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; 
}