Extended BSGS algorithm

Disclaimer: This article is a blogger original article, follow the CC 4.0 BY-SA copyright agreement, reproduced, please attach the original source link and this statement.

This link: https://blog.csdn.net/weixin_44316314/article/details/102747036

/*
计算a^x = b mod p，求x
*/
#include <cstdio>
#include <cmath>
#include <map>
using namespace std;

typedef long long ll;
ll q_pow(ll a,ll b,ll p)
{
	ll ans = 1;
	while( b )
	{
		if( b & 1 )
		{
			ans *= a;
			if( ans >= p ) ans %= p;
		}
		a *= a;
		if( a >= p ) a %= p;
		b >>= 1; 
	}
	return ans;
}

ll gcd(ll a,ll b)
{
	if( b == 0 ) return a;
	return gcd(b,a%b);
}

ll ext_BSGS(ll a,ll b,ll p)
{
	ll ans;
	if( b == 1 ) return 0;
	ll k = 0,d = 1,t;
	/*while( ( t = gcd(a,p) ) != 1 )   //扩展时加上这段代码 
	{
		if ( b % t ) return -1;
		k ++;
		b /= t;
		p /= t;
		d *= a / t;
		d %= p;
		if( d == b ) return k;
	}*/
	map<ll,ll> ma;
	ll m = sqrt(p*1.0) + 1,a_m = q_pow(a,m,p);
	ll mul = b;
	for (int r = 1; r <= m; r++)
	{
		mul = mul * a % p;
		ma[mul] = r;
	}
	for (int r = 1; r <= m; r++)
	{
		d = d * a_m % p;
		if( ma[d] )
		{
			ans = r * m - ma[d];
			ans += k;
			return ans;
		}
	}
	return -1;
}

int main()
{
	ll a,b,p;
	while( scanf("%lld%lld%lld",&p,&a,&b) != EOF )
	{
		ll res = ext_BSGS(a,b,p); 
		if( res == -1 ) printf("no solution\n");
		else printf("%lld\n",res);
	}
	return 0;
}