探索bsgs/exbsgs

Posted 2021-10-26 goto_1600

$$\begin{gathered} 众所周知bsg是处理a^x\equiv b(modp)求最小的幂次 \\ 如果(a,p)=1 \\ 我们可以发现a^x\equiv a^{x\%\varphi (p)}那么只需要开一个p的块 \\ 先预处理出来\sqrt{p}的余数，放进hash表里 \\ 注意这里hash表存的是右边，也就是带有b的p的幂次，每次查找的时候一整块一整块查询 \\ 复杂度O(\sqrt{p}) \\ 那么exgs能处理什么问题呢？处理(a,p)\ne 1的情况 \\ 先将a^x+kp\equiv b，如果a和p的gcd不能整除b的话那么就是无解 \\ 否则两边同÷d，也就是(a/d)\ast a^{x-1}+k(p/d)=b/d \\ (a/d)\ast a^{x-1}\equiv b/d(mod(p/d))，然后用exgcd找a/d在p/d下的逆元 \\ 也就是a^{x-1}\equiv (a/d{)}^{-1}\ast (b/d)(mod(p/d))) \\ 然后递归找exgs(a,b/d\ast inv(a/b),p/d)就可以了，找出来x后要记得+1 \end{gathered}$$

//#pragma GCC target("avx")
//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize("Ofast")
// created by myq 
#include<iostream>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<climits>
#include<cmath>
#include<cctype>
#include<stack>
#include<queue>
#include<list>
#include<vector>
#include<set>
#include<map>
#include<sstream>
#include<unordered_map>
#include<unordered_set>
using namespace std;
typedef long long ll;
#define x first
#define y second
typedef pair<int,int> pii;
const int N = 400010;
const int mod=998244353;
inline int read()
{
	int res=0;
	int f=1;
	char c=getchar();
	while(c>'9' ||c<'0')
	{
		if(c=='-')	f=-1;
		c=getchar();
	}
	while(c>='0'&&c<='9')
	{
		res=(res<<3)+(res<<1)+c-'0';
		c=getchar(); 
	}
	return res;
 } 
 const int INF=1e8;
 #define int long long
 int exgcd(int a,int b,int &x,int &y)
 {
 	if(!b){
 		x=1;
 		y=0;
 		return a;
 	}
 	int d=exgcd(b,a%b,y,x);
 	y-=a/b*x;
 	return d;
 }
 int bsgs(int a,int b,int p)
 {
 	if(b%p==1%p)	return 0;
 	int k=sqrt(p)+1;
 	map<int,int>mp;
 	int ak=1;
 	for(int i=0,j=b;i<k;i++,j=j*a%p)
 		mp[j]=i,ak=ak*a%p;
 	for(int i=1,t=ak;i<=k;i++,t=t*ak%p){
 		if(mp.count(t))
 		{
 			return i*k-mp[t];
 		}
 	}
 	
 	return -INF;
 		
 }
 int exgs(int a,int b,int p){
 	b=(b%p+p)%p;
 	if(b%p==1%p)	return 0;
 	if(__gcd(a,p)==1){
 		return bsgs(a,b,p);
 	}
 	else
 	{
 		int x,y;
 		int d=exgcd(a,p,x,y);
 		if(b%d)
 		{
 			return -INF;
 		}
 		exgcd(a/d,p/d,x,y);
 		return exgs(a,(b/d*x)%(p/d),p/d)+1;
 	}
 }
signed main() 
{ 
	// ios::sync_with_stdio(0);
	// cin.tie(0);
	// cout.tie(0);
	int a,b,p;
	while(scanf("%lld%lld%lld",&a,&p,&b),a||p||b){
		int res=exgs(a,b,p);
		if(res<0)
		puts("No Solution");
		else
		printf("%lld\\n",res);
		// cout<<res<<"\\n";
	}
	return 0;
	
}
/**
* In every life we have some trouble
* When you worry you make it double
* Don't worry,be happy.
**/