Codeforces 834E The Bakery【枚举+数位dp】

简介: E. Ever-Hungry Krakozyabra time limit per test:1 second memory limit per test:256 megabytes input:standard input output:stan...

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
Input
1 10
Output
9
Input
40 57
Output
17
Input
157 165
Output
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 }

 

目录
相关文章
|
6月前
不要62(数位dp)
不要62(数位dp)
40 0
|
6月前
|
存储
【题型总结】数位dp(一)
【题型总结】数位dp(一)
77 0
|
6月前
数位dp(计数问题)
数位dp(计数问题)
|
6月前
C. Unlucky Numbers(数位dp)
C. Unlucky Numbers(数位dp)
|
6月前
|
存储 网络架构
【题型总结】数位dp(二)
【题型总结】数位dp(二)
72 0
A. Codeforces Checking(打表枚举)
A. Codeforces Checking(打表枚举)
52 0
codeforces144——D. Missile Silos(最短路+枚举)
codeforces144——D. Missile Silos(最短路+枚举)
94 0
codeforces144——D. Missile Silos(最短路+枚举)
|
人工智能
codeforces455——A. Boredom(线性DP)
codeforces455——A. Boredom(线性DP)
123 0
codeforces455——A. Boredom(线性DP)
|
人工智能 vr&ar Perl
codeforces1509 C. The Sports Festival (区间DP)
codeforces1509 C. The Sports Festival (区间DP)
109 0