[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：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int maxn = 310000;
 5 const ll Inf = 1000000000LL;
 6 int d[maxn], dc[maxn][8], tot;
 7 int td, tc[8];
 8
 9 void dfs(int r, int s, ll num){
10      if (num > Inf) return;
11      if (r == 10 || s == 18){
12          d[tot] = num;
13          for (int i = 0;  i < 8; ++i)
14               dc[tot][i] = tc[i];
15          tot++;
16          return;
17      }
18      dfs(r + 1, s, num);
19      tc[r-2]++;
20      dfs(r, s + 1, num * r);
21      tc[r-2]--;
22 }
23
24 ll frac[25];
25 ll count(int m){
26    int left = m;
27    ll res = frac[m];
28    for (int i = 0; i < 8; ++i) left -= tc[i], res /= frac[tc[i]];
29    res /= frac[left];
30    return res;
31 }
32
33 int bit[20];
34
35 ll calculate(ll b, int p){
36      if (b <= 1) return 0;
37      int k = 0;
38      while (b > 0) bit[++k] = b % 10, b /= 10;
39      int size = 0;
40      for (int i = 0; i < 8; ++i)
41          tc[i] = dc[p][i], size += tc[i];
42      ll res = 0;
43      for (int i = 1; i < k; ++i)
44           if (i >= size) res += count(i);
45      for ( ;k > 0; --k){
46           if (bit[k] == 0 || size > k) break;
47           for (int i = 2; i < bit[k]; ++i) if (tc[i-2]){
48               --tc[i-2], --size;
49               if (k > size) res += count(k - 1);
50               ++tc[i-2], ++size;
51           }
52           if (bit[k] >= 2){
53               if (k > size) res += count(k-1);
54               if (tc[bit[k]-2] == 0) break;
55               --tc[bit[k]-2], --size;
56           }
57      }
58      return res;
59 }
60
61 ll calculate(ll x){
62    if (x <= 0) return 0;
63    ll res = 0;
64    for (int i = 0; i < tot; ++i)
65         res += calculate(x / d[i] + 1, i);
66    return res;
67 }
68
69 void pre_do(){
70     frac[0] = 1;
71     for (int i = 1; i <= 18; ++i) frac[i] = frac[i-1] * i;
72     memset(tc, 0, sizeof(tc));
73     tot = 0;
74     dfs(2, 0, 1);
75 }
76
77 int main(){
78  //   freopen("a.in", "r", stdin);
79  //   freopen("a.out", "w", stdout);
80     pre_do();
81     ll l, r;
82     while (cin >> l >> r){
83          ll ans  = calculate(r) - calculate(l - 1);
84          cout << ans << endl;
85     }
86     return 0;
87 }