1406: [AHOI2007]密码箱

1406: [AHOI2007]密码箱

https://www.lydsy.com/JudgeOnline/problem.php?id=1406

分析

$x^2 ≡ 1 mod n$
$x^2 = kn +１$
$x^2 - 1 = kn$
$(x + 1) ( x - 1) = kn$
设$n = a imes b$
$(x + 1) ( x - 1) = k imes a imes b$
那么有
$a | (x+1) , b|(x-1)$
$a | (x-1) , b|(x+1)$

所以枚举n个约数a,b，然后分两种情况讨论。

代码：

#include<bits/stdc++.h>
using namespace std;

set<int>s;

int main() {
    
    int n;cin >> n;
    if (n == 1) {
        printf("none");
        return 0;
    }
    for (int i=1,lim=sqrt(n); i<=lim; ++i) {
        if (n % i == 0) {
            int a = i, b = n / i;
            for (int x=1; x<=n; x+=b) { // b | x-1 , kb = x-1, x = kb+1
                if ((x + 1) % a == 0) s.insert(x);
            }
            for (int x=b-1; x<=n; x+=b) { // b | x+1 ,kb = x+1, x = kb - 1
                if ((x - 1) % a == 0) s.insert(x);
            }
        }
    }
    if (s.empty()) {
        printf("none");
        return 0;
    }
    set<int>:: iterator it;
    for (it=s.begin(); it!=s.end(); it++) {
        printf("%d
",*it);
    }    
    return 0;
}