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：

 #include<bits/stdc++.h>
 #define ll long long
 using namespace std;
 int a[],tmp[];
 ll dp[][][];
 ll dfs(int start,int pos,int ok,bool limit)
 {
     if(pos<)
         return ok;
     if(!limit&&dp[pos][ok][start]!=-)
         return dp[pos][ok][start];
     ll ans=;
     int up=limit?a[pos]:;
     for(int d=; d<=up; ++d)
     {
         tmp[pos]=d;
         if(start==pos&&d==)
             ans+=dfs(start-,pos-,ok,limit&&d==up);
         else if(ok&&pos<(start+)/)
             ans+=dfs(start,pos-,tmp[start-pos]==d,limit&&d==up);
         else
             ans+=dfs(start,pos-,ok,limit&&d==up);
     }
     if(!limit)
         dp[pos][ok][start]=ans;
     return ans;
 }
 ll solve(ll x)
 {
     memset(a,,sizeof(a));
     int cnt=;
     while(x!=)
     {
         a[cnt++]=x%;
         x/=;
     }
     return dfs(cnt-,cnt-,,);
 }
 int main()
 {
     memset(dp,-,sizeof(dp));
     int t,cnt=;
     scanf("%d",&t);
     while(t--)
     {
         ll x,y;
         scanf("%lld%lld",&x,&y);
         if(x>y)
             swap(x,y);
         printf("Case %d: %lld\n",cnt++,solve(y)-solve(x-));
     }
     return ;
 }