[violet2]sillyz

题意：定义S(n) = n*各数位之积，然后给定L<=R<=10^18，求有多少个n在[L,R]区间内

思路：

看了半天无从下手。。看完题解才豁然开朗。。

具体思路看vani神博客吧。讲的很清楚。。

code：

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 const int maxn = ;
 const ll Inf = 1000000000LL;
 int d[maxn], dc[maxn][], tot;
 int td, tc[];

 void dfs(int r, int s, ll num){
      if (num > Inf) return;
      if (r ==  || s == ){
          d[tot] = num;
          for (int i = ;  i < ; ++i)
               dc[tot][i] = tc[i];
          tot++;
          return;
      }
      dfs(r + , s, num);
      tc[r-]++;
      dfs(r, s + , num * r);
      tc[r-]--;
 }

 ll frac[];
 ll count(int m){
    int left = m;
    ll res = frac[m];
    for (int i = ; i < ; ++i) left -= tc[i], res /= frac[tc[i]];
    res /= frac[left];
    return res;
 }

 int bit[];

 ll calculate(ll b, int p){
      if (b <= ) return ;
      int k = ;
      while (b > ) bit[++k] = b % , b /= ;
      int size = ;
      for (int i = ; i < ; ++i)
          tc[i] = dc[p][i], size += tc[i];
      ll res = ;
      for (int i = ; i < k; ++i)
           if (i >= size) res += count(i);
      for ( ;k > ; --k){
           if (bit[k] ==  || size > k) break;
           for (int i = ; i < bit[k]; ++i) if (tc[i-]){
               --tc[i-], --size;
               if (k > size) res += count(k - );
               ++tc[i-], ++size;
           }
           if (bit[k] >= ){
               if (k > size) res += count(k-);
               if (tc[bit[k]-] == ) break;
               --tc[bit[k]-], --size;
           }
      }
      return res;
 }

 ll calculate(ll x){
    if (x <= ) return ;
    ll res = ;
    for (int i = ; i < tot; ++i)
         res += calculate(x / d[i] + , i);
    return res;
 }

 void pre_do(){
     frac[] = ;
     for (int i = ; i <= ; ++i) frac[i] = frac[i-] * i;
     memset(tc, , sizeof(tc));
     tot = ;
     dfs(, , );
 }

 int main(){
  //   freopen("a.in", "r", stdin);
  //   freopen("a.out", "w", stdout);
     pre_do();
     ll l, r;
     while (cin >> l >> r){
          ll ans  = calculate(r) - calculate(l - );
          cout << ans << endl;
     }
     return ;
 }