题意：

求 $[l, r]$ 中有多少个回文数。

思路：

裸的数位 $d p$ 啦，记 $d p [p o s] [p r e] [s t a t e]$ 表示到了第 $p o s$ 位，回文是从第 $p r e$ 位开始的，并且当前是否为回文串 $s t a t e$ 。这样就可以把要求的数字特征表示出来，让后还要记录一下选的数是多少，在 $p o s < = (p r e - 1) / 2$ 的时候就需要判断是否为回文串了。
好长时间没写数位 $d p$ 了， $s o l v e$ 传值的时候 $L L$ 传成了 $i n t$ 。

//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid (tr[u].l+tr[u].r>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;const int N=1010,mod=1e9+7,INF=0x3f3f3f3f;
const double eps=1e-6;LL l,r;
int a[N],num[N],tot;
LL f[30][30][2];LL dfs(int pos,int pre,int state,int flag)
{if(pos==-1) return state;if(flag&&f[pos][pre][state]!=-1) return f[pos][pre][state];int x=flag? 9:a[pos];LL ans=0;for(int i=0;i<=x;i++){num[pos]=i;if(i==0&&pos==pre) ans+=dfs(pos-1,pre-1,state,flag||i<x);else if(pos<=(pre-1)/2&&state) ans+=dfs(pos-1,pre,state&&(num[pre-pos]==i),flag||i<x);else ans+=dfs(pos-1,pre,state,flag||i<x);}if(flag) f[pos][pre][state]=ans;return ans;
}LL solve(LL x)
{tot=0;if(x<0) return 0;if(x==0) return 1;while(x){a[tot++]=x%10;x/=10;}return dfs(tot-1,tot-1,1,0);
}int main()
{
//	ios::sync_with_stdio(false);
//	cin.tie(0);memset(f,-1,sizeof(f));int _; scanf("%d",&_);for(int __=1;__<=_;__++){scanf("%lld%lld",&l,&r);if(l>r) swap(l,r);printf("Case %d: %lld\n",__,solve(r)-solve(l-1));}return 0;
}
/**/

带前导零，更好看点的代码：

// Problem: Palindromic Numbers
// Contest: Virtual Judge - LightOJ
// URL: https://vjudge.net/problem/LightOJ-1205
// Memory Limit: 65 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#include<random>
#include<cassert>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid ((tr[u].l+tr[u].r)>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;const int N=1000010,mod=1e9+7,INF=0x3f3f3f3f;
const double eps=1e-6;LL x,y;
int a[300],tot;
int f[30][30][2];
int num[30];LL dp(int pos,int st,int state,int lead,int flag) {if(pos==0) return state;if(f[pos][st][state]!=-1&&lead&&flag) return f[pos][st][state];LL ans=0;int x=flag? 9:a[pos];for(int i=0;i<=x;i++) {num[pos]=i;if(!lead) {ans+=dp(pos-1,i==0? st:pos,state,lead||i>0,flag||i<x);} else {if(pos<=st/2&&state) ans+=dp(pos-1,st,state&&(i==num[st-pos+1]),lead,flag||i<x);else ans+=dp(pos-1,st,state,lead,flag||i<x);}}if(lead&&flag) f[pos][st][state]=ans;return ans;
}LL solve(LL x) {if(x<0) return 0;if(x==0) return 1;tot=0;while(x) a[++tot]=x%10,x/=10;return dp(tot,0,1,0,0);
}int main()
{
//	ios::sync_with_stdio(false);
//	cin.tie(0);memset(f,-1,sizeof(f));int _; scanf("%d",&_);for(int i=1;i<=_;i++) {LL a,b;scanf("%lld%lld",&a,&b); if(a>b) swap(a,b);printf("Case %d: %lld\n",i,solve(b)-solve(a-1));}return 0;
}
/**/