BZOJ 4522: [Cqoi2016]密钥破解 exgcd+Pollard-Rho

挺简单的，正好能再复习一遍 $exgcd$~

按照题意一遍一遍模拟即可，注意一下 $pollard-rho$ 中的细节.

#include <ctime>
#include <cmath>
#include <cstdio>
#include <algorithm>
#define ll long long
#define ull unsigned long long
#define setIO(s) freopen(s".in","r",stdin), freopen(s".out","w",stdout)
using namespace std;
ll N,E,C,D,n,P,Q,R;
int array[20]={2,11,13,17,19};
ll exgcd(ll a,ll b,ll &x,ll &y)
{
	if(!b)
	{
		x=1,y=0;
		return a;
	}
	ll ans=exgcd(b,a%b,x,y),tmp=x;
	x=y,y=tmp-a/b*y;
	return ans;
}
ll mult(ll x,ll y,ll mod)
{
	ll tmp=(long double)x/mod*y;
	return ((ull)x*y-tmp*mod+mod)%mod;
}
ll qpow(ll base,ll k,ll mod)
{
	ll tmp=1;
	for(;k;k>>=1,base=mult(base,base,mod)) if(k&1) tmp=mult(tmp,base,mod);
	return tmp;
}
int isprime(ll x)
{
	if(x<=1) return 1;
	int i,j,k;
	ll pre,cur,a;
	for(cur=x-1,k=0;cur%2==0;cur>>=1) ++k;
	for(i=0;i<5;++i)
	{
		if(x==array[i]) return 1;
		a=pre=qpow(array[i],cur,x);
		for(j=1;j<=k;++j)
		{
			a=mult(pre,pre,x);
			if(a==1&&pre!=1&&pre!=x-1) return 0;
			pre=a;
		}
		if(a!=1) return 0;
	}
	return 1;
}
ll F(ll x,ll c,ll mod)
{
	return (mult(x,x,mod)+c)%mod;
}
ll pollard_rho(ll x)
{
	int step,k;
	ll s=0,t=0,c=rand()%(x-1)+1,val=1,d;
	for(k=1;;k<<=1,s=t,val=1)
	{
		for(step=1;step<=k;++step)
		{
			t=F(t,c,x);
			val=mult(val,abs(t-s),x);
			if(step%127==0)
			{
				d=__gcd(val,x);
				if(d>1) return d;
			}
		}
		d=__gcd(val,x);
		if(d>1) return d;
	}
}
void solve_d()
{
	for(P=N;P>=N;)
		P=pollard_rho(N);
	Q=N/P;
	R=(P-1)*(Q-1);
	ll x,y,gcd;
	gcd=exgcd(E,R,x,y);
	x=(x+R)%R;
	D=x;    

}
int main()
{
	int i,j;
	// setIO("input");
	srand((unsigned)time(NULL));
	scanf("%lld%lld%lld",&E,&N,&C);
	for(P=N;P>=N;)
		P=pollard_rho(N);
	Q=N/P;
	R=(P-1)*(Q-1);
	ll x,y,gcd;
	gcd=exgcd(E,R,x,y);
	x=(x+R)%R;
	D=x;
	n=qpow(C,D,N);
	printf("%lld %lld\n",D,n);
	return 0;
}