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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
1 /************************************************************** 2 Problem: 1406 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:36 ms 7 Memory:808 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <set> 12 13 using namespace std; 14 typedef long long ll; 15 typedef set <ll> :: iterator ITER; 16 17 set <ll> S; 18 ll n; 19 20 int main() { 21 int i, k; 22 ll a, b, x; 23 scanf("%lld\n", &n); 24 for (i = 1; i * i <= n; ++i) 25 if (n % i == 0) { 26 a = i, b = n / i; 27 for (k = 0; (x = b * k + 1) < n; ++k) 28 if ((x + 1) % a == 0) 29 S.insert(x); 30 for (k = 1; (x = b * k - 1) < n; ++k) 31 if ((x - 1) % a == 0) 32 S.insert(x); 33 } 34 for (ITER it = S.begin(); it != S.end(); ++it) 35 printf("%lld\n", *it); 36 return 0; 37 }
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原文地址：http://www.cnblogs.com/rausen/p/4294382.html
