BZOJ1406 [AHOI2007]密码箱

什么神奇的数论题。。。

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

我比较懒，直接存到set里去了

/**************************************************************
    Problem: 1406
    User: rausen
    Language: C++
    Result: Accepted
    Time:36 ms
    Memory:808 kb
****************************************************************/
 
#include <cstdio>
#include <set>
 
using namespace std;
typedef long long ll;
typedef set <ll> :: iterator ITER;
 
set <ll> S;
ll n;
 
int main() {
  int i, k;
  ll a, b, x;
  scanf("%lld
", &n);
  for (i = 1; i * i <= n; ++i)
    if (n % i == 0) {
      a = i, b = n / i;
      for (k = 0; (x = b * k + 1) < n; ++k)
    if ((x + 1) % a == 0)
      S.insert(x);
      for (k = 1; (x = b * k - 1) < n; ++k)
    if ((x - 1) % a == 0)
      S.insert(x);
    }
  for (ITER it = S.begin(); it != S.end(); ++it)
    printf("%lld
", *it);
  return 0;
}