POJ 3243 // HDU 2815（改下输出，加个判断）

A^x = B (mod C) 的模板题，不够要用扩展BSGS

（虽然AC，但完全理解不了模板0.0，以后学好数学在来慢慢理解555555）

#include <iostream>
#include <cstdio>
#include <ctime>
#include <cmath>
const int MAXN =  + ;
const int maxn = ;
const int INF = 0x7fffffff;
using namespace std;
typedef long long ll;
struct Hash{
       ll a,b,next;
}Hash[maxn << ];
ll flg[maxn + ];
ll top,idx;
void ins(ll a,ll b){
     ll k = b & maxn;
     if (flg[k] != idx){
        flg[k] = idx;
        Hash[k].next = -;
        Hash[k].a = a;
        Hash[k].b = b;
        return ;
     }
     while (Hash[k].next != -){
           if(Hash[k].b == b) return ;
           k = Hash[k].next;
     }
     Hash[k].next = ++ top;
     Hash[top].next = -;
     Hash[top].a = a;
     Hash[top].b = b;
}
ll Find(ll b){
    ll k = b & maxn;
    if (flg[k] != idx) return -;
    while (k != -){
          if(Hash[k].b == b) return Hash[k].a;
          k = Hash[k].next;
    }
    return -;
}
ll gcd(ll a,ll b) {return b == ? a: gcd(b, a % b);}
ll exgcd(ll a, ll b, ll &x, ll &y){
    if (b == ){x = ; y = ; return a;}
    ll tmp = exgcd(b, a % b, y, x);
    y -= x * (a / b);
    return tmp;
}
ll solve(ll a, ll b, ll c){
    ll x, y, Ans;
    ll tmp = exgcd(a, c, x, y);
    Ans = (ll)(x * b) % c;
    return Ans >=  ? Ans : Ans + c;
}
ll pow(ll a, ll b, ll c){
    ll ret = ;
    while(b)
    {
        if(b & ) ret = ret * a % c;
        a = a*a%c;
        b>>= ;
    }
    return ret;
}
ll BSGS(ll A, ll B, ll C){
    top = maxn;
    ++idx;
    ll buf =  % C, D = buf, K, tmp;
    for (ll i = ; i <= ; i++){
        if (buf == B) return i;
        buf = (buf * A) % C;
    }
    ll d = ;
    while ((tmp = gcd(A, C)) != ){
          if (B % tmp != ) return -;
          d++;
          B /= tmp;
          C /= tmp;
          D = D * A / tmp % C;
    }
    //hash表记录1-sqrt(c)的值
    ll M = (ll)ceil(sqrt(C * 1.0));
    buf =  % C;
    for (ll i = ; i <= M; i++){
        ins(i, buf);
        buf = (buf * A) % C;
    }
    K = pow(A, M, C);
    for (ll i = ; i <= M; i++){
        tmp = solve(D, B, C);
        ll w;
        if (tmp >=  && (w = Find(tmp)) != -) return i * M + w + d;
        D = (D * K) % C;
    }
    return -;
}

int main(){

    ll A, B, C;
    while (cin >> A >> C >> B && (A || B || C)){
          B %= C;
          ll tmp = BSGS(A, B, C); // A^x = B (mod C);
          if (tmp >= ) cout << tmp << endl;
          else cout << "No Solution\n";
    }
    return ;
}