扩展bsgs

#include<stdio.h>
#include<map>
#include<math.h>
#define ll long long
using namespace std;
struct H {
    static const int md=3999997;
    bool vis[md]; int dt[md], foo[md], s[md], top;
    void clr() { while(top) { int x=s[top--]; vis[x]=0, dt[x]=-1; } }
    void add(int a, int b) {
        int x=a%md;
        while(1) { if(!vis[x] || dt[x]==a) { if(!vis[x]) s[++top]=x; vis[x]=1; dt[x]=a; foo[x]=b; break; } ++x; if(x==md) x=0; }
    }
    int find(int a) {
        int x=a%md;
        while(1) { if(!vis[x]) return -1; if(dt[x]==a) return foo[x]; ++x; if(x==md) x=0; }
    }
}h;
int gcd(int a, int b) {return b ? gcd(b, a % b) : a;}
inline ll ksm(ll a, ll b, ll mod)
{
    ll ret = 1, h = a;
    while(b)
    {
        if(b & 1)
            ret *= h, ret %= mod;
        h *= h, h %= mod, b >>= 1;
    }
    return ret;
}
inline int exbsgs(int a, int b, int p)
{
    a %= p, b %= p;
    if(b == 1) return 0; 										//特判2
    int g, d = 0;
    ll c = 1;
    while((g = gcd(a, p)) > 1)
    {
        if(b % g) return -1;
        b /= g, p /= g, c = c * (a / g) % p, ++d;
        if(c == b) return d; 						//特判3（其实等价于特判2）
    }
//    h.clear();
    int t = int(sqrt(p * 1.0) + 1);
    ll base = b;		
    h.clr();							//计算右半部分的值
    for(int i = 0; i < t; i++)
        h.add(base,i), base = base * a % p;
    base = ksm(a, t, p); 							//用左半边的值验证
    ll now = c;
    for(int i = 1; i <= t + 1; i++)
    {
        now = now * base % p;
        if(h.find(now)!=-1)
            return i * t - h.find(now) + d;				//返回解
    }
    return -1;										//无解
}
int main()
{
    int a, b, p;
    while(~scanf("%d%d%d", &a, &p, &b) && a && b && p)
    {
        if(p == 1) {puts("0"); continue;} 			//特判1
        int ans = exbsgs(a, b, p);
        if(ans == -1) puts("No Solution");
        else printf("%d\n", ans);
    }
}