BL == N (mod P)

输入描述:

输出描述:

5 2 1
5 2 2
5 2 3
5 2 4
5 3 1
5 3 2
5 3 3
5 3 4
5 4 1
5 4 2
5 4 3
5 4 4
12345701 2 1111111
1111111121 65537 1111111111

0
1
3
2
0
3
1
2
0
no solution
no solution
1
9584351
462803587

备注:

  The solution to this problem requires a well known result in number theory that is probably expected of you for Putnam but not ACM competitions. It is Fermat's theorem that states 
 

 
   B(P-1) == 1 (mod P)
 
for any prime P and some other (fairly rare) numbers known as base-B pseudoprimes. A rarer subset of the base-B pseudoprimes, known as Carmichael numbers, are pseudoprimes for every base between 2 and P-1. A corollary to Fermat's theorem is that for any m 
 

 
   B(-m) == B(P-1-m) (mod P) .