【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=1;i<=(int)sqrt(n);i++)
        if (n%i==0)
        {
            LL x=n/i;
            for (LL j=1;j<=n;j+=x)
                if ((j+1)%i==0)
                    s.insert(j);
            for (LL j=x-1;j<=n;j+=x)
                if ((j-1)%i==0)
                    s.insert(j);
        }
    if (s.empty())
    {  
        printf("None
");
        return 0;
    }
    while (!s.empty())
    {
        printf("%lld
",*s.begin());
        s.erase(s.begin());
    }
    return 0;
}