
Input starts with an integer T (≤ 200), denoting the number of test cases.

Each case starts with a line containing two integers i j (0 ≤ i, j ≤ 1017).

Output 
For each case, print the case number and the total number of palindromic numbers between i and j (inclusive).

Sample Input 
4 
1 10 
100 1 
1 1000 
1 10000 
Output for Sample Input 
Case 1: 9 
Case 2: 18 
Case 3: 108 
Case 4: 198

代码：

 1 #include<bits/stdc++.h>
 2 #define ll long long
 3 using namespace std;
 4 int a[40],tmp[40];
 5 ll dp[40][100][100];
 6 ll dfs(int start,int pos,int ok,bool limit)
 7 {
 8     if(pos<0)
 9         return ok;
10     if(!limit&&dp[pos][ok][start]!=-1)
11         return dp[pos][ok][start];
12     ll ans=0;
13     int up=limit?a[pos]:9;
14     for(int d=0; d<=up; ++d)
15     {
16         tmp[pos]=d;
17         if(start==pos&&d==0)
18             ans+=dfs(start-1,pos-1,ok,limit&&d==up);
19         else if(ok&&pos<(start+1)/2)
20             ans+=dfs(start,pos-1,tmp[start-pos]==d,limit&&d==up);
21         else
22             ans+=dfs(start,pos-1,ok,limit&&d==up);
23     }
24     if(!limit)
25         dp[pos][ok][start]=ans;
26     return ans;
27 }
28 ll solve(ll x)
29 {
30     memset(a,0,sizeof(a));
31     int cnt=0;
32     while(x!=0)
33     {
34         a[cnt++]=x%10;
35         x/=10;
36     }
37     return dfs(cnt-1,cnt-1,1,1);
38 }
39 int main()
40 {
41     memset(dp,-1,sizeof(dp));
42     int t,cnt=1;
43     scanf("%d",&t);
44     while(t--)
45     {
46         ll x,y;
47         scanf("%lld%lld",&x,&y);
48         if(x>y)
49             swap(x,y);
50         printf("Case %d: %lld\n",cnt++,solve(y)-solve(x-1));
51     }
52     return 0;
53 }