Digital DP. sb. .
The leading 0 is influential and affects the choice of the first one, so remember it. Record the upper limit again, then memorize it when there is no limit.
//820kb 4ms
#include <cstdio>
#include <cstring>
#include <algorithm>
int L,R,A[12],f[12][10];
bool vis[12][10];
int DFS(int pos,int las,bool lim,bool lead)
{
if(!pos) return 1;
if(!lim && !lead && vis[pos][las]) return f[pos][las];
int res=0, up=lim?A[pos]:9;
for(int i=0; i<=up; ++i)
if(std::abs(i-las)>=2||lead) res+=DFS(pos-1,i,lim&&i==up,lead&&!i);//(lim&&)!
if(!lim && !lead) vis[pos][las]=1, f[pos][las]=res;
return res;
}
int Solve(int n)
{
for(A[0]=0; n; n/=10) A[++A[0]]=n%10;
memset(vis,0,sizeof vis);
return DFS(A[0],100,1,1);
}
int main()
{
scanf("%d%d",&L,&R);
printf("%d",Solve(R)-Solve(L-1));
return 0;
}