1406: [AHOI2007]密码箱

Posted 2020-12-24 mjtcn

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：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 
 4 set<int>s;
 5 
 6 int main() {
 7     
 8     int n;cin >> n;
 9     if (n == 1) {
10         printf("none");
11         return 0;
12     }
13     for (int i=1,lim=sqrt(n); i<=lim; ++i) {
14         if (n % i == 0) {
15             int a = i, b = n / i;
16             for (int x=1; x<=n; x+=b) { // b | x-1 , kb = x-1, x = kb+1
17                 if ((x + 1) % a == 0) s.insert(x);
18             }
19             for (int x=b-1; x<=n; x+=b) { // b | x+1 ,kb = x+1, x = kb - 1
20                 if ((x - 1) % a == 0) s.insert(x);
21             }
22         }
23     }
24     if (s.empty()) {
25         printf("none");
26         return 0;
27     }
28     set<int>:: iterator it;
29     for (it=s.begin(); it!=s.end(); it++) {
30         printf("%d
",*it);
31     }    
32     return 0;
33 }