HDU 4910 hdoj problem about GCD bestcoder #3 fourth question

First, when M = 1, ANS = 0
For M> 1
Given that numbers from 1 to 1 have a mass of M, a group is formed on the Multiplication Modulo operation of M.
Ai = (1 <= A <M & gcd (a, m) = 1)
Therefore, for a, B = a ^ (-1) = inv (a) makes a * B = 1 (mod m)
There are two cases
A! = B. Then, in the final concatenation formula, a B appears once, multiplied by 1.
A = B then only one A is displayed in the final concatenation
In fact, all the AI concatenation of a = inv (a) is the answer.
Continue to consider if gcd (a, m) = 1 then gcd (m-a, m) = 1
Note M-A =-A (mod m)
Then a * (-a) =-(A * A) =-1 (mod m)
M! = 2, m-! = A (otherwise a = m/2, gcd (M, M/2) = M/2! = 1)
So a and-A always appear in pairs.
So when the number of solutions for a ^ 2 = 1 (mod m)/2 is an odd number, the answer is-1. If it is an even number, the answer is 1.
When M = 2, the obtained answer is 1 (because 1 and-1 are equivalent, there is a special problem)

Therefore, for M> 2, only the number of solutions a ^ 2 = 1 (mod m) is required to be a multiple of 4.

 

A ^ 2 = 1 (mod m) Equivalent Transformation
(a + 1) (A-1) = 0 (mod m)
assume M = P0 ^ K0 * P1 ^ K1 *... * PI ^ KI (PI is a prime number)
the original equation based on the Chinese Remainder Theorem is equivalent to the
equations (a + 1) (A-1) = 0 (mod PI ^ KI)
consider a single equation first:
pi> 2, (A + 1) and (A-1) there must be a PI and PI (otherwise PI % 2 = 0)
so the solution of this equation is + 1 (mod PI ^ KI)
when Pi = 2,
when k = 1, the equation is resolved to 1 (mod 2)
when k = 2, the equation is solved as ± 1 (mod 4)
K> 2 equation is solved as ± 1, (2 ^ (k-1) + 1 ), (2 ^ (k-1)-1) (mod 2 ^ K)
when the equations have only one equation
consider multiple equations and merge them
according to the Chinese Remainder theorem, the values of each formula are as follows, all cases have only one solution in the range.
so the number of solutions of the equations is the product of the number of solutions of each equation.
m> 2, the number of solutions is not a multiple of 4 (that is, 2) Only the following
m = 2 ^ 2 = 4
M = P ^ K (P! = 2 and is a prime number)
m = 2 * P ^ K (P! = 2, and it is a prime number)