E. Ever-Hungry Krakozyabra
time limit per test：1 second
memory limit per test：256 megabytes
input：standard input
output：standard output
Recently, a wild Krakozyabra appeared at Jelly Castle. It is, truth to be said, always eager to have something for dinner.
Its favorite meal is natural numbers (typically served with honey sauce), or, to be more precise, the zeros in their corresponding decimal representations. As for other digits, Krakozyabra dislikes them; moreover, they often cause it indigestion! So, as a necessary precaution, Krakozyabra prefers to sort the digits of a number in non-descending order before proceeding to feast. Then, the leading zeros of the resulting number are eaten and the remaining part is discarded as an inedible tail.
For example, if Krakozyabra is to have the number 57040 for dinner, its inedible tail would be the number 457.
Slastyona is not really fond of the idea of Krakozyabra living in her castle. Hovewer, her natural hospitality prevents her from leaving her guest without food. Slastyona has a range of natural numbers from L to R, which she is going to feed the guest with. Help her determine how many distinct inedible tails are going to be discarded by Krakozyabra by the end of the dinner.
Input
In the first and only string, the numbers L and R are given – the boundaries of the range (1 ≤ L ≤ R ≤ 1018).
Output
Output the sole number – the answer for the problem.
Examples
1 10 9 40 57 17 157 165 9 Note
In the first sample case, the inedible tails are the numbers from 1 to 9. Note that 10 and 1 have the same inedible tail – the number 1.
In the second sample case, each number has a unique inedible tail, except for the pair 45, 54. The answer to this sample case is going to be (57 - 40 + 1) - 1 = 17.
题目链接：http://codeforces.com/contest/834/problem/E
官方题解：
下面给出AC代码：
1 #include <iostream>
2 #include <cstring>
3 #include <climits>
4
5 const int N = 19;
6
7 using LL = int64_t;
8
9 int a[N], b[N], c[10], cc[10];
10
11 bool check(int i, int need, bool gt, bool lt)
12 {
13 if (gt && lt) {
14 return need <= N - i;
15 }
16 if (i == N || need > N - i) {
17 return false;
18 }
19 for (int d = gt ? 0 : a[i]; d <= (lt ? 9 : b[i]); ++ d) {
20 cc[d] ++;
21 if (cc[d] <= c[d] && check(i + 1, need - !!d, gt || a[i] < d, lt || d < b[i])) {
22 return true;
23 }
24 cc[d] --;
25 }
26 return false;
27 }
28
29 int search(int d, int used)
30 {
31 if (d == 10) {
32 memset(cc, 0, sizeof(cc));
33 return check(0, used, false, false);
34 }
35 int result = 0;
36 for (c[d] = 0; used + c[d] <= N - 1; ++ c[d]) {
37 result += search(d + 1, used + c[d]);
38 }
39 return result;
40 }
41
42 int main()
43 {
44 #ifdef LOCAL_JUDGE
45 freopen("C.in", "r", stdin);
46 #endif
47 LL l, r;
48 while (std::cin >> l >> r) {
49 l --, r ++;
50 for (int i = N - 1; i >= 0; -- i) {
51 a[i] = l % 10, b[i] = r % 10;
52 l /= 10, r /= 10;
53 }
54 c[0] = INT_MAX;
55 printf("%d\n", search(1, 0));
56 }
57 } 作 者：Angel_Kitty
出 处：https://www.cnblogs.com/ECJTUACM-873284962/
关于作者：阿里云ACE，目前主要研究方向是Web安全漏洞以及反序列化。如有问题或建议，请多多赐教！
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