#include 
     
      #include 
      
       #include 
       
        #include 
        
         using namespace std;int t;long long dp[19][19][2005];long long l, r;int digit[20];long long dfs(int len,...int if4..., bool limit){    if (len == 0)        return 1; //个位    if (!limit && dp[len][...])        return dp[len][...];  //dp数组的内容应和dfs调用参数的内容相同，除了是否达到上限    long long cnt = 0;    int up_bound = (limit ? digit[len] : 9);    for (int i = 0; i <= up_bound; i++)    {        if(...) continue; //剪枝        ...;        cnt += dfs(len-1, ..., limit && i == up_bound);    }    if (!limit) //完整状态        dp[len][...] = cnt;    return cnt;}long long solve(long long x){    int k = 0;    while (x)    {        digit[++k] = x % 10;        x /= 10;    }    return dfs(k,...,1)}int main(){        memset(dp, 0, sizeof(dp));        scanf("%lld%lld", &l, &r);   //有些题目其实并不需要用到long long        printf("%lld\n", solve(r) - solve(l - 1)); //只有满足区间减法才能用    return 0;}