【bzoj1406】[AHOI2007]密码箱

x² ≡ 1 mod n => x² = k * n + 1 => n | (x + 1) * (x - 1)

令n = a * b，则 (a | x + 1 且 b | x - 1) 或 (a| x - 1 且 b | x + 1)

于是暴力枚举a ∈  [1, sqrt(n)] 就好了

然后直接丢到set里判重

 #include<algorithm>
 #include<iostream>
 #include<cstdlib>
 #include<cstring>
 #include<cstdio>
 #include<cmath>
 #include<queue>
 #include<set>
 using namespace std;

 typedef long long LL;

 LL n;

 set<LL>s;

 int main()
 {
     scanf("%lld",&n);
     for (int i=;i<=(int)sqrt(n);i++)
         if (n%i==)
         {
             LL x=n/i;
             for (LL j=;j<=n;j+=x)
                 if ((j+)%i==)
                     s.insert(j);
             for (LL j=x-;j<=n;j+=x)
                 if ((j-)%i==)
                     s.insert(j);
         }
     if (s.empty())
     {
         printf("None\n");
         return ;
     }
     while (!s.empty())
     {
         printf("%lld\n",*s.begin());
         s.erase(s.begin());
     }
     return ;
 }