题目大意：给出一个p进制数的乘法表，第i行第j组（一组有两个数）表示i*j结果的高位和低位，例如样例中：
4
2 3 1 1 3 2 1 0
1 1 1 1 1 1 1 1
3 2 1 1 3 1 1 2
1 0 1 1 1 2 1 3
0
第二行的“2 3”表示0*0的结果为23，后面的“1 1”表示0*1的结果为11，依此类推。
写出九九乘法表之后你会发现当i*j中，所有高位的数字为0-i-1。然后就可以哈希统计出来高位有多少个，那这一行就是几了啊。
Multiplication table
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 736 Accepted Submission(s): 335
Problem Description Teacher Mai has a multiplication table in base p.
For example, the following is a multiplication table in base 4:
* 0 1 2 3
0 00 00 00 00
1 00 01 02 03
2 00 02 10 12
3 00 03 12 21
But a naughty kid maps numbers 0..p-1 into another permutation and shuffle the multiplication table.
For example Teacher Mai only can see:
1*1=11 1*3=11 1*2=11 1*0=11
3*1=11 3*3=13 3*2=12 3*0=10
2*1=11 2*3=12 2*2=31 2*0=32
0*1=11 0*3=10 0*2=32 0*0=23
Teacher Mai wants you to recover the multiplication table. Output the permutation number 0..p-1 mapped into.
It's guaranteed the solution is unique.
Input There are multiple test cases, terminated by a line "0".
For each test case, the first line contains one integer p(2<=p<=500).
In following p lines, each line contains 2*p integers.
The (2*j+1)-th number x and (2*j+2)-th number y in the i-th line indicates equation i*j=xy in the shuffled multiplication table.
Warning: Large IO!
Output For each case, output one line.
First output "Case #k:", where k is the case number counting from 1. The following are p integers, indicating the permutation number 0..p-1 mapped into.
Sample Input
4
2 3 1 1 3 2 1 0
1 1 1 1 1 1 1 1
3 2 1 1 3 1 1 2
1 0 1 1 1 2 1 3
0
Sample Output
Case #1: 1 3 2 0
Author xudyh
Source 2014 Multi-University Training Contest 8
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