BZOJ2480 Spoj3105 Mod

乍一看题面：$$a^x \equiv b \ (mod \ m)$$

是一道BSGS，但是很可惜$m$不是质数，而且$(m, a) \not= 1$，这个叫扩展BSGS【额......

于是我们需要通过变换使得$(m, a) = 1$

首先令$g = (a, m)$，则原式等价于：$$a ^ x + k * m = b, k \in \mathbb{Z}$$

移项可得：$$\frac{a} {g} * a ^ {x - 1} + k * \frac {m} {g} = \frac {b} {g}$$

此时如果$b \not \equiv 0 (mod\ g)$则无解

令$m' = \frac {m} {g}, b' = \frac {b} {g} * (\frac{a} {g}) ^ {-1}$

于是得到新式：$$a ^ {x - 1} = b' (mod\ m')$$

于是可以一直迭代到$(m, a) = 1$，然后用BSGS来计算答案即可

 /**************************************************************
     Problem: 2480
     User: rausen
     Language: C++
     Result: Accepted
     Time:3256 ms
     Memory:1568 kb
 ****************************************************************/

 #include <cstdio>
 #include <algorithm>
 #include <ext/pb_ds/assoc_container.hpp>
 #include <ext/pb_ds/hash_policy.hpp>

 using namespace std;
 using namespace std;
 typedef long long ll;
 typedef __gnu_pbds::cc_hash_table <int, int> hash;

 inline int read();

 int a, b, m, ans;
 hash h;

 inline int pow(ll x, ll y, ll mod) {
     static ll res;
     res = ;
     while (y) {
         if (y & ) res = res * x % mod;
         x = x * x % mod, y >>= ;
     }
     return (int) res;
 }

 inline int BSGS(int a, int b, int p, ll now) {
     static int m, i;
     static ll base;
     m = (int) ceil(sqrt(p)), base = b;
     h.clear();
     for (i = ; i < m; ++i)
         h[base] = i, base = base * a % p;

     base = pow(a, m, p);
     for (i = ; i <= m + ; ++i) {
         now = now * base % p;
         if (h.find(now) != h.end()) return i * m - h[now];
     }
     return -;
 }

 int extend_BSGS(int a, int b, int m) {
     static int cnt, g, res;
     static ll t;
     a %= m, b %= m;
     if (b == ) return ;
     cnt = , g = __gcd(a, m), t = ;
     while (g != ) {
         if (b % g) return -;
         m /= g, b /= g, t = t * a / g % m;
         ++cnt;
         if (b == t) return cnt;
         g = __gcd(a, m);
     }
     res = BSGS(a, b, m, t);
     return ~res ? res + cnt : res;
 }

 int main() {
     while () {
         a = read(), m = read(), b = read();
         if (!a && !m && !b) return ;
         ans = extend_BSGS(a, b, m);
         if (!~ans) puts("No Solution");
         else printf("%d\n", ans);
     }
     return ;
 }

 inline int read() {
     static int x;
     static char ch;
     x = , ch = getchar();
     while (ch < '' || '' < ch)
         ch = getchar();
     while ('' <= ch && ch <= '') {
         x = x *  + ch - '';
         ch = getchar();
     }
     return x;
 }