F Find 3-friendly Integers

数位DP，到每个position时，都可以从这个position开始，也可以继承前面的数。

#include <bits/stdc++.h>

using namespace std;

#define File(x) freopen("(x)","r",stdin)
#define pf printf
#define ull unsigned long long
#define db double
#define ll long long
#define MAXN 
#define MAXM 
#define P 

ll dp[21][2][2][2][2];
int a[21];
int len;

ll dfs(int pos,int ans,int has0,int has1,int has2,int lead,int limit){
    if(!pos)return ans;
    if(dp[pos][ans][has0][has1][has2]!=-1&&!lead&&!limit)
    return dp[pos][ans][has0][has1][has2];
    ll re=0;
    int up=limit?a[pos]:9;
    for(int i=0;i<=up;i++){
        int d0=0,d1=0,d2=0,dans=0;
        if(i%3==0)d0=1;
        else if(i%3==1)d1=1;
        else d2=1;
        if(has1){
            if((i+1)%3==0)d0=1;
            else if((i+1)%3==1)d1=1;
            else d2=1;
        
        }
        if(has2){
            if((i+2)%3==0)d0=1;
            else if((i+2)%3==1)d1=1;
            else d2=1;
        }
        if(d0&&(!lead||lead&&i))dans=1;
        re+=dfs(pos-1,ans||dans,d0,d1,d2,!i&&lead,limit&&i==up);
    }
    if(!lead&&!limit)dp[pos][ans][has0][has1][has2]=re;
  //  cout<<has0<<" "<<has1<<" "<<has2<<endl;
    return re;
}

ll calc(ll x){
    len=0;
    while (x)
    {
        a[++len]=x%10;
        x/=10;
    }
    memset(dp,-1,sizeof(dp));
    return dfs(len,0,1,0,0,1,1);
    
}

int main(){

   int T;
   cin>>T;
   while (T--)
   {
       ll l,r;
       cin>>l>>r;
       cout<<calc(r)-calc(l-1)<<endl;
   }
   
   return 0;
}