BZOJ 1406: [AHOI2007]密码箱

二次联通门 : BZOJ 1406: [AHOI2007]密码箱

/*
    BZOJ 1406: [AHOI2007]密码箱

    数论
    要求 x^2 ≡ 1 (mod n)
    可以转换为 x ^ 2 - k *n = 1
    (x + 1) * (x - 1) = k * n
    设 n = a * b
    则 a * b | (x + 1) * (x - 1)
       那么枚举b即可
*/
#include <cstdio>
#include <cmath>
#include <set>
typedef long long LL;
#define Set std :: set <LL> 

int main ()
{
    LL N, a, b; scanf ("%lld", &N); register LL i, j;
    int L = sqrt (N); Set Answer;
    for (i = ; i <= L; ++ i)
        if (N % i == )
        {
            a = i, b = N / i;
            for (j = ; j <= N; j += b)
            {
                if ((j + ) % a ==  && j +  < N) Answer.insert (j + );
                   if ((j - ) % a ==  && j -  >= ) Answer.insert (j - );
            }
        }
    if (Answer.size () == ) return printf ("None"), ;
    for (Set :: iterator i = Answer.begin (); i != Answer.end (); ++ i)
        printf ("%lld\n", *i);
    return ;
}