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 = 1;
    while (y) {
        if (y & 1) res = res * x % mod;
        x = x * x % mod, y >>= 1;
    }
    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 = 0; i < m; ++i)
        h[base] = i, base = base * a % p;
         
    base = pow(a, m, p);
    for (i = 1; i <= m + 1; ++i) {
        now = now * base % p;
        if (h.find(now) != h.end()) return i * m - h[now];
    }
    return -1;
}
 
int extend_BSGS(int a, int b, int m) {
    static int cnt, g, res;
    static ll t;
    a %= m, b %= m;
    if (b == 1) return 0;
    cnt = 0, g = __gcd(a, m), t = 1;
    while (g != 1) {
        if (b % g) return -1;
        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 (1) {
        a = read(), m = read(), b = read();
        if (!a && !m && !b) return 0;
        ans = extend_BSGS(a, b, m);
        if (!~ans) puts("No Solution");
        else printf("%d\n", ans);
    }
    return 0;
}
 
inline int read() {
    static int x;
    static char ch;
    x = 0, ch = getchar();
    while (ch < '0' || '9' < ch)
        ch = getchar();
    while ('0' <= ch && ch <= '9') {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x;
}