传送门：http://poj.org/problem?id=3872
K-equivalence
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 117 | Accepted: 32 |
Description
Consider a set K of positive integers.
Let p and q be two non-zero decimal digits. Call them K-equivalent if the following condition applies:
For every n 
K, if you replace one digit p with q or one digit q with p in the
decimal notation of n then the resulting number will be an element of K.
For example, when K is the set of integers divisible by 3, the
digits 1, 4, and 7 are K-equivalent. Indeed, replacing a 1 with a 4 in
the decimal notation of a number never changes its divisibility by 3.
It can be seen that K-equivalence is an equivalence relation (it is reflexive, symmetric and transitive).
You are given a finite set K in form of a union of disjoint finite intervals of positive integers.
Your task is to find the equivalence classes of digits 1 to 9.
Input
The first line contains n, the number of intervals composing the set K (1 <= n <= 10 000).
Each of the next n lines contains two positive integers ai and bi
that describe the interval [ai, bi] (i. e. the set of positive integers
between ai and bi, inclusive), where 1 <= ai <= bi <= 1018. Also, for i [2..n]: ai >= b(i-1) + 2.
Output
Represent each equivalence class as a concatenation of its elements, in ascending order.
Output all the equivalence classes of digits 1 to 9, one at a line, sorted lexicographically.
Sample Input
Sample Input #1: 1 1 566 Sample Input #2: 1 30 75
Sample Output
Sample Output #1: 1234 5 6 789 Sample Output #2: 12 345 6 7 89
Source
