自己瞎搞了一个 ExBSGS ,求卡
思路如下
AxAilim−jAilim≡B (mod p)≡B (mod p)≡BAj (mod p)预处理Ailim,枚举 j 来做,但是由于可能没有逆元,所以一组 i 和 j ,不一定能找到一个合法的 x,所以特判一下是否合法
#include<bits/stdc++.h>
#define ll long long
#define ull unsigned ll
#define uint unsigned int
#define db double
#define ld long double
#define pint pair<int,int>
#define fi first
#define se second
#define mk make_pair
#define pb push_back
#define eb emplace_back
#define ins insert
#define Rep(x,y,z) for(int x=(y);x<=(z);x++)
#define Red(x,y,z) for(int x=(y);x>=(z);x--)
using namespace std;
const int Mod=998244353,Mod1=Mod-1;
const int MAXN=1e4+5;
inline int read(){
int x;scanf("%d",&x);return x;
}
char n[MAXN];int x,m,len,tmp;
unordered_map<int,int>vis;
inline int Fp(int x,int k){int ans=1;for(;k;k>>=1,x=1ll*x*x%Mod)if(k&1)ans=1ll*ans*x%Mod;return ans;}
inline int Fpow(int x,int k,int Mod){int ans=1;for(;k;k>>=1,x=1ll*x*x%Mod)if(k&1)ans=1ll*ans*x%Mod;return ans;}
namespace Task2{
inline int Solve(){
int lim=sqrt(m),tmpp=1;vis.clear();
for(int i=0;i<lim;i++)tmpp=1ll*tmpp*tmp%m;
int tmppp=tmpp;
for(int i=1;i<=lim;i++){
if(!vis.count(tmppp))vis[tmppp]=i;
tmppp=1ll*tmppp*tmpp%m;
}
int ans=-1,s=x;
if(x==1)ans=0;
for(int i=0;i<lim;i++){
if(vis.count(x)){
if(Fpow(tmp,vis[x]*lim-i,m)==s){
if(ans==-1)ans=vis[x]*lim-i;
else ans=min(ans,vis[x]*lim-i);
}
}x=1ll*x*tmp%m;
}return ans;
}
}
int main(){
// freopen("problem.in","r",stdin);
// freopen("problem.out","w",stdout);
for(;scanf("%d%d%d",&tmp,&m,&x)!=EOF;){
if(m==0&&tmp==0&&x==0)break;
int ans=Task2::Solve();
if(ans==-1)puts("No Solution");
else cout<<ans<<"\n";
}
return 0;
}