比較基本的數位DP，限制都較為簡單；

#include<bits/stdc++.h>
#define rep(i,k,n) for(int i=k;i<=n;i++)
#define rep2(i,k,n) for(int i=k;i>=n;i--)
using namespace std;
typedef long long ll;
const int N=13;
void upd(ll& x,ll y){x+=y;}
ll f[N][N][3][3][2][2],L,R,a[N],top;
//f[位][首位][連續][4,8][是否已經有連續的三個][上界0,1]
 

ll solve(ll x){
    memset(f,0,sizeof(f));
    top=0;for(;x;a[++top]=x%10,x/=10);
    rep(i,0,9)f[0][i][0][0][0][0]=f[0][i][0][0][0][1]=1;
    rep(i,0,top-1){
        rep(now,0,9){
            rep(ne,0,9){
                int op=(ne==now);
                int ss=(ne==4)|((ne==8)<<1);
                rep(len,0,2
 
){
                    if(!len && !op)continue;
                    int ok=((len==2) && op);
                    if(len==2 && !op)op=-1;
                    rep(s,0,2){
                        if((ss|s)==3)continue;
                        rep(_3,0,1){
                            ll res0=f[i][now][len][s][_3][0
 
];
                            ll res1=f[i][now][len][s][_3][1];

                            ll& goal0=f[i+1][ne][min(2,len+op)][ss|s][_3|ok][0];
                            ll& goal1=f[i+1][ne][min(2,len+op)][ss|s][_3|ok][1];

                            upd(goal0,res0);
                            if(ne<a[i+1])upd(goal1,res0);
                            else if(ne==a[i+1])upd(goal1,res1);
                        }
                    }
                }
            }
        }
    }
    ll ans=0;
    rep(i,0,9)rep(len,0,2)rep(s,0,2)upd(ans,f[top][i][len][s][1][1]);
    return ans;
}
int main(){
    scanf("%lld%lld",&L,&R);
    ll lim=10000000000;
    ll res1=solve(R);
    ll res2=solve(max(lim,L-1));
    printf("%lld\n",res1-res2+(L==lim));
}